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Chapter 9 Differential Equations
Welcome to the solutions guide for Chapter 9: Differential Equations. This chapter introduces a profoundly important area of mathematics that deals with equations involving functions and their derivatives. While algebraic equations relate variables, differential equations (DEs) relate a function to its rates of change. This makes them exceptionally powerful tools for modeling dynamic systems and processes where change is inherent. From population growth and radioactive decay in biology and physics to circuit analysis in engineering, chemical reaction rates, and economic modeling, differential equations provide the mathematical language to describe and predict the behavior of countless real-world phenomena. This chapter lays the groundwork for understanding what differential equations are, how they are classified, and, most importantly, how to find the functions that satisfy these equations – the process known as solving the differential equation.
The solutions begin by establishing the fundamental definition of a differential equation as an equation containing derivatives (or differentials) of one or more dependent variables with respect to one or more independent variables. We then learn how to classify these equations based on two key characteristics:
- The Order of a DE is the order of the highest derivative appearing in the equation.
- The Degree of a DE is the power (exponent) of the highest order derivative, provided the equation is expressed as a polynomial in its derivatives (after clearing any radicals or fractional powers involving derivatives).
Solutions demonstrate how to determine the order and degree for various given differential equations, which helps in categorizing them and choosing appropriate solution methods.
Understanding the nature of solutions is crucial. A general solution to a differential equation of order $n$ typically involves $n$ arbitrary constants. It represents a family of functions satisfying the DE. In contrast, a particular solution is a specific solution obtained from the general solution by using given initial conditions or boundary conditions to determine the values of the arbitrary constants. The solutions also cover the inverse process: the formation of differential equations from a given family of curves (defined by an equation with arbitrary constants) by differentiating the equation sufficiently many times and then eliminating the constants.
The core focus of this chapter, and its solutions, lies in developing techniques for solving first-order, first-degree differential equations. Three primary methods are detailed:
- Variables Separable Method: This technique applies when the differential equation can be algebraically rearranged such that all terms involving the dependent variable $y$ and its differential $dy$ are on one side, and all terms involving the independent variable $x$ and its differential $dx$ are on the other side. This results in an equation of the form $f(y) dy = g(x) dx$. The general solution is then found by integrating both sides: $\int f(y) \, dy = \int g(x) \, dx + C$. The solutions also address equations that can be reduced to this form through suitable substitutions.
- Homogeneous Differential Equations: These are equations that can be expressed in the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ (or $\frac{dx}{dy} = G\left(\frac{x}{y}\right)$). The key characteristic is that the function on the right-hand side depends only on the ratio $\frac{y}{x}$ (or $\frac{x}{y}$). Solutions demonstrate verifying homogeneity and then employing the substitution $y = vx$ (which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$) to transform the homogeneous equation into a variable separable equation involving $v$ and $x$, which can then be solved.
- Linear Differential Equations: These are equations of the specific form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ alone (or constants). The solutions detail the standard method involving the calculation of the Integrating Factor (IF), given by $IF = e^{\int P(x) dx}$. Multiplying the entire equation by the IF makes the left side the derivative of $(y \times IF)$. The general solution is then given by the formula: $y \times (\text{IF}) = \int [Q(x) \times (\text{IF})] dx + C$. The analogous form $\frac{dx}{dy} + P(y)x = Q(y)$ is solved similarly using $IF = e^{\int P(y) dy}$.
The solutions provide clear, step-by-step procedures for identifying which type of first-order, first-degree DE is presented and applying the corresponding solution technique. Finding particular solutions using initial conditions and exploring applications like simple growth and decay models are also frequently covered.
Example 1 (Before Exercise 9.1)
Example 1: Find the order and degree, if defined, of each of the following differential equations:
(i) $\frac{dy}{dx} - \cos x = 0$
(ii) $xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$
(iii) $y’’’ + y^2 + e^{y’} = 0$
Answer:
Solution:
The order of a differential equation is the order of the highest order derivative appearing in the equation.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
(i) $\frac{dy}{dx} - \cos x = 0$
The highest order derivative is $\frac{dy}{dx}$.
The order of the highest order derivative is 1.
The equation is a polynomial in the derivative $\frac{dy}{dx}$.
The power of the highest order derivative ($\frac{dy}{dx}$) is 1.
Therefore, the order is 1 and the degree is 1.
(ii) $xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$
The highest order derivative is $\frac{d^2y}{dx^2}$.
The order of the highest order derivative is 2.
The equation is a polynomial in the derivatives $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$.
The power of the highest order derivative ($\frac{d^2y}{dx^2}$) is 1.
Therefore, the order is 2 and the degree is 1.
(iii) $y''' + y^2 + e^{y'} = 0$
The equation can be written as $\frac{d^3y}{dx^3} + y^2 + e^{\frac{dy}{dx}} = 0$.
The highest order derivative is $\frac{d^3y}{dx^3}$.
The order of the highest order derivative is 3.
The term $e^{\frac{dy}{dx}}$ involves the derivative $\frac{dy}{dx}$ in the exponent. This means the equation is not a polynomial in its derivatives.
Therefore, the order is 3, but the degree is not defined.
Exercise 9.1
Determine order and degree (if defined) of differential equations given in Exercises 1 to 10.
Question 1. $\frac{d^4y}{dx^4} + \sin (y’’’) = 0$
Answer:
Given:
The differential equation is $\frac{d^4y}{dx^4} + \sin (y''') = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation is $\frac{d^4y}{dx^4} + \sin \left(\frac{d^3y}{dx^3}\right) = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $\frac{d^4y}{dx^4}$ and $\frac{d^3y}{dx^3}$.
The highest order derivative is $\frac{d^4y}{dx^4}$.
The order of $\frac{d^4y}{dx^4}$ is 4.
So, the order of the differential equation is 4.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
In the given equation, the term $\sin(y''')$ or $\sin(\frac{d^3y}{dx^3})$ makes the equation non-polynomial in its derivatives because the derivative $\frac{d^3y}{dx^3}$ is inside the sine function.
Therefore, the degree of the differential equation is not defined.
Question 2. $y' + 5y = 0$
Answer:
Given:
The differential equation is $y' + 5y = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be written as $\frac{dy}{dx} + 5y = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the only derivative present is $\frac{dy}{dx}$ or $y'$.
The order of $\frac{dy}{dx}$ is 1.
So, the order of the differential equation is 1.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The equation $\frac{dy}{dx} + 5y = 0$ is a polynomial in the derivative $\frac{dy}{dx}$.
The highest order derivative is $\frac{dy}{dx}$, and its power is 1.
Therefore, the degree of the differential equation is 1.
Question 3. $\left( \frac{ds}{dt} \right)^4 + 3s \frac{d^2s}{dt^2} = 0$
Answer:
Given:
The differential equation is $\left( \frac{ds}{dt} \right)^4 + 3s \frac{d^2s}{dt^2} = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation involves derivatives of the dependent variable $s$ with respect to the independent variable $t$.
The derivatives present are $\frac{ds}{dt}$ and $\frac{d^2s}{dt^2}$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
The order of $\frac{ds}{dt}$ is 1.
The order of $\frac{d^2s}{dt^2}$ is 2.
The highest order derivative is $\frac{d^2s}{dt^2}$, which has an order of 2.
So, the order of the differential equation is 2.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The equation $\left( \frac{ds}{dt} \right)^4 + 3s \frac{d^2s}{dt^2} = 0$ is a polynomial in the derivatives $\frac{ds}{dt}$ and $\frac{d^2s}{dt^2}$.
The highest order derivative is $\frac{d^2s}{dt^2}$. Its power in the term $3s \frac{d^2s}{dt^2}$ is 1.
Therefore, the degree of the differential equation is 1.
The order is 2 and the degree is 1.
Question 4. $\left( \frac{d^2y}{dx^2} \right)^2 + \cos \left( \frac{dx}{dy} \right) = 0$
Answer:
Given:
The differential equation is $\left( \frac{d^2y}{dx^2} \right)^2 + \cos \left( \frac{dx}{dy} \right) = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The order of a differential equation is the order of the highest order derivative appearing in the equation.
The derivatives present in the equation are $\frac{d^2y}{dx^2}$ and $\frac{dx}{dy}$.
The term $\frac{d^2y}{dx^2}$ is a second-order derivative of $y$ with respect to $x$.
The term $\frac{dx}{dy}$ is a first-order derivative of $x$ with respect to $y$. While it's possible to relate this to $\frac{dy}{dx}$, the equation as given contains derivatives of different forms. Considering $y$ as the dependent variable and $x$ as the independent variable (which is suggested by the $\frac{d^2y}{dx^2}$ term), the highest order derivative of $y$ with respect to $x$ is $\frac{d^2y}{dx^2}$.
The order of $\frac{d^2y}{dx^2}$ is 2.
So, the order of the differential equation is 2.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
In the given equation, the term $\cos \left( \frac{dx}{dy} \right)$ involves a derivative inside a trigonometric function (cosine). This means the equation is not a polynomial in its derivatives (whether we consider derivatives of $y$ w.r.t. $x$ or derivatives of $x$ w.r.t. $y$).
Therefore, the degree of the differential equation is not defined.
The order is 2 and the degree is not defined.
Question 5. $\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$
Answer:
Given:
The differential equation is $\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be rewritten as $\frac{d^2y}{dx^2} - \cos 3x - \sin 3x = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the only derivative present is $\frac{d^2y}{dx^2}$.
The order of $\frac{d^2y}{dx^2}$ is 2.
So, the order of the differential equation is 2.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The equation $\frac{d^2y}{dx^2} - \cos 3x - \sin 3x = 0$ is a polynomial in the derivative $\frac{d^2y}{dx^2}$. The terms $\cos 3x$ and $\sin 3x$ depend only on the independent variable $x$ and do not affect the polynomial nature with respect to the derivatives.
The highest order derivative is $\frac{d^2y}{dx^2}$, and its power is 1.
Therefore, the degree of the differential equation is 1.
The order is 2 and the degree is 1.
Question 6. $(y’’’)^2 + (y’’)^3 + (y’)^4 + y^5 = 0$
Answer:
Given:
The differential equation is $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be written using Leibniz notation as $\left( \frac{d^3y}{dx^3} \right)^2 + \left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^4 + y^5 = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $y'''$ (third order), $y''$ (second order), and $y'$ (first order).
The highest order derivative is $y'''$ or $\frac{d^3y}{dx^3}$, which has an order of 3.
So, the order of the differential equation is 3.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The given equation is $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$. This equation is a polynomial in the derivatives $y'''$, $y''$, and $y'$.
The highest order derivative is $y'''$. The term involving the highest order derivative is $(y''')^2$.
The power of the highest order derivative ($y'''$) in this polynomial equation is 2.
Therefore, the degree of the differential equation is 2.
The order is 3 and the degree is 2.
Question 7. $y’’’ + 2y’’ + y’ = 0$
Answer:
Given:
The differential equation is $y''' + 2y'' + y' = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be written using Leibniz notation as $\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $y'''$ (third order), $y''$ (second order), and $y'$ (first order).
The highest order derivative is $y'''$ or $\frac{d^3y}{dx^3}$, which has an order of 3.
So, the order of the differential equation is 3.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The given equation is $y''' + 2y'' + y' = 0$. This equation is a polynomial in the derivatives $y'''$, $y''$, and $y'$.
The highest order derivative is $y'''$. The term involving the highest order derivative is $y'''$.
The power of the highest order derivative ($y'''$) in this polynomial equation is 1.
Therefore, the degree of the differential equation is 1.
The order is 3 and the degree is 1.
Question 8. $y’ + y = e^x$
Answer:
Given:
The differential equation is $y' + y = e^x$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be written using Leibniz notation as $\frac{dy}{dx} + y = e^x$, or $\frac{dy}{dx} + y - e^x = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the only derivative present is $y'$ or $\frac{dy}{dx}$.
The order of $\frac{dy}{dx}$ is 1.
So, the order of the differential equation is 1.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The equation $\frac{dy}{dx} + y - e^x = 0$ is a polynomial in the derivative $\frac{dy}{dx}$. The terms $y$ and $e^x$ do not involve derivatives in a non-polynomial way (like being inside a trigonometric function, exponential function, etc.).
The highest order derivative is $\frac{dy}{dx}$, and its power is 1.
Therefore, the degree of the differential equation is 1.
The order is 1 and the degree is 1.
Question 9. $y’’ + (y’)^2 + 2y = 0$
Answer:
Given:
The differential equation is $y'' + (y')^2 + 2y = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be written using Leibniz notation as $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 2y = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $y''$ (second order) and $y'$ (first order).
The highest order derivative is $y''$ or $\frac{d^2y}{dx^2}$, which has an order of 2.
So, the order of the differential equation is 2.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The given equation is $y'' + (y')^2 + 2y = 0$. This equation is a polynomial in the derivatives $y''$ and $y'$.
The highest order derivative is $y''$. The term involving the highest order derivative is $y''$.
The power of the highest order derivative ($y''$) in this polynomial equation is 1.
Therefore, the degree of the differential equation is 1.
The order is 2 and the degree is 1.
Question 10. $y’’ + 2y’ + \sin y = 0$
Answer:
Given:
The differential equation is $y'' + 2y' + \sin y = 0$.
To Determine:
The order and degree (if defined) of the given differential equation.
Solution:
The given differential equation can be written using Leibniz notation as $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \sin y = 0$.
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $y''$ (second order) and $y'$ (first order).
The highest order derivative is $y''$ or $\frac{d^2y}{dx^2}$, which has an order of 2.
So, the order of the differential equation is 2.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
The given equation is $y'' + 2y' + \sin y = 0$. This equation is a polynomial in the derivatives $y''$ and $y'$. The presence of $\sin y$ where $y$ is the dependent variable does not make the equation non-polynomial in the derivatives themselves.
The highest order derivative is $y''$. The term involving the highest order derivative is $y''$.
The power of the highest order derivative ($y''$) in this polynomial equation is 1.
Therefore, the degree of the differential equation is 1.
The order is 2 and the degree is 1.
Question 11. The degree of the differential equation
$\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin \left( \frac{dy}{dx} \right) + 1 = 0$ is
(A) 3
(B) 2
(C) 1
(D) not defined
Answer:
Given:
The differential equation is $\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin \left( \frac{dy}{dx} \right) + 1 = 0$.
To Find:
The degree of the given differential equation.
Solution:
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $\frac{d^2y}{dx^2}$ (second order) and $\frac{dy}{dx}$ (first order).
The highest order derivative is $\frac{d^2y}{dx^2}$. Its order is 2.
So, the order of the differential equation is 2.
The degree of a differential equation is the power of the highest order derivative when the equation is a polynomial in its derivatives. If the equation is not a polynomial in its derivatives, the degree is not defined.
Let $Y'' = \frac{d^2y}{dx^2}$ and $Y' = \frac{dy}{dx}$. The equation can be written as $(Y'')^3 + (Y')^2 + \sin(Y') + 1 = 0$.
For the degree to be defined, the differential equation must be a polynomial equation in the derivatives $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, ..., $\frac{d^ny}{dx^n}$.
The given equation contains the term $\sin \left( \frac{dy}{dx} \right)$. Since the derivative $\frac{dy}{dx}$ appears inside the sine function, the equation is not a polynomial in its derivatives.
Therefore, the degree of the differential equation is not defined.
Comparing this with the given options, we find that option (D) is correct.
The correct answer is (D) not defined.
Question 12. The order of the differential equation
$2x^2 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0$ is
(A) 2
(B) 1
(C) 0
(D) not defined
Answer:
Given:
The differential equation is $2x^2 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0$.
To Find:
The order of the given differential equation.
Solution:
The order of a differential equation is the order of the highest order derivative appearing in the equation.
In the given equation, the derivatives present are $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$.
The order of $\frac{d^2y}{dx^2}$ is 2.
The order of $\frac{dy}{dx}$ is 1.
The highest order derivative is $\frac{d^2y}{dx^2}$.
The order of the highest order derivative is 2.
So, the order of the differential equation is 2.
Comparing this with the given options, we find that option (A) is correct.
The correct answer is (A) 2.
Example 2 & 3 (Before Exercise 9.2)
Example 2: Verify that the function y = e–3x is a solution of the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.
Answer:
Given:
The function is $y = e^{-3x}$.
The differential equation is $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.
To Verify:
Verify that the given function $y = e^{-3x}$ is a solution of the given differential equation.
Solution:
To verify if $y = e^{-3x}$ is a solution to the differential equation, we need to calculate the first and second derivatives of the function with respect to $x$ and substitute them, along with the function itself, into the differential equation.
Given the function:
$y = e^{-3x}$
Calculate the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{d}{dx}(e^{-3x})$
Using the chain rule, $\frac{d}{dx}(e^{ku}) = ke^{ku} \frac{du}{dx}$. Here $u = -3x$, so $\frac{du}{dx} = -3$.
$\frac{dy}{dx} = e^{-3x} \cdot (-3) = -3e^{-3x}$
Calculate the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-3e^{-3x})$
$\frac{d^2y}{dx^2} = -3 \frac{d}{dx}(e^{-3x}) = -3 (-3e^{-3x}) = 9e^{-3x}$
Now substitute $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the left-hand side (LHS) of the given differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$:
LHS = $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y$
LHS = $(9e^{-3x}) + (-3e^{-3x}) - 6(e^{-3x})$
LHS = $9e^{-3x} - 3e^{-3x} - 6e^{-3x}$
LHS = $(9 - 3 - 6)e^{-3x}$
LHS = $(6 - 6)e^{-3x}$
LHS = $0 \cdot e^{-3x}$
LHS = 0
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = e^{-3x}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.
Example 3: Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.
Answer:
Given:
The function is $y = a \cos x + b \sin x$, where $a, b \in R$.
The differential equation is $\frac{d^2y}{dx^2} + y = 0$.
To Verify:
Verify that the given function $y = a \cos x + b \sin x$ is a solution of the given differential equation.
Solution:
To verify if $y = a \cos x + b \sin x$ is a solution to the differential equation, we need to calculate the first and second derivatives of the function with respect to $x$ and substitute them, along with the function itself, into the differential equation.
Given the function:
$y = a \cos x + b \sin x$
Calculate the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{d}{dx}(a \cos x + b \sin x)$
$\frac{dy}{dx} = a \frac{d}{dx}(\cos x) + b \frac{d}{dx}(\sin x)$
$\frac{dy}{dx} = a (-\sin x) + b (\cos x)$
$\frac{dy}{dx} = -a \sin x + b \cos x$
Calculate the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-a \sin x + b \cos x)$
$\frac{d^2y}{dx^2} = -a \frac{d}{dx}(\sin x) + b \frac{d}{dx}(\cos x)$
$\frac{d^2y}{dx^2} = -a (\cos x) + b (-\sin x)$
$\frac{d^2y}{dx^2} = -a \cos x - b \sin x$
Now substitute $y$ and $\frac{d^2y}{dx^2}$ into the left-hand side (LHS) of the given differential equation $\frac{d^2y}{dx^2} + y = 0$:
LHS = $\frac{d^2y}{dx^2} + y$
LHS = $(-a \cos x - b \sin x) + (a \cos x + b \sin x)$
LHS = $-a \cos x - b \sin x + a \cos x + b \sin x$
LHS = $(-a + a) \cos x + (-b + b) \sin x$
LHS = $0 \cdot \cos x + 0 \cdot \sin x$
LHS = $0 + 0$
LHS = 0
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = a \cos x + b \sin x$, where $a, b \in R$, is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.
Exercise 9.2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
Question 1.
$y = e^x + 1$
:
$y'' - y' = 0$
Answer:
Given:
The function is $y = e^x + 1$.
The differential equation is $y'' - y' = 0$.
To Verify:
Verify that the given function $y = e^x + 1$ is a solution of the given differential equation $y'' - y' = 0$.
Solution:
To verify if $y = e^x + 1$ is a solution to the differential equation, we need to calculate the first and second derivatives of the function with respect to $x$ and substitute them into the differential equation.
Given the function:
$y = e^x + 1$
Calculate the first derivative $y' = \frac{dy}{dx}$:
$y' = \frac{d}{dx}(e^x + 1)$
$y' = \frac{d}{dx}(e^x) + \frac{d}{dx}(1)$
$y' = e^x + 0$
$y' = e^x$
Calculate the second derivative $y'' = \frac{d^2y}{dx^2}$:
$y'' = \frac{d}{dx}(y') = \frac{d}{dx}(e^x)$
$y'' = e^x$
Now substitute $y'$ and $y''$ into the left-hand side (LHS) of the given differential equation $y'' - y' = 0$:
LHS = $y'' - y'$
LHS = $(e^x) - (e^x)$
LHS = $e^x - e^x$
LHS = $0$
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = e^x + 1$ is a solution of the differential equation $y'' - y' = 0$.
Question 2.
$y = x^2 + 2x + C$
:
$y' - 2x - 2 = 0$
Answer:
Given:
The function is $y = x^2 + 2x + C$, where C is a constant.
The differential equation is $y' - 2x - 2 = 0$.
To Verify:
Verify that the given function $y = x^2 + 2x + C$ is a solution of the given differential equation $y' - 2x - 2 = 0$.
Solution:
To verify if $y = x^2 + 2x + C$ is a solution to the differential equation, we need to calculate the first derivative of the function with respect to $x$ and substitute it into the differential equation.
Given the function:
$y = x^2 + 2x + C$
Calculate the first derivative $y' = \frac{dy}{dx}$:
$y' = \frac{d}{dx}(x^2 + 2x + C)$
$y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(C)$
$y' = 2x + 2 \cdot 1 + 0$
$y' = 2x + 2$
Now substitute $y'$ into the left-hand side (LHS) of the given differential equation $y' - 2x - 2 = 0$:
LHS = $y' - 2x - 2$
LHS = $(2x + 2) - 2x - 2$
LHS = $2x + 2 - 2x - 2$
LHS = $(2x - 2x) + (2 - 2)$
LHS = $0 + 0$
LHS = $0$
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = x^2 + 2x + C$ is a solution of the differential equation $y' - 2x - 2 = 0$.
Question 3.
$y = \cos x + C$
:
$y' + \sin x = 0$
Answer:
Given:
The function is $y = \cos x + C$, where C is a constant.
The differential equation is $y' + \sin x = 0$.
To Verify:
Verify that the given function $y = \cos x + C$ is a solution of the given differential equation $y' + \sin x = 0$.
Solution:
To verify if $y = \cos x + C$ is a solution to the differential equation, we need to calculate the first derivative of the function with respect to $x$ and substitute it into the differential equation.
Given the function:
$y = \cos x + C$
Calculate the first derivative $y' = \frac{dy}{dx}$:
$y' = \frac{d}{dx}(\cos x + C)$
$y' = \frac{d}{dx}(\cos x) + \frac{d}{dx}(C)$
$y' = -\sin x + 0$
$y' = -\sin x$
Now substitute $y'$ into the left-hand side (LHS) of the given differential equation $y' + \sin x = 0$:
LHS = $y' + \sin x$
LHS = $(-\sin x) + \sin x$
LHS = $-\sin x + \sin x$
LHS = $0$
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = \cos x + C$ is a solution of the differential equation $y' + \sin x = 0$.
Question 4.
$y = \sqrt{1+x^2}$
:
$y' = \frac{xy}{1+x^2}$
Answer:
Given:
The function is $y = \sqrt{1+x^2}$.
The differential equation is $y' = \frac{xy}{1+x^2}$.
To Verify:
Verify that the given function $y = \sqrt{1+x^2}$ is a solution of the given differential equation $y' = \frac{xy}{1+x^2}$.
Solution:
To verify if $y = \sqrt{1+x^2}$ is a solution to the differential equation, we need to calculate the first derivative of the function with respect to $x$ and substitute it, along with the function itself, into the differential equation.
Given the function:
$y = \sqrt{1+x^2} = (1+x^2)^{1/2}$
Calculate the first derivative $y' = \frac{dy}{dx}$ using the chain rule:
$y' = \frac{d}{dx}((1+x^2)^{1/2})$
$y' = \frac{1}{2}(1+x^2)^{-1/2} \cdot \frac{d}{dx}(1+x^2)$
$y' = \frac{1}{2}(1+x^2)^{-1/2} \cdot (2x)$
$y' = x (1+x^2)^{-1/2}$
$y' = \frac{x}{\sqrt{1+x^2}}$
Now, consider the right-hand side (RHS) of the differential equation and substitute the given function $y = \sqrt{1+x^2}$:
RHS = $\frac{xy}{1+x^2}$
RHS = $\frac{x(\sqrt{1+x^2})}{1+x^2}$
Since $1+x^2 = (\sqrt{1+x^2})^2$, we can write:
RHS = $\frac{x \sqrt{1+x^2}}{(\sqrt{1+x^2})^2}$
Cancel out one factor of $\sqrt{1+x^2}$ from the numerator and the denominator:
RHS = $\frac{x}{\sqrt{1+x^2}}$
We found that the calculated left-hand side (LHS), $y'$, is $\frac{x}{\sqrt{1+x^2}}$.
We also found that the right-hand side (RHS) of the differential equation, after substituting $y$, is $\frac{x}{\sqrt{1+x^2}}$.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = \sqrt{1+x^2}$ is a solution of the differential equation $y' = \frac{xy}{1+x^2}$.
Question 5.
$y = Ax$
:
$xy' = y \quad (x \neq 0)$
Answer:
Given:
The function is $y = Ax$, where A is a constant.
The differential equation is $xy' = y$, with the condition $x \neq 0$.
To Verify:
Verify that the given function $y = Ax$ is a solution of the given differential equation $xy' = y$.
Solution:
To verify if $y = Ax$ is a solution to the differential equation, we need to calculate the first derivative of the function with respect to $x$ and substitute it, along with the function itself, into the differential equation.
Given the function:
$y = Ax$
Calculate the first derivative $y' = \frac{dy}{dx}$:
$y' = \frac{d}{dx}(Ax)$
$y' = A \frac{d}{dx}(x)$
$y' = A \cdot 1$
$y' = A$
Now substitute $y'$ into the left-hand side (LHS) of the given differential equation $xy' = y$:
LHS = $xy'$
LHS = $x(A)$
LHS = $Ax$
The right-hand side (RHS) of the differential equation is $y$. Substitute the given function $y = Ax$ into the RHS:
RHS = $y$
RHS = $Ax$
We found that LHS = $Ax$ and RHS = $Ax$.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = Ax$ is a solution of the differential equation $xy' = y$ (for $x \neq 0$).
Question 6.
$y = x \sin x$
:
$xy' = y + x \sqrt{x^2 - y^2} \quad (x \neq 0 \text{ and } x > y \text{ or } x < -y)$
Answer:
Given:
The function is $y = x \sin x$.
The differential equation is $xy' = y + x \sqrt{x^2 - y^2}$, with conditions $x \neq 0$ and $x > y$ or $x < -y$.
To Verify:
Verify that the given function $y = x \sin x$ is a solution of the given differential equation $xy' = y + x \sqrt{x^2 - y^2}$.
Solution:
To verify if $y = x \sin x$ is a solution to the differential equation, we need to calculate the first derivative of the function with respect to $x$ and substitute it, along with the function itself, into the differential equation.
Given the function:
$y = x \sin x$
Calculate the first derivative $y' = \frac{dy}{dx}$ using the product rule:
$y' = \frac{d}{dx}(x \sin x)$
$y' = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x)$
$y' = 1 \cdot \sin x + x \cdot \cos x$
$y' = \sin x + x \cos x$
Now substitute $y'$ into the left-hand side (LHS) of the given differential equation $xy' = y + x \sqrt{x^2 - y^2}$:
LHS = $xy'$
LHS = $x(\sin x + x \cos x)$
LHS = $x \sin x + x^2 \cos x$
Now consider the right-hand side (RHS) of the differential equation and substitute the given function $y = x \sin x$:
RHS = $y + x \sqrt{x^2 - y^2}$
Substitute $y = x \sin x$ into the RHS:
RHS = $(x \sin x) + x \sqrt{x^2 - (x \sin x)^2}$
RHS = $x \sin x + x \sqrt{x^2 - x^2 \sin^2 x}$
Factor out $x^2$ from the term under the square root:
RHS = $x \sin x + x \sqrt{x^2(1 - \sin^2 x)}$
Use the trigonometric identity $1 - \sin^2 x = \cos^2 x$:
RHS = $x \sin x + x \sqrt{x^2 \cos^2 x}$
Simplify the square root: $\sqrt{A^2} = |A|$.
RHS = $x \sin x + x |x \cos x|$
The condition $x > y$ or $x < -y$ implies $x^2 > y^2$, which is $x^2 > x^2 \sin^2 x$, or $x^2(1-\sin^2 x) > 0$. Since $x \neq 0$, $x^2 > 0$, so $1-\sin^2 x > 0$, which means $\cos^2 x > 0$. This implies $\cos x \neq 0$. The conditions also ensure $x^2 - y^2 > 0$, so the square root is real and non-zero.
The verification requires LHS = RHS, which means:
$x \sin x + x^2 \cos x = x \sin x + x |x \cos x|$
Subtract $x \sin x$ from both sides:
$x^2 \cos x = x |x \cos x|$
We know that $|x \cos x| = |x| |\cos x|$.
If $x > 0$, $|x| = x$, so $x^2 \cos x = x (x |\cos x|) = x^2 |\cos x|$. Since $x \neq 0$, $\cos x = |\cos x|$, which implies $\cos x \ge 0$.
If $x < 0$, $|x| = -x$, so $x^2 \cos x = x (-x |\cos x|) = -x^2 |\cos x|$. Since $x \neq 0$, $\cos x = -|\cos x|$, which implies $\cos x \le 0$.
The equality $x^2 \cos x = x |x \cos x|$ holds if and only if $x \cos x \ge 0$.
Assuming the domain is restricted such that $x \cos x \ge 0$ (in addition to the stated conditions which ensure $x^2-y^2>0$), the verification holds.
Thus, substituting the calculated $y'$ and the given $y$ into the differential equation, we get:
LHS = $x(\sin x + x \cos x)$
RHS = $x \sin x + x \sqrt{x^2 \cos^2 x} = x \sin x + x |x \cos x|$
For $x \cos x \ge 0$, $|x \cos x| = x \cos x$.
RHS = $x \sin x + x (x \cos x) = x \sin x + x^2 \cos x$
In this case, LHS = RHS.
Therefore, the function $y = x \sin x$ is a solution of the differential equation $xy' = y + x \sqrt{x^2 - y^2}$ for values of $x$ where $x \neq 0$, $x^2 - y^2 > 0$, and $x \cos x \ge 0$. The given conditions $x \neq 0$ and $x > y$ or $x < -y$ ensure $x^2 - y^2 > 0$.
Question 7.
$xy = \log y + C$
:
$y' = \frac{y^2}{1 - xy} \quad (xy \neq 1)$
Answer:
Given:
The implicit function is $xy = \log y + C$, where C is a constant.
The differential equation is $y' = \frac{y^2}{1 - xy}$, with the condition $xy \neq 1$.
To Verify:
Verify that the given implicit function $xy = \log y + C$ is a solution of the given differential equation $y' = \frac{y^2}{1 - xy}$.
Solution:
To verify if the implicit function $xy = \log y + C$ is a solution to the differential equation, we need to implicitly differentiate the function with respect to $x$ to find $\frac{dy}{dx} = y'$ and then compare it with the given differential equation.
Given the implicit function:
$xy = \log y + C$
Differentiate both sides with respect to $x$:
$\frac{d}{dx}(xy) = \frac{d}{dx}(\log y) + \frac{d}{dx}(C)$
Using the product rule on the left side and the chain rule on the $\log y$ term on the right side:
$\left( \frac{d}{dx}(x) \right) y + x \left( \frac{d}{dx}(y) \right) = \frac{1}{y} \frac{dy}{dx} + 0$
$1 \cdot y + x \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$
$y + xy' = \frac{1}{y} y'$
Now, we need to solve this equation for $y'$:
Subtract $\frac{1}{y} y'$ from both sides:
$y = \frac{1}{y} y' - xy'$
Factor out $y'$ from the terms on the right side:
$y = y' \left( \frac{1}{y} - x \right)$
Simplify the expression in the parenthesis by finding a common denominator:
$y = y' \left( \frac{1 - xy}{y} \right)$
Now, isolate $y'$ by multiplying both sides by $\frac{y}{1 - xy}$ (assuming $1 - xy \neq 0$, which is given as $xy \neq 1$):
$y \cdot \frac{y}{1 - xy} = y'$
$y' = \frac{y^2}{1 - xy}$
The calculated derivative $y'$ is $\frac{y^2}{1 - xy}$.
The given differential equation is $y' = \frac{y^2}{1 - xy}$.
Since the calculated $y'$ from the implicit function matches the $y'$ in the given differential equation, the function is a solution.
Therefore, the implicit function $xy = \log y + C$ is a solution of the differential equation $y' = \frac{y^2}{1 - xy}$ (for $xy \neq 1$).
Question 8.
$y - \cos y = x$
:
$(y \sin y + \cos y + x) y' = y$
Answer:
Given:
The implicit function is $y - \cos y = x$.
The differential equation is $(y \sin y + \cos y + x) y' = y$.
To Verify:
Verify that the given implicit function $y - \cos y = x$ is a solution of the given differential equation $(y \sin y + \cos y + x) y' = y$.
Solution:
To verify if $y - \cos y = x$ is a solution to the differential equation, we need to implicitly differentiate the function with respect to $x$ to find $\frac{dy}{dx} = y'$ and then substitute this into the differential equation.
Given the implicit function:
$y - \cos y = x$
Differentiate both sides with respect to $x$:
$\frac{d}{dx}(y - \cos y) = \frac{d}{dx}(x)$
Using the chain rule for the terms involving $y$:
$\frac{dy}{dx} - \frac{d}{dy}(\cos y) \frac{dy}{dx} = 1$
$y' - (-\sin y) y' = 1$
$y' + \sin y \cdot y' = 1$
Factor out $y'$ from the terms on the left side:
$y' (1 + \sin y) = 1$
Solve for $y'$:
$y' = \frac{1}{1 + \sin y}$
Now substitute the expression for $y'$ and the original function $x = y - \cos y$ into the left-hand side (LHS) of the given differential equation $(y \sin y + \cos y + x) y' = y$:
LHS = $(y \sin y + \cos y + x) y'$
Substitute $x = y - \cos y$ into the first parenthesis:
LHS = $(y \sin y + \cos y + (y - \cos y)) y'$
Simplify the expression inside the parenthesis:
LHS = $(y \sin y + \cos y + y - \cos y) y'$
LHS = $(y \sin y + y) y'$
Factor out $y$ from the first parenthesis:
LHS = $y (\sin y + 1) y'$
Now substitute the expression for $y' = \frac{1}{1 + \sin y}$ into the LHS:
LHS = $y (1 + \sin y) \left( \frac{1}{1 + \sin y} \right)$
Assuming $1 + \sin y \neq 0$, we can cancel the term $(1 + \sin y)$:
LHS = $y$
The right-hand side (RHS) of the differential equation is $y$.
Since LHS = RHS, the given implicit function satisfies the differential equation.
Therefore, the implicit function $y - \cos y = x$ is a solution of the differential equation $(y \sin y + \cos y + x) y' = y$ (provided $1 + \sin y \neq 0$).
Question 9.
$x + y = \tan^{-1} y$
:
$y^2 y' + y^2 + 1 = 0$
Answer:
Given:
The implicit function is $x + y = \tan^{-1} y$.
The differential equation is $y^2 y' + y^2 + 1 = 0$.
To Verify:
Verify that the given implicit function $x + y = \tan^{-1} y$ is a solution of the given differential equation $y^2 y' + y^2 + 1 = 0$.
Solution:
To verify if $x + y = \tan^{-1} y$ is a solution to the differential equation, we need to implicitly differentiate the function with respect to $x$ to find $\frac{dy}{dx} = y'$ and then substitute this into the differential equation.
Given the implicit function:
$x + y = \tan^{-1} y$
Differentiate both sides with respect to $x$:
$\frac{d}{dx}(x + y) = \frac{d}{dx}(\tan^{-1} y)$
Using the chain rule for the terms involving $y$:
$\frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dy}(\tan^{-1} y) \cdot \frac{dy}{dx}$
$1 + \frac{dy}{dx} = \frac{1}{1+y^2} \cdot \frac{dy}{dx}$
$1 + y' = \frac{1}{1+y^2} y'$
Now, we need to rearrange this equation to solve for $y'$:
$1 = \frac{1}{1+y^2} y' - y'$
$1 = y' \left( \frac{1}{1+y^2} - 1 \right)$
$1 = y' \left( \frac{1 - (1+y^2)}{1+y^2} \right)$
$1 = y' \left( \frac{1 - 1 - y^2}{1+y^2} \right)$
$1 = y' \left( \frac{-y^2}{1+y^2} \right)$
Multiply both sides by $\frac{1+y^2}{-y^2}$ to isolate $y'$ (assuming $-y^2 \neq 0$, i.e., $y \neq 0$):
$y' = \frac{1+y^2}{-y^2} = -\frac{1+y^2}{y^2}$
Now substitute the expression for $y' = -\frac{1+y^2}{y^2}$ into the left-hand side (LHS) of the given differential equation $y^2 y' + y^2 + 1 = 0$:
LHS = $y^2 y' + y^2 + 1$
Substitute the expression for $y'$:
LHS = $y^2 \left( -\frac{1+y^2}{y^2} \right) + y^2 + 1$
Assuming $y^2 \neq 0$, we can cancel the $y^2$ terms:
LHS = $-(1+y^2) + y^2 + 1$
LHS = $-1 - y^2 + y^2 + 1$
LHS = $(-y^2 + y^2) + (-1 + 1)$
LHS = $0 + 0$
LHS = $0$
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given implicit function satisfies the differential equation (for $y \neq 0$).
Therefore, the implicit function $x + y = \tan^{-1} y$ is a solution of the differential equation $y^2 y' + y^2 + 1 = 0$ (provided $y \neq 0$).
Question 10.
$y = \sqrt{a^2 - x^2} \quad x \in (-a, a)$
:
$x + y \frac{dy}{dx} = 0 \quad (y \neq 0)$
Answer:
Given:
The function is $y = \sqrt{a^2 - x^2}$, with $x \in (-a, a)$.
The differential equation is $x + y \frac{dy}{dx} = 0$, with the condition $y \neq 0$.
To Verify:
Verify that the given function $y = \sqrt{a^2 - x^2}$ is a solution of the given differential equation $x + y \frac{dy}{dx} = 0$.
Solution:
To verify if $y = \sqrt{a^2 - x^2}$ is a solution to the differential equation, we need to calculate the first derivative of the function with respect to $x$ and substitute it, along with the function itself, into the differential equation.
Given the function:
$y = \sqrt{a^2 - x^2} = (a^2 - x^2)^{1/2}$
Calculate the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}((a^2 - x^2)^{1/2})$
$\frac{dy}{dx} = \frac{1}{2}(a^2 - x^2)^{-1/2} \cdot \frac{d}{dx}(a^2 - x^2)$
$\frac{dy}{dx} = \frac{1}{2}(a^2 - x^2)^{-1/2} \cdot (-2x)$
$\frac{dy}{dx} = -x (a^2 - x^2)^{-1/2}$
$\frac{dy}{dx} = -\frac{x}{\sqrt{a^2 - x^2}}$
Now substitute the calculated derivative $\frac{dy}{dx} = -\frac{x}{\sqrt{a^2 - x^2}}$ and the given function $y = \sqrt{a^2 - x^2}$ into the left-hand side (LHS) of the differential equation $x + y \frac{dy}{dx} = 0$:
LHS = $x + y \frac{dy}{dx}$
LHS = $x + (\sqrt{a^2 - x^2}) \left( -\frac{x}{\sqrt{a^2 - x^2}} \right)$
Since $x \in (-a, a)$, $a^2 - x^2 > 0$, so $\sqrt{a^2 - x^2}$ is a real positive number. Also, the condition $y \neq 0$ implies $\sqrt{a^2-x^2} \neq 0$, which means $a^2 - x^2 \neq 0$. So we can cancel the term $\sqrt{a^2 - x^2}$ from the numerator and denominator:
LHS = $x + \cancel{\sqrt{a^2 - x^2}} \left( -\frac{x}{\cancel{\sqrt{a^2 - x^2}}} \right)$
LHS = $x + (-x)$
LHS = $x - x$
LHS = $0$
The right-hand side (RHS) of the differential equation is 0.
Since LHS = RHS, the given function satisfies the differential equation.
Therefore, the function $y = \sqrt{a^2 - x^2}$ is a solution of the differential equation $x + y \frac{dy}{dx} = 0$ (for $x \in (-a, a)$ and $y \neq 0$).
Question 11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
Answer:
Given:
A differential equation of fourth order.
To Find:
The number of arbitrary constants in the general solution.
Solution:
The general solution of a differential equation of order $n$ contains $n$ arbitrary constants.
The given differential equation is of fourth order.
Here, the order of the differential equation is $n = 4$.
Therefore, the number of arbitrary constants in its general solution is 4.
Comparing this with the given options, we find that option (D) is correct.
The correct answer is (D) 4.
Question 12. The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
Answer:
Given:
A differential equation of third order.
To Find:
The number of arbitrary constants in the particular solution.
Solution:
A general solution of a differential equation of order $n$ contains $n$ arbitrary constants.
A particular solution is obtained from the general solution by giving specific values to the arbitrary constants.
Once specific values are assigned to the constants, they are no longer arbitrary. Therefore, a particular solution contains no arbitrary constants.
The order of the differential equation in this question is 3.
While the general solution would have 3 arbitrary constants, the particular solution, by definition, has these constants replaced by specific determined values.
Thus, the number of arbitrary constants in the particular solution is 0.
Comparing this with the given options, we find that option (D) is correct.
The correct answer is (D) 0.
Example 4 to 8 (Before Exercise 9.3)
Example 4: Form the differential equation representing the family of curves y = mx, where, m is arbitrary constant.
Answer:
Given:
The family of curves is represented by the equation $y = mx$, where $m$ is an arbitrary constant.
To Find:
Form the differential equation representing this family of curves.
Solution:
The given equation is:
$y = mx$
The number of arbitrary constants in the given equation is one (which is $m$). Therefore, the order of the differential equation will be 1. We need to eliminate the constant $m$ by differentiating the equation with respect to $x$.
Differentiate both sides of the equation $y = mx$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(mx)$
$\frac{dy}{dx} = m \frac{d}{dx}(x)$
$\frac{dy}{dx} = m \cdot 1$
$y' = m$
Now we have an expression for $m$ in terms of $y'$. Substitute this expression for $m$ back into the original equation $y = mx$ to eliminate the constant $m$:
$y = (y')x$
This equation relates $y$, $x$, and $y'$, and it does not contain the arbitrary constant $m$. This is the required differential equation.
We can rearrange the equation into a standard form:
$y = xy'$
or
$xy' - y = 0$
or
$y - xy' = 0$
The differential equation representing the family of curves $y = mx$ is $xy' - y = 0$.
Example 5: Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.
Answer:
Given:
The family of curves is represented by the equation $y = a \sin (x + b)$, where $a$ and $b$ are arbitrary constants.
To Find:
Form the differential equation representing this family of curves.
Solution:
The given equation is:
$y = a \sin (x + b)$
There are two arbitrary constants ($a$ and $b$). Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate the given equation with respect to $x$ to find the first derivative, $y'$:
$y' = \frac{dy}{dx} = \frac{d}{dx}(a \sin (x + b))$
$y' = a \cos (x + b) \cdot \frac{d}{dx}(x+b)$
$y' = a \cos (x + b) \cdot 1$
$y' = a \cos (x + b)$
Differentiate the first derivative equation with respect to $x$ to find the second derivative, $y''$:
$y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}(a \cos (x + b))$
$y'' = a \frac{d}{dx}(\cos (x + b)) \cdot \frac{d}{dx}(x+b)$
$y'' = a (-\sin (x + b)) \cdot 1$
$y'' = -a \sin (x + b)$
Now we have the following equations:
$y = a \sin (x + b)$
$y' = a \cos (x + b)$
$y'' = -a \sin (x + b)$
From the third equation, we can see that $y'' = - (a \sin (x + b))$.
Substitute the expression for $a \sin (x + b)$ from the first equation ($y = a \sin (x + b)$) into this equation:
$y'' = -y$
Rearrange the equation to get the differential equation in the standard form:
$y'' + y = 0$
This equation relates $y$ and its derivatives and does not contain the arbitrary constants $a$ and $b$. This is the required differential equation.
The differential equation representing the family of curves $y = a \sin (x + b)$ is $y'' + y = 0$.
Example 6: Form the differential equation representing the family of ellipses having foci on x-axis and centre at the origin.
Answer:
Given:
The family of ellipses has foci on the x-axis and the centre at the origin.
To Find:
Form the differential equation representing this family of ellipses.
Solution:
The standard equation of an ellipse with foci on the x-axis and centre at the origin is given by:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
where $a$ and $b$ are the semi-major and semi-minor axes, respectively. Since the foci are on the x-axis, $a > b$. The constants $a^2$ and $b^2$ are arbitrary constants that determine the specific ellipse in the family.
There are two arbitrary constants ($a^2$ and $b^2$) in the equation. Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate the equation with respect to $x$:
$\frac{d}{dx}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = \frac{d}{dx}(1)$
$\frac{1}{a^2} \frac{d}{dx}(x^2) + \frac{1}{b^2} \frac{d}{dx}(y^2) = 0$
$\frac{1}{a^2} (2x) + \frac{1}{b^2} (2y \frac{dy}{dx}) = 0$
Divide by 2:
$\frac{x}{a^2} + \frac{y}{b^2} \frac{dy}{dx} = 0$
We can write $\frac{dy}{dx}$ as $y'$:
$\frac{x}{a^2} + \frac{y y'}{b^2} = 0$
Rearrange to isolate the terms with $a^2$ and $b^2$ on opposite sides:
$\frac{x}{a^2} = -\frac{y y'}{b^2}$
... (1)
Now, differentiate equation (1) with respect to $x$. Use the quotient rule on the LHS and the product rule on the RHS:
$\frac{d}{dx}\left(\frac{x}{a^2}\right) = \frac{d}{dx}\left(-\frac{y y'}{b^2}\right)$
$\frac{1}{a^2} \frac{d}{dx}(x) = -\frac{1}{b^2} \frac{d}{dx}(y y')$
$\frac{1}{a^2} (1) = -\frac{1}{b^2} \left( \frac{d}{dx}(y) y' + y \frac{d}{dx}(y') \right)$
$\frac{1}{a^2} = -\frac{1}{b^2} (y' \cdot y' + y \cdot y'')$
$\frac{1}{a^2} = -\frac{1}{b^2} ((y')^2 + y y'')$
$\frac{1}{a^2} = -\frac{(y')^2 + y y''}{b^2}$
... (2)
Now we have two equations involving $\frac{1}{a^2}$ and $\frac{1}{b^2}$ (or their ratio).
From equation (1), we can write:
$\frac{1}{a^2} = -\frac{y y'}{x b^2}$
Substitute this expression for $\frac{1}{a^2}$ into equation (2):
$-\frac{y y'}{x b^2} = -\frac{(y')^2 + y y''}{b^2}$
Multiply both sides by $-b^2$ (assuming $b^2 \neq 0$):
$\frac{y y'}{x} = (y')^2 + y y''$
Multiply both sides by $x$ (assuming $x \neq 0$):
$y y' = x (y')^2 + x y y''$
Rearrange the terms to set the equation to zero:
$x y y'' + x (y')^2 - y y' = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$.
The differential equation representing the family of ellipses having foci on the x-axis and centre at the origin is $x y y'' + x (y')^2 - y y' = 0$ (for $x \neq 0$).
Example 7 Form the differential equation of the family of circles touching the x-axis at origin.
Answer:
Given:
The family of circles touching the x-axis at the origin.
To Find:
Form the differential equation representing this family of circles.
Solution:
A circle touching the x-axis at the origin must have its centre on the y-axis. Let the coordinates of the centre be $(0, a)$. Since the circle touches the x-axis at the origin $(0, 0)$, the radius of the circle is the distance from the centre $(0, a)$ to the origin $(0, 0)$, which is $|a|$. For simplicity, we can consider the case where the centre is at $(0, a)$ with radius $a$, where $a$ is a non-zero arbitrary constant. The equation of such a circle is:
$(x - 0)^2 + (y - a)^2 = a^2$
$x^2 + y^2 - 2ay + a^2 = a^2$
Simplifying the equation, we get:
$x^2 + y^2 - 2ay = 0$
This equation represents the family of circles touching the x-axis at the origin. The arbitrary constant is $a$. Since there is one arbitrary constant, the order of the required differential equation will be 1.
We need to eliminate the constant $a$ by differentiating the equation with respect to $x$.
Differentiate the equation $x^2 + y^2 - 2ay = 0$ with respect to $x$:
$\frac{d}{dx}(x^2 + y^2 - 2ay) = \frac{d}{dx}(0)$
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ay) = 0$
Using the power rule and chain rule for differentiation:
$2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$
$2x + 2yy' - 2ay' = 0$
Now we need to eliminate $a$ from this equation and the original equation $x^2 + y^2 - 2ay = 0$.
From the original equation, we can express $2a$ in terms of $x$ and $y$ (assuming $y \neq 0$):
$2ay = x^2 + y^2$
$2a = \frac{x^2 + y^2}{y}$
Substitute this expression for $2a$ into the differentiated equation $2x + 2yy' - 2ay' = 0$:
$2x + 2yy' - \left(\frac{x^2 + y^2}{y}\right) y' = 0$
To clear the denominator, multiply the entire equation by $y$ (assuming $y \neq 0$):
$2xy + 2y^2y' - (x^2 + y^2)y' = 0$
$2xy + 2y^2y' - x^2y' - y^2y' = 0$
Combine the terms containing $y'$:
$2xy + (2y^2 - x^2 - y^2)y' = 0$
$2xy + (y^2 - x^2)y' = 0$
This equation relates $x$, $y$, and $y'$ and does not contain the arbitrary constant $a$. This is the required differential equation.
The differential equation representing the family of circles touching the x-axis at the origin is $(y^2 - x^2) y' + 2xy = 0$.
Example 8: Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Answer:
Given:
The family of parabolas has vertex at the origin and axis along the positive direction of the x-axis.
To Find:
Form the differential equation representing this family of parabolas.
Solution:
The standard equation of a parabola with vertex at the origin $(0,0)$ and axis along the positive x-axis is of the form:
$y^2 = 4ax$
where $a$ is a non-zero arbitrary constant (specifically $a > 0$ for the axis to be along the positive x-axis). This equation represents the family of such parabolas.
There is one arbitrary constant ($a$) in the equation. Therefore, the order of the required differential equation will be 1. We need to eliminate the constant $a$ by differentiating the equation with respect to $x$.
Differentiate the given equation $y^2 = 4ax$ implicitly with respect to $x$:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)$
Using the chain rule on the left side and the constant multiple rule on the right side:
$2y \frac{dy}{dx} = 4a \frac{d}{dx}(x)$
$2y y' = 4a \cdot 1$
$2y y' = 4a$
Now we need to eliminate $a$ using this equation and the original equation $y^2 = 4ax$.
From the differentiated equation, we can express $4a$ directly:
$4a = 2yy'$
Substitute this expression for $4a$ back into the original equation $y^2 = 4ax$:
$y^2 = (4a)x$
$y^2 = (2yy')x$
$y^2 = 2xyy'$
Rearrange the equation to express it as a differential equation:
$y^2 - 2xyy' = 0$
or
$2xyy' - y^2 = 0$
This equation relates $x$, $y$, and $y'$ and does not contain the arbitrary constant $a$. This is the required differential equation.
We can also write it as:
$y(y - 2xy') = 0$
This implies either $y=0$ (the axis, which is a part of the closure of the parabola family, but typically not the intended differential equation for the curves themselves) or $y - 2xy' = 0$.
For points on the parabola where $y \neq 0$, we can write:
$2xy' = y$
$y' = \frac{y}{2x}$
This form is also a valid differential equation for the family, valid for $x \neq 0$ and $y \neq 0$. The form $2xyy' - y^2 = 0$ includes the points where $y=0$ (which is just the origin for $a>0$) without explicit division by $y$.
The differential equation representing the family of parabolas having vertex at origin and axis along the positive direction of x-axis is $2xyy' - y^2 = 0$.
Exercise 9.3
In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
Question 1. $\frac{x}{a} + \frac{y}{b} = 1$
Answer:
Given:
The family of curves is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are arbitrary constants.
To Form:
Form the differential equation representing this family of curves by eliminating the arbitrary constants $a$ and $b$.
Solution:
The given equation is:
$\frac{x}{a} + \frac{y}{b} = 1$
There are two arbitrary constants ($a$ and $b$) in the equation. Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate the given equation with respect to $x$:
$\frac{d}{dx}\left(\frac{x}{a} + \frac{y}{b}\right) = \frac{d}{dx}(1)$
$\frac{1}{a} \frac{d}{dx}(x) + \frac{1}{b} \frac{d}{dx}(y) = 0$
$\frac{1}{a} (1) + \frac{1}{b} \frac{dy}{dx} = 0$
$\frac{1}{a} + \frac{1}{b} y' = 0$
... (1)
Now, differentiate equation (1) with respect to $x$:
$\frac{d}{dx}\left(\frac{1}{a} + \frac{1}{b} y'\right) = \frac{d}{dx}(0)$
$\frac{d}{dx}\left(\frac{1}{a}\right) + \frac{1}{b} \frac{d}{dx}(y') = 0$
Since $a$ and $b$ are constants, $\frac{d}{dx}(\frac{1}{a}) = 0$. The derivative of $y'$ is $y''$.
$0 + \frac{1}{b} y'' = 0$
$\frac{1}{b} y'' = 0$
... (2)
From equation (2), assuming $b$ is a non-zero arbitrary constant (which is required for the term $\frac{y}{b}$ to be well-defined and represent a family of non-vertical lines), we can multiply by $b$:
$y'' = 0$
This differential equation relates the second derivative of $y$ with respect to $x$ to zero. It does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of curves $\frac{x}{a} + \frac{y}{b} = 1$ is $y'' = 0$.
Question 2. y2 = a (b2 – x2)
Answer:
Given:
The family of curves is $y^2 = a (b^2 - x^2)$, where $a$ and $b$ are arbitrary constants.
To Form:
Form the differential equation representing this family of curves by eliminating the arbitrary constants $a$ and $b$.
Solution:
The given equation is:
$y^2 = a (b^2 - x^2)$
Rearrange the equation:
$y^2 = ab^2 - ax^2$
There are two arbitrary constants ($a$ and $b$). We can consider $ab^2$ and $-a$ as two independent constants. Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate the given equation with respect to $x$ implicitly:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(ab^2 - ax^2)$
$2y \frac{dy}{dx} = 0 - a(2x)$
$2y y' = -2ax$
Divide both sides by 2:
$y y' = -ax$
... (1)
Now, differentiate equation (1) with respect to $x$ using the product rule on the left side:
$\frac{d}{dx}(y y') = \frac{d}{dx}(-ax)$
$\left(\frac{d}{dx}(y)\right) y' + y \left(\frac{d}{dx}(y')\right) = -a \frac{d}{dx}(x)$
$y' \cdot y' + y \cdot y'' = -a \cdot 1$
$(y')^2 + y y'' = -a$
... (2)
Now we have two equations involving the constants $a$ and $b$ (though $b$ has already been eliminated), or involving expressions that help eliminate $a$. From equation (2), we have an expression for $-a$. Substitute this expression for $-a$ into equation (1):
$y y' = (-a)x$
$y y' = ((y')^2 + y y'')x$
Expand the right side:
$y y' = x(y')^2 + xy y''$
Rearrange the terms to form the differential equation:
$xy y'' + x(y')^2 - y y' = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of curves $y^2 = a (b^2 - x^2)$ is $xy y'' + x(y')^2 - y y' = 0$.
Question 3. y = a e3x + b e–2x
Answer:
Given:
The family of curves is $y = a e^{3x} + b e^{-2x}$, where $a$ and $b$ are arbitrary constants.
To Form:
Form the differential equation representing this family of curves by eliminating the arbitrary constants $a$ and $b$.
Solution:
The given equation is:
y = a e$^{3x}$ + b e$^{–2x}$
... (1)
There are two arbitrary constants ($a$ and $b$). Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate equation (1) with respect to $x$ to find the first derivative, $y'$:
$y' = \frac{d}{dx}(a e^{3x} + b e^{-2x})$
$y' = a \frac{d}{dx}(e^{3x}) + b \frac{d}{dx}(e^{-2x})$
$y' = a (3e^{3x}) + b (-2e^{-2x})$
y' = 3a e$^{3x}$ - 2b e$^{–2x}$
... (2)
Differentiate equation (2) with respect to $x$ to find the second derivative, $y''$:
$y'' = \frac{d}{dx}(3a e^{3x} - 2b e^{-2x})$
$y'' = 3a \frac{d}{dx}(e^{3x}) - 2b \frac{d}{dx}(e^{-2x})$
$y'' = 3a (3e^{3x}) - 2b (-2e^{-2x})$
y'' = 9a e$^{3x}$ + 4b e$^{–2x}$
... (3)
Now we have three equations:
(1) y = a e$^{3x}$ + b e$^{–2x}$
(2) y' = 3a e$^{3x}$ - 2b e$^{–2x}$
(3) y'' = 9a e$^{3x}$ + 4b e$^{–2x}$
We can eliminate $a e^{3x}$ and $b e^{-2x}$ from these equations. Multiply equation (1) by 2 and add it to equation (2):
$2 \times (1) \implies 2y = 2a e^{3x} + 2b e^{-2x}$
$(2y) + (y') = (2a e^{3x} + 2b e^{-2x}) + (3a e^{3x} - 2b e^{-2x})$
$2y + y' = 5a e^{3x}$
5a e$^{3x} = 2y + y'$
... (4)
Multiply equation (1) by 3 and subtract equation (2) from it:
$3 \times (1) \implies 3y = 3a e^{3x} + 3b e^{-2x}$
$(3y) - (y') = (3a e^{3x} + 3b e^{-2x}) - (3a e^{3x} - 2b e^{-2x})$
$3y - y' = 5b e^{-2x}$
5b e$^{–2x} = 3y - y'$
... (5)
From equation (4), $a e^{3x} = \frac{2y + y'}{5}$. From equation (5), $b e^{-2x} = \frac{3y - y'}{5}$.
Now substitute these expressions into equation (3):
$y'' = 9a e^{3x} + 4b e^{-2x}$
$y'' = 9 \left( \frac{2y + y'}{5} \right) + 4 \left( \frac{3y - y'}{5} \right)$
Multiply both sides by 5:
$5y'' = 9(2y + y') + 4(3y - y')$
$5y'' = 18y + 9y' + 12y - 4y'$
$5y'' = (18y + 12y) + (9y' - 4y')$
$5y'' = 30y + 5y'$
Divide both sides by 5:
$y'' = 6y + y'$
Rearrange the terms to form the differential equation:
$y'' - y' - 6y = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of curves $y = a e^{3x} + b e^{-2x}$ is $y'' - y' - 6y = 0$.
Question 4. y = e2x (a + bx)
Answer:
Given:
The family of curves is $y = e^{2x} (a + bx)$, where $a$ and $b$ are arbitrary constants.
To Form:
Form the differential equation representing this family of curves by eliminating the arbitrary constants $a$ and $b$.
Solution:
The given equation is:
y = e$^{2x}$ (a + bx)
... (1)
There are two arbitrary constants ($a$ and $b$). Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate equation (1) with respect to $x$ to find the first derivative, $y'$, using the product rule:
$y' = \frac{d}{dx}(e^{2x} (a + bx))$
$y' = \left(\frac{d}{dx}(e^{2x})\right) (a + bx) + e^{2x} \left(\frac{d}{dx}(a + bx)\right)$
$y' = (2e^{2x}) (a + bx) + e^{2x} (b)$
$y' = 2e^{2x}(a + bx) + be^{2x}$
Notice that $e^{2x}(a+bx)$ is equal to $y$ from equation (1).
$y' = 2y + be^{2x}$
Rearrange this equation to isolate the term containing $b$:
be$^{2x} = y' - 2y$
... (2)
Now, differentiate equation (2) with respect to $x$ to find the second derivative, $y''$:
$\frac{d}{dx}(y') = \frac{d}{dx}(2y + be^{2x})$
$y'' = \frac{d}{dx}(2y) + \frac{d}{dx}(be^{2x})$
$y'' = 2\frac{dy}{dx} + b\frac{d}{dx}(e^{2x})$
$y'' = 2y' + b(2e^{2x})$
y'' = 2y' + 2be$^{2x}$
... (3)
Now we use equation (2) to eliminate $be^{2x}$ from equation (3).
Substitute $be^{2x} = y' - 2y$ from (2) into (3):
$y'' = 2y' + 2(y' - 2y)$
Expand the right side:
$y'' = 2y' + 2y' - 4y$
Combine the terms with $y'$:
$y'' = 4y' - 4y$
Rearrange the terms to form the differential equation:
$y'' - 4y' + 4y = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of curves $y = e^{2x} (a + bx)$ is $y'' - 4y' + 4y = 0$.
Question 5. y = ex (a cos x + b sin x)
Answer:
Given:
The family of curves is $y = e^x (a \cos x + b \sin x)$, where $a$ and $b$ are arbitrary constants.
To Form:
Form the differential equation representing this family of curves by eliminating the arbitrary constants $a$ and $b$.
Solution:
The given equation is:
y = e$^x$ (a cos x + b sin x)
... (1)
There are two arbitrary constants ($a$ and $b$). Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate equation (1) with respect to $x$ using the product rule $\frac{d}{dx}(uv) = u'v + uv'$:
$y' = \frac{dy}{dx} = \frac{d}{dx}(e^x (a \cos x + b \sin x))$
$y' = \left(\frac{d}{dx}(e^x)\right) (a \cos x + b \sin x) + e^x \left(\frac{d}{dx}(a \cos x + b \sin x)\right)$
$y' = e^x (a \cos x + b \sin x) + e^x (-a \sin x + b \cos x)$
From equation (1), we know that $e^x (a \cos x + b \sin x) = y$. Substitute this into the expression for $y'$:
y' = y + e$^x$ (-a sin x + b cos x)
... (2)
Rearrange equation (2) to isolate the exponential term multiplied by constants:
$y' - y = e^x (-a \sin x + b \cos x)$
Now differentiate equation (2) with respect to $x$:
$y'' = \frac{d}{dx}(y + e^x (-a \sin x + b \cos x))$
$y'' = \frac{dy}{dx} + \frac{d}{dx}(e^x (-a \sin x + b \cos x))$
Using the product rule on the second term:
$y'' = y' + \left(\frac{d}{dx}(e^x)\right) (-a \sin x + b \cos x) + e^x \left(\frac{d}{dx}(-a \sin x + b \cos x)\right)$
$y'' = y' + e^x (-a \sin x + b \cos x) + e^x (-a \cos x - b \sin x)$
From equation (2), we know that $e^x (-a \sin x + b \cos x) = y' - y$. Substitute this into the equation for $y''$:
$y'' = y' + (y' - y) + e^x (-a \cos x - b \sin x)$
$y'' = 2y' - y - e^x (a \cos x + b \sin x)$
From equation (1), we know that $e^x (a \cos x + b \sin x) = y$. Substitute this into the equation for $y''$:
$y'' = 2y' - y - y$
$y'' = 2y' - 2y$
Rearrange the terms to form the differential equation:
$y'' - 2y' + 2y = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of curves $y = e^x (a \cos x + b \sin x)$ is $y'' - 2y' + 2y = 0$.
Question 6. Form the differential equation of the family of circles touching the y-axis at origin.
Answer:
Given:
The family of circles touching the y-axis at the origin.
To Form:
Form the differential equation representing this family of circles.
Solution:
A circle touching the y-axis at the origin $(0,0)$ must have its centre on the x-axis. Let the coordinates of the centre be $(a, 0)$. Since the circle touches the y-axis at the origin, the radius of the circle is the distance from the centre $(a, 0)$ to the origin $(0, 0)$, which is $|a|$. We can consider the case where the centre is at $(a, 0)$ with radius $a$, where $a$ is a non-zero arbitrary constant. The equation of such a circle is:
$(x - a)^2 + (y - 0)^2 = a^2$
$x^2 - 2ax + a^2 + y^2 = a^2$
Simplifying the equation, we get:
x$^2$ + y$^2$ - 2ax = 0
... (1)
This equation represents the family of circles touching the y-axis at the origin. The arbitrary constant is $a$. Since there is one arbitrary constant, the order of the required differential equation will be 1. We need to eliminate the constant $a$ by differentiating the equation with respect to $x$.
Differentiate equation (1) with respect to $x$:
$\frac{d}{dx}(x^2 + y^2 - 2ax) = \frac{d}{dx}(0)$
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ax) = 0$
Using the power rule and chain rule for differentiation:
$2x + 2y \frac{dy}{dx} - 2a \frac{d}{dx}(x) = 0$
$2x + 2yy' - 2a \cdot 1 = 0$
2x + 2yy' - 2a = 0
... (2)
Now we need to eliminate $a$ from equation (2) and the original equation (1).
From equation (1), we can express $2a$ in terms of $x$ and $y$ (assuming $x \neq 0$):
$2ax = x^2 + y^2$
$2a = \frac{x^2 + y^2}{x}$
Substitute this expression for $2a$ into equation (2):
$2x + 2yy' - \left(\frac{x^2 + y^2}{x}\right) = 0$
To clear the denominator, multiply the entire equation by $x$ (assuming $x \neq 0$):
$x(2x) + x(2yy') - x \left(\frac{x^2 + y^2}{x}\right) = 0$
$2x^2 + 2xyy' - (x^2 + y^2) = 0$
$2x^2 + 2xyy' - x^2 - y^2 = 0$
Combine the terms:
$(2x^2 - x^2) + 2xyy' - y^2 = 0$
$x^2 + 2xyy' - y^2 = 0$
Rearrange the terms to form the differential equation:
$2xyy' + x^2 - y^2 = 0$
This equation relates $x$, $y$, and $y'$ and does not contain the arbitrary constant $a$. This is the required differential equation.
The differential equation representing the family of circles touching the y-axis at the origin is $2xyy' + x^2 - y^2 = 0$ (for $x \neq 0$).
Question 7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Answer:
Given:
The family of parabolas has vertex at the origin and axis along the positive y-axis.
To Form:
Form the differential equation representing this family of parabolas.
Solution:
The standard equation of a parabola with vertex at the origin $(0,0)$ and axis along the positive y-axis is of the form:
x$^2$ = 4ay
... (1)
where $a$ is a non-zero arbitrary constant (specifically $a > 0$ for the axis to be along the positive y-axis). This equation represents the family of such parabolas.
There is one arbitrary constant ($a$) in the equation. Therefore, the order of the required differential equation will be 1. We need to eliminate the constant $a$ by differentiating the equation with respect to $x$.
Differentiate equation (1) with respect to $x$ implicitly:
$\frac{d}{dx}(x^2) = \frac{d}{dx}(4ay)$
Using the power rule and chain rule for differentiation:
$2x = 4a \frac{dy}{dx}$
2x = 4ay'
... (2)
Now we need to eliminate $a$ using equation (1) and equation (2).
From equation (1), we can express $4a$ in terms of $x$ and $y$ (assuming $y \neq 0$):
$4a = \frac{x^2}{y}$
Substitute this expression for $4a$ into equation (2):
$2x = \left(\frac{x^2}{y}\right) y'$
Multiply both sides by $y$ (assuming $y \neq 0$):
$2xy = x^2 y'$
Rearrange the terms to form the differential equation:
$x^2 y' - 2xy = 0$
This equation relates $x$, $y$, and $y'$ and does not contain the arbitrary constant $a$. This is the required differential equation.
Note that we can factor out $x$ from the equation: $x(xy' - 2y) = 0$. This implies $x=0$ or $xy' - 2y = 0$. The line $x=0$ is the y-axis, which contains the axis of the parabola. The equation $xy' - 2y = 0$ is valid for points on the parabola where $x \neq 0$. The form $x^2 y' - 2xy = 0$ includes both cases.
The differential equation representing the family of parabolas having vertex at origin and axis along the positive y-axis is $x^2 y' - 2xy = 0$.
Question 8. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.
Answer:
Given:
The family of ellipses has foci on the y-axis and the centre at the origin.
To Form:
Form the differential equation representing this family of ellipses.
Solution:
The standard equation of an ellipse with foci on the y-axis and centre at the origin is given by:
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
where $a$ is the semi-major axis and $b$ is the semi-minor axis. Since the foci are on the y-axis, $a > b$. The constants $a^2$ and $b^2$ are arbitrary constants that determine the specific ellipse in the family.
There are two arbitrary constants ($a^2$ and $b^2$) in the equation. Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate the equation with respect to $x$:
$\frac{d}{dx}\left(\frac{x^2}{b^2} + \frac{y^2}{a^2}\right) = \frac{d}{dx}(1)$
$\frac{1}{b^2} \frac{d}{dx}(x^2) + \frac{1}{a^2} \frac{d}{dx}(y^2) = 0$
$\frac{1}{b^2} (2x) + \frac{1}{a^2} (2y \frac{dy}{dx}) = 0$
Divide by 2:
$\frac{x}{b^2} + \frac{y}{a^2} \frac{dy}{dx} = 0$
We can write $\frac{dy}{dx}$ as $y'$:
$\frac{x}{b^2} + \frac{y y'}{a^2} = 0$
Rearrange to isolate the terms with $a^2$ and $b^2$ on opposite sides:
$\frac{x}{b^2} = -\frac{y y'}{a^2}$
... (1)
Now, differentiate equation (1) with respect to $x$. Use the quotient rule on the LHS (conceptually, treating $b^2$ as a constant) and the product rule on the RHS:
$\frac{d}{dx}\left(\frac{x}{b^2}\right) = \frac{d}{dx}\left(-\frac{y y'}{a^2}\right)$
$\frac{1}{b^2} \frac{d}{dx}(x) = -\frac{1}{a^2} \frac{d}{dx}(y y')$
$\frac{1}{b^2} (1) = -\frac{1}{a^2} \left( \frac{d}{dx}(y) y' + y \frac{d}{dx}(y') \right)$
$\frac{1}{b^2} = -\frac{1}{a^2} (y' \cdot y' + y \cdot y'')$
$\frac{1}{b^2} = -\frac{(y')^2 + y y''}{a^2}$
... (2)
Now we have two equations involving $\frac{1}{a^2}$ and $\frac{1}{b^2}$.
From equation (1), we can write:
$\frac{1}{b^2} = -\frac{y y'}{x a^2}$
Substitute this expression for $\frac{1}{b^2}$ into equation (2):
$-\frac{y y'}{x a^2} = -\frac{(y')^2 + y y''}{a^2}$
Multiply both sides by $-a^2$ (assuming $a^2 \neq 0$):
$\frac{y y'}{x} = (y')^2 + y y''$
Multiply both sides by $x$ (assuming $x \neq 0$):
$y y' = x (y')^2 + x y y''$
Rearrange the terms to set the equation to zero:
$x y y'' + x (y')^2 - y y' = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of ellipses having foci on the y-axis and centre at the origin is $x y y'' + x (y')^2 - y y' = 0$ (for $x \neq 0$).
Question 9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Answer:
Given:
The family of hyperbolas has foci on the x-axis and the centre at the origin.
To Form:
Form the differential equation representing this family of hyperbolas.
Solution:
The standard equation of a hyperbola with foci on the x-axis and centre at the origin is given by:
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
where $a$ and $b$ are arbitrary constants related to the semi-transverse and semi-conjugate axes. The constants $a^2$ and $b^2$ are arbitrary constants that determine the specific hyperbola in the family.
There are two arbitrary constants ($a^2$ and $b^2$) in the equation. Therefore, the order of the required differential equation will be 2. We need to differentiate the equation twice with respect to $x$ to eliminate these two constants.
Differentiate the equation with respect to $x$:
$\frac{d}{dx}\left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) = \frac{d}{dx}(1)$
$\frac{1}{a^2} \frac{d}{dx}(x^2) - \frac{1}{b^2} \frac{d}{dx}(y^2) = 0$
$\frac{1}{a^2} (2x) - \frac{1}{b^2} (2y \frac{dy}{dx}) = 0$
Divide by 2:
$\frac{x}{a^2} - \frac{y}{b^2} \frac{dy}{dx} = 0$
We can write $\frac{dy}{dx}$ as $y'$:
$\frac{x}{a^2} - \frac{y y'}{b^2} = 0$
Rearrange to move the terms with $a^2$ and $b^2$ to opposite sides:
$\frac{x}{a^2} = \frac{y y'}{b^2}$
... (1)
Now, differentiate equation (1) with respect to $x$. Use the quotient rule on the LHS (conceptually, treating $a^2$ as a constant) and the product rule on the RHS:
$\frac{d}{dx}\left(\frac{x}{a^2}\right) = \frac{d}{dx}\left(\frac{y y'}{b^2}\right)$
$\frac{1}{a^2} \frac{d}{dx}(x) = \frac{1}{b^2} \frac{d}{dx}(y y')$
$\frac{1}{a^2} (1) = \frac{1}{b^2} \left( \frac{d}{dx}(y) y' + y \frac{d}{dx}(y') \right)$
$\frac{1}{a^2} = \frac{1}{b^2} (y' \cdot y' + y \cdot y'')$
$\frac{1}{a^2} = \frac{(y')^2 + y y''}{b^2}$
... (2)
Now we have two equations involving $\frac{1}{a^2}$ and $\frac{1}{b^2}$.
From equation (1), we can write $\frac{1}{a^2}$ in terms of $\frac{1}{b^2}$ (assuming $x \neq 0$):
$\frac{1}{a^2} = \frac{y y'}{x b^2}$
Substitute this expression for $\frac{1}{a^2}$ into equation (2):
$\frac{y y'}{x b^2} = \frac{(y')^2 + y y''}{b^2}$
Multiply both sides by $b^2$ (assuming $b^2 \neq 0$):
$\frac{y y'}{x} = (y')^2 + y y''$
Multiply both sides by $x$ (assuming $x \neq 0$):
$y y' = x (y')^2 + x y y''$
Rearrange the terms to set the equation to zero:
$x y y'' + x (y')^2 - y y' = 0$
This equation is a differential equation of order 2, and it does not contain the arbitrary constants $a$ or $b$. This is the required differential equation.
The differential equation representing the family of hyperbolas having foci on the x-axis and centre at the origin is $x y y'' + x (y')^2 - y y' = 0$ (for $x \neq 0$).
Question 10. Form the differential equation of the family of circles having centre on y-axis and radius 3 units.
Answer:
Given:
The family of circles having centre on y-axis and radius 3 units.
To Form:
Form the differential equation representing this family of circles by eliminating the arbitrary constant.
Solution:
A circle with its centre on the y-axis has coordinates $(0, a)$, where $a$ is an arbitrary constant. The radius of the circle is given as 3 units.
The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the given information, $h=0$, $k=a$, and $r=3$, the equation of the family of circles is:
$(x - 0)^2 + (y - a)^2 = 3^2$
x$^2$ + (y - a)$^2$ = 9
... (1)
There is one arbitrary constant ($a$) in the equation. Therefore, the order of the required differential equation will be 1. We need to eliminate the constant $a$ by differentiating the equation with respect to $x$.
Differentiate equation (1) implicitly with respect to $x$:
$\frac{d}{dx}(x^2 + (y - a)^2) = \frac{d}{dx}(9)$
$\frac{d}{dx}(x^2) + \frac{d}{dx}((y - a)^2) = 0$
Using the power rule and chain rule for differentiation:
$2x + 2(y - a) \frac{d}{dx}(y - a) = 0$
$2x + 2(y - a) \left(\frac{dy}{dx} - \frac{da}{dx}\right) = 0$
Since $a$ is a constant with respect to $x$, $\frac{da}{dx} = 0$. Also, let $y' = \frac{dy}{dx}$.
$2x + 2(y - a) y' = 0$
Divide both sides by 2:
x + (y - a) y' = 0
... (2)
Now we need to eliminate $a$ using equation (1) and equation (2).
From equation (2), assuming $y' \neq 0$, we can express $(y - a)$:
$(y - a) y' = -x$
$y - a = -\frac{x}{y'}$
Substitute this expression for $(y - a)$ into equation (1):
$x^2 + \left(-\frac{x}{y'}\right)^2 = 9$
$x^2 + \frac{x^2}{(y')^2} = 9$
To eliminate the denominator, multiply the entire equation by $(y')^2$ (assuming $y' \neq 0$):
$x^2 (y')^2 + \frac{x^2}{(y')^2} (y')^2 = 9 (y')^2$
$x^2 (y')^2 + x^2 = 9 (y')^2$
Rearrange the terms to form the differential equation:
$x^2 (y')^2 - 9 (y')^2 + x^2 = 0$
Factor out $(y')^2$ from the first two terms:
$(x^2 - 9) (y')^2 + x^2 = 0$
This equation relates $x$, $y$, and $y'$ and does not contain the arbitrary constant $a$. This is the required differential equation.
Alternatively, we can write it as $(9 - x^2)(y')^2 = x^2$.
Note that if $y'=0$, equation (2) gives $x=0$. Substituting $x=0$ into equation (1) gives $(y-a)^2 = 9$, so $y-a = \pm 3$. This corresponds to the points $(0, a \pm 3)$ where the tangent is horizontal ($y'=0$). Let's check if $(x, y') = (0, 0)$ satisfies the differential equation: $(0^2 - 9)(0)^2 + 0^2 = -9 \cdot 0 + 0 = 0$. It is satisfied.
The differential equation representing the family of circles having centre on y-axis and radius 3 units is $(x^2 - 9)(y')^2 + x^2 = 0$ or $(9 - x^2)(y')^2 = x^2$.
Question 11. Which of the following differential equations has y = c1 ex + c2 e–x as the general solution?
(A) $\frac{d^2y}{dx^2} + y = 0$
(B) $\frac{d^2y}{dx^2} - y = 0$
(C) $\frac{d^2y}{dx^2} +1 = 0$
(D) $\frac{d^2y}{dx^2} - 1 = 0$
Answer:
Given:
The general solution is $y = c_1 e^x + c_2 e^{-x}$, where $c_1$ and $c_2$ are arbitrary constants.
To Find:
The differential equation that has the given general solution.
Solution:
To find the differential equation, we need to differentiate the given general solution with respect to $x$ as many times as there are arbitrary constants (which is 2 in this case), and then eliminate the constants from the resulting equations.
Given the general solution:
y = c$_1$ e$^x$ + c$_2$ e$^{–x}$
... (1)
Calculate the first derivative $\frac{dy}{dx}$:
$y' = \frac{d}{dx}(c_1 e^x + c_2 e^{-x})$
$y' = c_1 \frac{d}{dx}(e^x) + c_2 \frac{d}{dx}(e^{-x})$
y' = c$_1$ e$^x$ - c$_2$ e$^{–x}$
... (2)
Calculate the second derivative $\frac{d^2y}{dx^2}$:
$y'' = \frac{d}{dx}(y') = \frac{d}{dx}(c_1 e^x - c_2 e^{-x})$
$y'' = c_1 \frac{d}{dx}(e^x) - c_2 \frac{d}{dx}(e^{-x})$
$y'' = c_1 e^x - c_2 (-e^{-x})$
y'' = c$_1$ e$^x$ + c$_2$ e$^{–x}$
... (3)
Now, observe equation (1) and equation (3).
From equation (1), we have $y = c_1 e^x + c_2 e^{-x}$.
From equation (3), we have $y'' = c_1 e^x + c_2 e^{-x}$.
Comparing (1) and (3), we see that $y'' = y$.
Rearranging this equation, we get:
$y'' - y = 0$
or
$\frac{d^2y}{dx^2} - y = 0$
This is the differential equation corresponding to the given general solution.
Comparing this with the given options, we find that option (B) is correct.
The correct answer is (B) $\frac{d^2y}{dx^2} - y = 0$.
Question 12. Which of the following differential equations has y = x as one of its particular solution?
(A) $\frac{d^2y}{dx^2} - x^2 \frac{dy}{dx} + xy = x$
(B) $\frac{d^2y}{dx^2} + x \frac{dy}{dx} + xy = x$
(C) $\frac{d^2y}{dx^2} - x^2 \frac{dy}{dx} + xy = 0$
(D) $\frac{d^2y}{dx^2} + x \frac{dy}{dx} + xy = 0$
Answer:
Given:
A proposed particular solution $y = x$.
Four differential equations.
To Find:
Which of the given differential equations has $y = x$ as one of its particular solutions.
Solution:
For $y = x$ to be a particular solution of a differential equation, it must satisfy the equation for all values of $x$ in the domain where the equation is defined.
First, we find the necessary derivatives of the given function $y = x$:
Calculate the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$
So, $y' = 1$.
Calculate the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(1) = 0$
So, $y'' = 0$.
Now, we substitute $y=x$, $y'=1$, and $y''=0$ into each of the given differential equations:
(A) $\frac{d^2y}{dx^2} - x^2 \frac{dy}{dx} + xy = x$
Substitute $y''=0$, $y'=1$, $y=x$:
$0 - x^2 (1) + x(x) = x$
$0 - x^2 + x^2 = x$
$0 = x$
This equation $0 = x$ is only true for $x=0$. Since it is not true for all values of $x$, $y=x$ is not a solution for differential equation (A).
(B) $\frac{d^2y}{dx^2} + x \frac{dy}{dx} + xy = x$
Substitute $y''=0$, $y'=1$, $y=x$:
$0 + x (1) + x(x) = x$
$0 + x + x^2 = x$
$x + x^2 = x$
$x^2 = 0$
This equation $x^2 = 0$ is only true for $x=0$. Since it is not true for all values of $x$, $y=x$ is not a solution for differential equation (B).
(C) $\frac{d^2y}{dx^2} - x^2 \frac{dy}{dx} + xy = 0$
Substitute $y''=0$, $y'=1$, $y=x$:
$0 - x^2 (1) + x(x) = 0$
$0 - x^2 + x^2 = 0$
$0 = 0$
This equation $0 = 0$ is true for all values of $x$. Since the substitution satisfies the equation identically, $y=x$ is a solution for differential equation (C).
(D) $\frac{d^2y}{dx^2} + x \frac{dy}{dx} + xy = 0$
Substitute $y''=0$, $y'=1$, $y=x$:
$0 + x (1) + x(x) = 0$
$0 + x + x^2 = 0$
$x + x^2 = 0$
$x(1+x) = 0$
This equation $x(1+x) = 0$ is only true for $x=0$ or $x=-1$. Since it is not true for all values of $x$, $y=x$ is not a solution for differential equation (D).
The only differential equation that is satisfied by $y=x$ for all values of $x$ is option (C).
Therefore, $y=x$ is a particular solution of the differential equation in option (C).
The correct answer is (C) $\frac{d^2y}{dx^2} - x^2 \frac{dy}{dx} + xy = 0$.
Example 9 to 14 (Before Exercise 9.4)
Example 9: Find the general solution of the differential equation $\frac{dy}{dx} = \frac{x + 1}{2 − y}$ , (y ≠ 2).
Answer:
Given:
The differential equation is $\frac{dy}{dx} = \frac{x + 1}{2 − y}$, with $y \neq 2$.
To Find:
The general solution of the given differential equation.
Solution:
The given differential equation is a first-order differential equation. We can check if it is a separable differential equation.
The equation is $\frac{dy}{dx} = \frac{x + 1}{2 − y}$.
We can separate the variables by multiplying both sides by $(2-y)$ and by $dx$:
$(2 - y) dy = (x + 1) dx$
Now, integrate both sides of the separated equation:
$\int (2 - y) dy = \int (x + 1) dx$
Evaluate the integral on the left side with respect to $y$:
$\int (2 - y) dy = \int 2 dy - \int y dy = 2y - \frac{y^2}{2} + C_1$
Evaluate the integral on the right side with respect to $x$:
$\int (x + 1) dx = \int x dx + \int 1 dx = \frac{x^2}{2} + x + C_2$
Equating the results of the integration on both sides:
$2y - \frac{y^2}{2} + C_1 = \frac{x^2}{2} + x + C_2$
Combine the constants of integration into a single constant $C$, where $C = C_2 - C_1$:
2y - $\frac{y^2}{2}$ = $\frac{x^2}{2}$ + x + C
... (1)
We can rearrange the equation to a more standard form, for example, by multiplying by 2 to clear the denominators:
$2 \left( 2y - \frac{y^2}{2} \right) = 2 \left( \frac{x^2}{2} + x + C \right)$
$4y - y^2 = x^2 + 2x + 2C$
Let $C' = -2C$ (since $C$ is arbitrary, $C'$ is also an arbitrary constant).
$x^2 + y^2 + 2x - 4y + 2C = 0$
$x^2 + y^2 + 2x - 4y - C' = 0$
A common way to express the general solution is to move all terms involving $x$ and $y$ to one side and the constant to the other, or keep the constant on one side.
Using equation (1):
$2y - \frac{y^2}{2} - \frac{x^2}{2} - x = C$
Multiply by $-2$ to make the $x^2$ and $y^2$ terms positive:
$-4y + y^2 + x^2 + 2x = -2C$
Let the new arbitrary constant be $K = -2C$.
$x^2 + y^2 + 2x - 4y = K$
The general solution of the differential equation is $x^2 + y^2 + 2x - 4y = K$, where $K$ is an arbitrary constant.
Another valid form is $2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C$.
Example 10: Find the general solution of the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$ .
Answer:
Given:
The differential equation is $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$.
To Find:
The general solution of the given differential equation.
Solution:
The given differential equation is $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$.
This is a separable variable type differential equation. We can rearrange it to separate the variables $x$ and $y$ on different sides by dividing both sides by $(1+y^2)$ and multiplying by $dx$:
$\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$
... (1)
Now, integrate both sides of equation (1) with respect to their respective variables:
$\int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2}$
Using the standard integral formula $\int \frac{du}{1 + u^2} = \tan^{-1}(u) + C$:
$\tan^{-1}(y) + C_1 = \tan^{-1}(x) + C_2$
Where $C_1$ and $C_2$ are constants of integration.
Combine the constants into a single arbitrary constant $C = C_2 - C_1$:
$\tan^{-1}(y) = \tan^{-1}(x) + C$
... (2)
This equation represents the general solution of the given differential equation.
The general solution of the differential equation is $\tan^{-1}(y) = \tan^{-1}(x) + C$, where $C$ is an arbitrary constant.
Example 11: Find the particular solution of the differential equation $\frac{dy}{dx} = −4xy^2$ given that y = 1, when x = 0.
Answer:
Given:
The differential equation is $\frac{dy}{dx} = -4xy^2$.
The initial condition is $y = 1$ when $x = 0$.
To Find:
The particular solution of the given differential equation that satisfies the initial condition.
Solution:
The given differential equation is $\frac{dy}{dx} = -4xy^2$.
This is a separable variable type differential equation. We can separate the variables $x$ and $y$ on different sides by dividing both sides by $y^2$ (assuming $y \neq 0$) and multiplying by $dx$:
$\frac{dy}{y^2} = -4x \, dx$
... (1)
Now, integrate both sides of equation (1) with respect to their respective variables:
$\int \frac{dy}{y^2} = \int -4x \, dx$
Integrate the left side with respect to $y$:
$\int y^{-2} dy = \frac{y^{-2+1}}{-2+1} + C_1 = \frac{y^{-1}}{-1} + C_1 = -\frac{1}{y} + C_1$
Integrate the right side with respect to $x$:
$\int -4x \, dx = -4 \int x \, dx = -4 \frac{x^{1+1}}{1+1} + C_2 = -4 \frac{x^2}{2} + C_2 = -2x^2 + C_2$
Equating the results of the integration on both sides:
$-\frac{1}{y} + C_1 = -2x^2 + C_2$
Combine the constants of integration into a single arbitrary constant $C = C_2 - C_1$:
$-\frac{1}{y} = -2x^2 + C$
... (2)
This is the general solution. Now, we use the given initial condition $y = 1$ when $x = 0$ to find the value of the arbitrary constant $C$.
Substitute $x=0$ and $y=1$ into the general solution (2):
$-\frac{1}{1} = -2(0)^2 + C$
$-1 = -2(0) + C$
$-1 = 0 + C$
$C = -1$
Now, substitute the value of $C = -1$ back into the general solution (2) to obtain the particular solution:
$-\frac{1}{y} = -2x^2 + (-1)$
$-\frac{1}{y} = -2x^2 - 1$
Multiply both sides by $-1$:
$\frac{1}{y} = 2x^2 + 1$
We can express $y$ explicitly by taking the reciprocal of both sides:
$y = \frac{1}{2x^2 + 1}$
Note that the case $y=0$ was excluded when separating variables. If $y=0$ identically, then $\frac{dy}{dx}=0$ and $-4xy^2 = -4x(0)^2 = 0$, so $y=0$ is a solution. However, the initial condition $y(0)=1$ requires a non-zero solution.
The particular solution of the differential equation satisfying the given initial condition is $y = \frac{1}{2x^2 + 1}$.
Example 12: Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx (x ≠ 0).
Answer:
Given:
The differential equation is $x \, dy = (2x^2 + 1) \, dx$, with $x \neq 0$.
The curve passes through the point (1, 1).
To Find:
The equation of the curve passing through the given point (the particular solution) that satisfies the differential equation.
Solution:
The given differential equation is $x \, dy = (2x^2 + 1) \, dx$.
This is a first-order differential equation. We can rearrange it to separate the variables $x$ and $y$. Divide both sides by $x$ (given $x \neq 0$) and by $dx$:
$\frac{dy}{dx} = \frac{2x^2 + 1}{x}$
Split the fraction on the right side:
$\frac{dy}{dx} = \frac{2x^2}{x} + \frac{1}{x}$
$\frac{dy}{dx} = 2x + \frac{1}{x}$
Now, separate the variables by multiplying both sides by $dx$:
dy = $\left(2x + \frac{1}{x}\right) dx$
... (1)
Now, integrate both sides of equation (1):
$\int dy = \int \left(2x + \frac{1}{x}\right) dx$
Integrate the left side with respect to $y$ and the right side with respect to $x$:
$\int dy = y + C_1$
$\int \left(2x + \frac{1}{x}\right) dx = \int 2x \, dx + \int \frac{1}{x} \, dx = 2 \cdot \frac{x^2}{2} + \ln|x| + C_2 = x^2 + \ln|x| + C_2$
Equating the results of the integration on both sides:
$y + C_1 = x^2 + \ln|x| + C_2$
Combine the constants of integration into a single arbitrary constant $C = C_2 - C_1$:
y = x$^2$ + $\ln|x|$ + C
... (2)
This is the general solution of the differential equation.
Now, we use the given information that the curve passes through the point (1, 1) to find the value of the arbitrary constant $C$. This means when $x=1$, $y=1$.
Substitute $x=1$ and $y=1$ into the general solution (2):
$1 = (1)^2 + \ln|1| + C$
$1 = 1 + 0 + C$
$1 = 1 + C$
$C = 1 - 1$
$C = 0$
Now, substitute the value of $C = 0$ back into the general solution (2) to obtain the particular solution (the equation of the curve):
$y = x^2 + \ln|x| + 0$
$y = x^2 + \ln|x|$
Since the point (1, 1) has a positive x-coordinate ($x=1$), the curve we are interested in is likely in the region where $x > 0$. In this region, $|x| = x$.
So, the particular solution can be written as:
$y = x^2 + \ln x$
The equation of the curve passing through the point (1, 1) with the given differential equation is $y = x^2 + \ln x$.
Example 13: Find the equation of a curve passing through the point (–2, 3), given that the slope of the tangent to the curve at any point (x, y) is $\frac{2x}{y^2}$ .
Answer:
Given:
The curve passes through the point (-2, 3).
The slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.
To Find:
The equation of the curve passing through the given point.
Solution:
The slope of the tangent to a curve at any point $(x, y)$ is given by the first derivative $\frac{dy}{dx}$.
So, the given information translates to the differential equation:
$\frac{dy}{dx} = \frac{2x}{y^2}$
This is a first-order separable differential equation. We can separate the variables by multiplying both sides by $y^2$ and by $dx$:
y$^2$ dy = 2x dx
... (1)
Now, integrate both sides of equation (1) with respect to their respective variables:
$\int y^2 dy = \int 2x dx$
Integrate the left side with respect to $y$:
$\int y^2 dy = \frac{y^{2+1}}{2+1} + C_1 = \frac{y^3}{3} + C_1$
Integrate the right side with respect to $x$:
$\int 2x dx = 2 \int x dx = 2 \frac{x^{1+1}}{1+1} + C_2 = 2 \frac{x^2}{2} + C_2 = x^2 + C_2$
Equating the results of the integration on both sides:
$\frac{y^3}{3} + C_1 = x^2 + C_2$
Combine the constants of integration into a single arbitrary constant $C = C_2 - C_1$:
$\frac{y^3}{3} = x^2 + C$
... (2)
This is the general solution of the differential equation.
Now, we use the given information that the curve passes through the point (-2, 3) to find the value of the arbitrary constant $C$. This means when $x=-2$, $y=3$.
Substitute $x=-2$ and $y=3$ into the general solution (2):
$\frac{(3)^3}{3} = (-2)^2 + C$
$\frac{27}{3} = 4 + C$
$9 = 4 + C$
$C = 9 - 4$
$C = 5$
Now, substitute the value of $C = 5$ back into the general solution (2) to obtain the particular solution (the equation of the curve):
$\frac{y^3}{3} = x^2 + 5$
We can multiply by 3 to clear the denominator:
$y^3 = 3(x^2 + 5)$
$y^3 = 3x^2 + 15$
The equation of the curve passing through the point (-2, 3) is $y^3 = 3x^2 + 15$.
Example 14: In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?
Answer:
Given:
The rate of increase of the principal P is 5% per year, continuously.
Initial principal = Rs 1000.
To Find:
The number of years it takes for the principal to double itself.
Solution:
Let $P$ be the principal at time $t$ years.
The rate of increase of the principal is given as 5% per year, continuously. This can be expressed as a differential equation:
$\frac{dP}{dt} \propto P$
$\frac{dP}{dt} = k P$
The rate constant $k$ is given as 5% per year, which is 0.05.
$\frac{dP}{dt} = 0.05 P$
... (1)
This is a first-order separable differential equation. We can separate the variables $P$ and $t$:
$\frac{dP}{P} = 0.05 \, dt$
Now, integrate both sides:
$\int \frac{dP}{P} = \int 0.05 \, dt$
$\ln|P| = 0.05t + C_1$
Since the principal $P$ is positive, we can remove the absolute value:
$\ln P = 0.05t + C_1$
Exponentiate both sides to solve for $P$:
$P(t) = e^{0.05t + C_1} = e^{C_1} e^{0.05t}$
Let $A = e^{C_1}$, where $A$ is an arbitrary constant ($A > 0$ since $P > 0$).
P(t) = A e$^{0.05t}$
... (2)
This is the general solution for the principal at time $t$.
We are given that the initial principal is Rs 1000. This means at $t = 0$, $P = 1000$. We use this initial condition to find the value of the constant $A$.
Substitute $t=0$ and $P=1000$ into equation (2):
$1000 = A e^{0.05 \cdot 0}$
$1000 = A e^0$
$1000 = A \cdot 1$
$A = 1000$
Substitute the value of $A$ back into the general solution (2) to get the particular solution for this problem:
P(t) = 1000 e$^{0.05t}$
... (3)
We want to find the time $t$ when the principal doubles itself. This means $P(t) = 2 \times 1000 = 2000$.
Set $P(t) = 2000$ in equation (3):
$2000 = 1000 e^{0.05t}$
Divide both sides by 1000:
$2 = e^{0.05t}$
Take the natural logarithm of both sides:
$\ln(2) = \ln(e^{0.05t})$
Using the logarithm property $\ln(e^u) = u$:
$\ln(2) = 0.05t$
Solve for $t$:
$t = \frac{\ln(2)}{0.05}$
$t = \frac{\ln(2)}{1/20}$
$t = 20 \ln(2)$
The time taken for the principal to double is $20 \ln(2)$ years.
If we need a numerical value, we can use the approximate value $\ln(2) \approx 0.6931$:
$t \approx 20 \times 0.6931$
$t \approx 13.862$ years
The principal doubles itself in $20 \ln(2)$ years.
Exercise 9.4
For each of the differential equations in Exercises 1 to 10, find the general solution:
Question 1. $\frac{dy}{dx} = \frac{1 − \cos x}{1 + \cos x}$
Answer:
Given:
The differential equation is $\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}$.
Solution:
The given differential equation is:
$\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}$
We use the half-angle trigonometric identities for $1 - \cos x$ and $1 + \cos x$:
$1 - \cos x = 2 \sin^2 \frac{x}{2}$
$1 + \cos x = 2 \cos^2 \frac{x}{2}$
Substitute these identities into the right-hand side of the differential equation:
$\frac{dy}{dx} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}}$
Simplify the expression:
$\frac{dy}{dx} = \left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)^2 = \tan^2 \frac{x}{2}$
Now, use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$ for $\theta = \frac{x}{2}$:
$\frac{dy}{dx} = \sec^2 \frac{x}{2} - 1$
To find the general solution, we separate the variables and integrate both sides of the equation:
$dy = \left(\sec^2 \frac{x}{2} - 1\right) dx$
$\int\limits dy = \int\limits \left(\sec^2 \frac{x}{2} - 1\right) dx$
Integrate the right-hand side. We can split the integral:
$\int\limits \left(\sec^2 \frac{x}{2} - 1\right) dx = \int\limits \sec^2 \frac{x}{2} dx - \int\limits 1 dx$
For the integral $\int\limits \sec^2 \frac{x}{2} dx$, let $u = \frac{x}{2}$. Then, the differential $du = \frac{1}{2} dx$, which means $dx = 2 du$.
So, $\int\limits \sec^2 \frac{x}{2} dx = \int\limits \sec^2 u (2 du) = 2 \int\limits \sec^2 u du$.
The integral of $\sec^2 u$ with respect to $u$ is $\tan u$.
$2 \int\limits \sec^2 u du = 2 \tan u + C_1$
Substitute back $u = \frac{x}{2}$:
$\int\limits \sec^2 \frac{x}{2} dx = 2 \tan \frac{x}{2} + C_1$
The second integral is straightforward:
$\int\limits 1 dx = x + C_2$
Combining the results for the right-hand side integral:
$\int\limits \left(\sec^2 \frac{x}{2} - 1\right) dx = \left(2 \tan \frac{x}{2} + C_1\right) - (x + C_2) = 2 \tan \frac{x}{2} - x + (C_1 - C_2)$
Let $C = C_1 - C_2$ be the constant of integration.
Integrating the left side $\int\limits dy$ gives $y$.
Equating the integrated sides:
$y = 2 \tan \frac{x}{2} - x + C$
The general solution of the given differential equation is:
$\mathbf{y = 2 \tan \frac{x}{2} - x + C}$
Question 2. $\frac{dy}{dx} = \sqrt{4 − y^2} \;\;(-2 < y < 2)$
Answer:
Given:
The differential equation is $\frac{dy}{dx} = \sqrt{4 − y^2}$ with the condition $-2 < y < 2$.
Solution:
The given differential equation is:
$\frac{dy}{dx} = \sqrt{4 − y^2}$
We can separate the variables by moving the $y$ terms to the left side and the $x$ terms to the right side. Since $-2 < y < 2$, $\sqrt{4 - y^2} \neq 0$, so we can divide by it.
$\frac{dy}{\sqrt{4 − y^2}} = dx$
Now, integrate both sides of the equation:
$\int\limits \frac{dy}{\sqrt{4 − y^2}} = \int\limits dx$
The integral on the left side is a standard integral of the form $\int \frac{dz}{\sqrt{a^2 - z^2}} = \arcsin\left(\frac{z}{a}\right) + C$. Here, $z = y$ and $a^2 = 4$, so $a = 2$.
$\int\limits \frac{dy}{\sqrt{4 − y^2}} = \arcsin\left(\frac{y}{2}\right)$
The integral on the right side is:
$\int\limits dx = x$
Equating the results of the integration and adding the constant of integration $C$:
$\arcsin\left(\frac{y}{2}\right) = x + C$
To express the general solution explicitly in terms of $y$, we can take the sine of both sides:
$\frac{y}{2} = \sin(x + C)$
$y = 2 \sin(x + C)$
The general solution of the given differential equation is:
$\mathbf{y = 2 \sin(x + C)}$
Question 3. $\frac{dy}{dx} + y = 1$ (y ≠ 1)
Answer:
Given:
The differential equation is $\frac{dy}{dx} + y = 1$, with the condition $y \neq 1$.
Solution:
The given differential equation can be rewritten as:
$\frac{dy}{dx} = 1 - y$
Since the condition $y \neq 1$ is given, we know that $1 - y \neq 0$. This allows us to separate the variables:
$\frac{dy}{1 - y} = dx$
Now, integrate both sides of the equation:
$\int\limits \frac{dy}{1 - y} = \int\limits dx$
For the integral on the left side, let $u = 1 - y$. Then $du = -dy$, so $dy = -du$.
$\int\limits \frac{-du}{u} = - \int\limits \frac{du}{u} = - \ln|u| + C_1$
Substituting back $u = 1 - y$:
$\int\limits \frac{dy}{1 - y} = - \ln|1 - y| + C_1$
The integral on the right side is:
$\int\limits dx = x + C_2$
Equating the results of the integration:
$- \ln|1 - y| + C_1 = x + C_2$
$- \ln|1 - y| = x + C_2 - C_1$
Let $K = C_2 - C_1$, which is an arbitrary constant.
$- \ln|1 - y| = x + K$
$\ln|1 - y| = -(x + K)$
Exponentiate both sides with base $e$:
$|1 - y| = e^{-(x + K)} = e^{-x} e^{-K}$
Let $A = e^{-K}$. Since $K$ is an arbitrary constant, $A$ is an arbitrary positive constant ($A > 0$).
$|1 - y| = A e^{-x}$
This implies $1 - y = \pm A e^{-x}$.
Let $C = \pm A$. Since $A$ is any positive constant, $C$ can be any non-zero real constant ($C \neq 0$).
$1 - y = C e^{-x}$
Solving for $y$:
$y = 1 - C e^{-x}$
Given the condition $y \neq 1$, the constant $C$ must be non-zero.
The general solution of the given differential equation under the condition $y \neq 1$ is:
$\mathbf{y = 1 - C e^{-x}, \text{\$ } C \neq 0}$
Question 4. sec2 x tan y dx + sec2 y tan x dy = 0
Answer:
Given:
The differential equation is $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$.
Solution:
The given differential equation is:
$\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$
We can rewrite the equation to separate the variables. Move the term involving $y$ and $dy$ to the other side:
$\sec^2 x \tan y \, dx = - \sec^2 y \tan x \, dy$
Now, divide both sides by $\tan x \tan y$ (assuming $\tan x \neq 0$ and $\tan y \neq 0$) to separate the variables $x$ and $y$:
$\frac{\sec^2 x}{\tan x} \, dx = - \frac{\sec^2 y}{\tan y} \, dy$
Now, integrate both sides of the equation:
$\int\limits \frac{\sec^2 x}{\tan x} \, dx = \int\limits - \frac{\sec^2 y}{\tan y} \, dy$
To evaluate the integrals, we use the substitution method. For an integral of the form $\int \frac{f'(z)}{f(z)} dz$, the result is $\ln|f(z)|$. Here, the derivative of $\tan z$ is $\sec^2 z$.
Applying this to both sides of our separated equation:
$\int\limits \frac{\sec^2 x}{\tan x} \, dx = \ln|\tan x| + C_1$
$\int\limits - \frac{\sec^2 y}{\tan y} \, dy = - \int\limits \frac{\sec^2 y}{\tan y} \, dy = - (\ln|\tan y| + C_2) = - \ln|\tan y| - C_2$
Equating the integrated results:
$\ln|\tan x| + C_1 = - \ln|\tan y| - C_2$
Rearrange the terms to group the logarithm terms together:
$\ln|\tan x| + \ln|\tan y| = - C_1 - C_2$
Let $C' = - C_1 - C_2$ be the constant of integration.
$\ln|\tan x| + \ln|\tan y| = C'$
Using the logarithm property $\ln A + \ln B = \ln(AB)$:
$\ln|\tan x \tan y| = C'$
Exponentiate both sides with base $e$:
$|\tan x \tan y| = e^{C'}$
Let $C = e^{C'}$. Since $C'$ is an arbitrary constant, $C$ is an arbitrary positive constant ($C > 0$).
$|\tan x \tan y| = C$
This implies $\tan x \tan y = \pm C$. Let $K = \pm C$. Since $C$ is any positive constant, $K$ can be any non-zero constant.
$\tan x \tan y = K, \text{\$ } K \neq 0$
Note that the cases where $\tan x = 0$ or $\tan y = 0$ (which correspond to $x = n\pi$ or $y = m\pi$ for integers $n, m$) also satisfy the original differential equation. These solutions are included in the form $\tan x \tan y = C$ if we allow the constant $C$ to be zero.
Thus, the general solution is $\tan x \tan y = C$, where $C$ is an arbitrary constant.
The general solution of the given differential equation is:
$\mathbf{\tan x \tan y = C}$
Question 5. (ex + e–x) dy – (ex – e–x) dx = 0
Answer:
Given:
The differential equation is $(e^x + e^{–x}) dy – (e^x – e^{–x}) dx = 0$.
Solution:
The given differential equation is:
$(e^x + e^{-x}) dy – (e^x – e^{-x}) dx = 0$
We can separate the variables by moving the term with $dx$ to the right side:
$(e^x + e^{-x}) dy = (e^x – e^{-x}) dx$
Now, isolate $dy$ and $dx$ terms on opposite sides by dividing by $(e^x + e^{-x})$ (note that $e^x + e^{-x} > 0$ for all real $x$):
$dy = \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$
Now, integrate both sides of the equation:
$\int\limits dy = \int\limits \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$
The integral on the left side is straightforward:
$\int\limits dy = y + C_1$
For the integral on the right side, $\int\limits \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$, observe that the numerator $(e^x - e^{-x})$ is the derivative of the denominator $(e^x + e^{-x})$.
This integral is of the form $\int \frac{f'(x)}{f(x)} dx$, which integrates to $\ln|f(x)|$.
Here, $f(x) = e^x + e^{-x}$.
So, $\int\limits \frac{e^x – e^{-x}}{e^x + e^{-x}} dx = \ln|e^x + e^{-x}| + C_2$.
Since $e^x$ is always positive and $e^{-x}$ is always positive for real $x$, their sum $e^x + e^{-x}$ is always positive. Thus, we can remove the absolute value.
$\int\limits \frac{e^x – e^{-x}}{e^x + e^{-x}} dx = \ln(e^x + e^{-x}) + C_2$.
Equating the integrated results:
$y + C_1 = \ln(e^x + e^{-x}) + C_2$
Solving for $y$:
$y = \ln(e^x + e^{-x}) + C_2 - C_1$
Let $C = C_2 - C_1$ be the arbitrary constant of integration.
$y = \ln(e^x + e^{-x}) + C$
The general solution of the given differential equation is:
$\mathbf{y = \ln(e^x + e^{-x}) + C}$
Question 6. $\frac{dy}{dx} = (1 + x^2) (1 + y^2)$
Answer:
Given:
The differential equation is $\frac{dy}{dx} = (1 + x^2) (1 + y^2)$.
Solution:
The given differential equation is:
$\frac{dy}{dx} = (1 + x^2) (1 + y^2)$
This is a separable differential equation. We can separate the variables by dividing both sides by $(1 + y^2)$ and multiplying both sides by $dx$. Note that $1+y^2$ is always positive, so we don't need to worry about division by zero.
$\frac{dy}{1 + y^2} = (1 + x^2) dx$
Now, integrate both sides of the equation:
$\int\limits \frac{dy}{1 + y^2} = \int\limits (1 + x^2) dx$
The integral on the left side is a standard integral:
$\int\limits \frac{dy}{1 + y^2} = \arctan(y) + C_1$
The integral on the right side can be split into two parts:
$\int\limits (1 + x^2) dx = \int\limits 1 dx + \int\limits x^2 dx$
$\int\limits 1 dx = x + C_2$
$\int\limits x^2 dx = \frac{x^{2+1}}{2+1} + C_3 = \frac{x^3}{3} + C_3$
Combining the integrals on the right side:
$\int\limits (1 + x^2) dx = x + \frac{x^3}{3} + (C_2 + C_3)$
Equating the integrated results from both sides:
$\arctan(y) + C_1 = x + \frac{x^3}{3} + C_2 + C_3$
Move the constant $C_1$ to the right side and combine the constants:
$\arctan(y) = x + \frac{x^3}{3} + C_2 + C_3 - C_1$
Let $C = C_2 + C_3 - C_1$ be the arbitrary constant of integration.
$\arctan(y) = x + \frac{x^3}{3} + C$
This gives the general solution in implicit form. We can also write the explicit form by taking the tangent of both sides:
$y = \tan\left(x + \frac{x^3}{3} + C\right)$
The general solution of the given differential equation is:
$\mathbf{\arctan(y) = x + \frac{x^3}{3} + C}$ or $\mathbf{y = \tan\left(x + \frac{x^3}{3} + C\right)}$
Question 7. y log y dx – x dy = 0
Answer:
Given:
The differential equation is $y \log y \, dx – x \, dy = 0$.
Solution:
The given differential equation is:
$y \log y \, dx – x \, dy = 0$
We can separate the variables. Move the term involving $dy$ to the right side:
$y \log y \, dx = x \, dy$
Now, divide both sides by $x y \log y$ (assuming $x \neq 0$ and $y \log y \neq 0$) to separate the variables:
$\frac{dx}{x} = \frac{dy}{y \log y}$
Now, integrate both sides of the equation:
$\int\limits \frac{dx}{x} = \int\limits \frac{dy}{y \log y}$
The integral on the left side is:
$\int\limits \frac{dx}{x} = \ln|x| + C_1$
For the integral on the right side, $\int\limits \frac{dy}{y \log y}$, we use substitution. Let $u = \log y$. Then the differential $du = \frac{1}{y} dy$.
The integral becomes:
$\int\limits \frac{1}{\log y} \left(\frac{1}{y} dy\right) = \int\limits \frac{1}{u} du$
Integrating with respect to $u$:
$\int\limits \frac{1}{u} du = \ln|u| + C_2$
Substitute back $u = \log y$:
$\int\limits \frac{dy}{y \log y} = \ln|\log y| + C_2$
Equating the integrated results from both sides:
$\ln|x| + C_1 = \ln|\log y| + C_2$
Rearrange the terms:
$\ln|\log y| - \ln|x| = C_1 - C_2$
Let $C' = C_1 - C_2$ be the arbitrary constant. Using logarithm properties $\ln A - \ln B = \ln(A/B)$:
$\ln\left|\frac{\log y}{x}\right| = C'$
Exponentiate both sides with base $e$:
$\left|\frac{\log y}{x}\right| = e^{C'}$
Let $K = \pm e^{C'}$. $K$ is a non-zero constant.
$\frac{\log y}{x} = K$
$\log y = Kx$
Exponentiate both sides with base $e$ again:
$y = e^{Kx}$
Note that the case where $y=1$ (i.e., $\log y = 0$) is a singular solution to the original equation ($1 \cdot 0 \cdot dx - x \cdot 0 \cdot dy = 0$). This solution $y=1$ is obtained from $y=e^{Kx}$ by setting $K=0$. Therefore, the constant $K$ can be any real number.
The general solution of the given differential equation is:
$\mathbf{y = e^{Kx}}$
where $K$ is an arbitrary constant.
Question 8. $x^5 \frac{dy}{dx} = -y^5$
Answer:
Given:
The differential equation is $x^5 \frac{dy}{dx} = -y^5$.
Solution:
The given differential equation is:
$x^5 \frac{dy}{dx} = -y^5$
This is a separable differential equation. We can separate the variables by dividing both sides by $x^5$ and $y^5$ (assuming $x \neq 0$ and $y \neq 0$):
$\frac{dy}{y^5} = \frac{-dx}{x^5}$
Rewrite the terms using negative exponents:
$y^{-5} dy = -x^{-5} dx$
Now, integrate both sides of the equation:
$\int\limits y^{-5} dy = \int\limits -x^{-5} dx$
$\int\limits y^{-5} dy = - \int\limits x^{-5} dx$
Using the power rule for integration $\int z^n dz = \frac{z^{n+1}}{n+1} + C$ (for $n \neq -1$):
$\frac{y^{-5+1}}{-5+1} = - \left(\frac{x^{-5+1}}{-5+1}\right) + C$
$\frac{y^{-4}}{-4} = - \left(\frac{x^{-4}}{-4}\right) + C$
$-\frac{1}{4y^4} = \frac{1}{4x^4} + C$
Rearrange the equation to group terms:
$-\frac{1}{4y^4} - \frac{1}{4x^4} = C$
Multiply the entire equation by $-4$:
$\frac{1}{y^4} + \frac{1}{x^4} = -4C$
Let $K = -4C$ be the new arbitrary constant. Since $C$ is arbitrary, $K$ is also an arbitrary constant.
$\frac{1}{y^4} + \frac{1}{x^4} = K$
This can also be written by finding a common denominator on the left side:
$\frac{x^4 + y^4}{x^4 y^4} = K$
$x^4 + y^4 = K x^4 y^4$
Note that $y=0$ is also a solution to the original differential equation ($x^5 \cdot 0 = -0^5 \implies 0=0$). This singular solution is not included in the general solution $x^4 + y^4 = K x^4 y^4$ unless $K$ is such that it forces $y=0$ (which is not generally the case). However, the question asks for the general solution obtained by integrating the separated form.
The general solution of the given differential equation is:
$\mathbf{\frac{1}{y^4} + \frac{1}{x^4} = K}$ or $\mathbf{x^4 + y^4 = K x^4 y^4}$
where $K$ is an arbitrary constant.
Question 9. $\frac{dy}{dx} = \sin^{-1} x$
Answer:
Given:
The differential equation is $\frac{dy}{dx} = \sin^{-1} x$.
Solution:
The given differential equation is:
$\frac{dy}{dx} = \sin^{-1} x$
This is a separable differential equation. We can write it as:
$dy = \sin^{-1} x \, dx$
Now, integrate both sides of the equation:
$\int\limits dy = \int\limits \sin^{-1} x \, dx$
The integral on the left side is:
$\int\limits dy = y$
To evaluate the integral on the right side, $\int\limits \sin^{-1} x \, dx$, we use integration by parts. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.
Let $u = \sin^{-1} x$ and $dv = dx$.
Then, $du = \frac{d}{dx}(\sin^{-1} x) \, dx = \frac{1}{\sqrt{1 - x^2}} \, dx$.
And, $v = \int dv = \int dx = x$.
Applying the integration by parts formula:
$\int\limits \sin^{-1} x \, dx = x \sin^{-1} x - \int\limits x \left(\frac{1}{\sqrt{1 - x^2}}\right) dx$
$\int\limits \sin^{-1} x \, dx = x \sin^{-1} x - \int\limits \frac{x}{\sqrt{1 - x^2}} dx$
Now, we need to evaluate the integral $\int\limits \frac{x}{\sqrt{1 - x^2}} dx$. We can use substitution.
Let $t = 1 - x^2$.
Then, $dt = \frac{d}{dx}(1 - x^2) \, dx = -2x \, dx$.
So, $x \, dx = -\frac{1}{2} dt$.
The integral becomes:
$\int\limits \frac{x}{\sqrt{1 - x^2}} dx = \int\limits \frac{-\frac{1}{2} dt}{\sqrt{t}} = -\frac{1}{2} \int\limits t^{-1/2} dt$
Integrating $t^{-1/2}$ with respect to $t$:
$-\frac{1}{2} \left(\frac{t^{-1/2 + 1}}{-1/2 + 1}\right) + C'$
$-\frac{1}{2} \left(\frac{t^{1/2}}{1/2}\right) + C'$
$-\frac{1}{2} (2 t^{1/2}) + C'$
$-t^{1/2} + C' = -\sqrt{t} + C'$
Substitute back $t = 1 - x^2$:
$\int\limits \frac{x}{\sqrt{1 - x^2}} dx = -\sqrt{1 - x^2} + C'$
Substitute this result back into the integration by parts equation for $\int\limits \sin^{-1} x \, dx$:
$\int\limits \sin^{-1} x \, dx = x \sin^{-1} x - (-\sqrt{1 - x^2}) + C$
$\int\limits \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C$
Equating the integrated results for $y$:
$y = x \sin^{-1} x + \sqrt{1 - x^2} + C$
The general solution of the given differential equation is:
$\mathbf{y = x \sin^{-1} x + \sqrt{1 - x^2} + C}$
where $C$ is an arbitrary constant.
Question 10. ex tan y dx + (1 – ex) sec2 y dy = 0
Answer:
Given:
The differential equation is $e^x \tan y \, dx + (1 – e^x) \sec^2 y \, dy = 0$.
Solution:
The given differential equation is:
$e^x \tan y \, dx + (1 – e^x) \sec^2 y \, dy = 0$
This is a separable differential equation. We can rearrange the terms to separate the variables:
$e^x \tan y \, dx = - (1 – e^x) \sec^2 y \, dy$
$e^x \tan y \, dx = (e^x - 1) \sec^2 y \, dy$
Now, divide both sides by $(e^x - 1)$ and $\tan y$ to separate the $x$ and $y$ terms. We assume $e^x - 1 \neq 0$ (i.e., $x \neq 0$) and $\tan y \neq 0$.
$\frac{e^x}{e^x - 1} dx = \frac{\sec^2 y}{\tan y} dy$
Now, integrate both sides of the equation:
$\int\limits \frac{e^x}{e^x - 1} dx = \int\limits \frac{\sec^2 y}{\tan y} dy$
To evaluate the integral on the left side, $\int\limits \frac{e^x}{e^x - 1} dx$, let $u = e^x - 1$. Then $du = e^x dx$.
The integral becomes $\int\limits \frac{du}{u} = \ln|u| + C_1 = \ln|e^x - 1| + C_1$.
To evaluate the integral on the right side, $\int\limits \frac{\sec^2 y}{\tan y} dy$, let $v = \tan y$. Then $dv = \sec^2 y dy$.
The integral becomes $\int\limits \frac{dv}{v} = \ln|v| + C_2 = \ln|\tan y| + C_2$.
Equating the results of the integration:
$\ln|e^x - 1| + C_1 = \ln|\tan y| + C_2$
Rearrange the terms:
$\ln|e^x - 1| - \ln|\tan y| = C_2 - C_1$
Let $C = C_2 - C_1$ be the arbitrary constant of integration. Using logarithm properties, $\ln A - \ln B = \ln(A/B)$:
$\ln\left|\frac{e^x - 1}{\tan y}\right| = C$
Exponentiate both sides with base $e$:
$\left|\frac{e^x - 1}{\tan y}\right| = e^C$
Let $K = \pm e^C$. Since $e^C > 0$, $K$ is an arbitrary non-zero constant.
$\frac{e^x - 1}{\tan y} = K$
Rearrange to solve for $\tan y$:
$\tan y = \frac{e^x - 1}{K}$
Let $C' = 1/K$. Since $K$ is any non-zero constant, $C'$ is also any non-zero constant.
$\tan y = C'(e^x - 1), \text{\$ } C' \neq 0$
We should also consider the cases where we divided by zero, i.e., $e^x - 1 = 0$ or $\tan y = 0$.
Case 1: $e^x - 1 = 0 \implies x = 0$. Substituting $x=0$ into the original differential equation:
$e^0 \tan y \, dx + (1 - e^0) \sec^2 y \, dy = 0$
$1 \cdot \tan y \, dx + (1 - 1) \sec^2 y \, dy = 0$
$\tan y \, dx = 0$
This equation holds if $\tan y = 0$, i.e., $y = n\pi$ for some integer $n$. So $(x=0, y=n\pi)$ are points where the equation holds. The line $x=0$ itself is not a solution.
Case 2: $\tan y = 0 \implies y = n\pi$ for some integer $n$. Substituting $\tan y = 0$ and $\sec^2 y = 1 + \tan^2 y = 1 + 0^2 = 1$ into the original differential equation:
$e^x \cdot 0 \, dx + (1 - e^x) \cdot 1 \, dy = 0$
$(1 - e^x) dy = 0$
This equation holds if $1 - e^x = 0$ (i.e., $x = 0$) or $dy = 0$. If $dy = 0$, then $y$ is a constant, $y=n\pi$. So the lines $y = n\pi$ are solutions to the differential equation for all $x$.
The solution $y = n\pi$ corresponds to $\tan y = 0$. Our general solution $\tan y = C'(e^x - 1)$ yields $\tan y = 0$ if and only if $C'(e^x - 1) = 0$. If $C' \neq 0$, this only happens when $e^x - 1 = 0$, i.e., $x=0$. However, if we allow $C' = 0$, the general solution becomes $\tan y = 0 \cdot (e^x - 1) = 0$, which gives $y = n\pi$. Therefore, the constant $C'$ can be any real number, including zero.
The general solution of the given differential equation is:
$\mathbf{\tan y = C(e^x - 1)}$
where $C$ is an arbitrary constant.
For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:
Question 11. $(x^3 + x^2 + x + 1) \frac{dy}{dx} = 2x^2 + x$ ; y = 1 when x = 0
Answer:
Given:
The differential equation is $(x^3 + x^2 + x + 1) \frac{dy}{dx} = 2x^2 + x$.
The initial condition is $y = 1$ when $x = 0$.
Solution:
The given differential equation is:
$(x^3 + x^2 + x + 1) \frac{dy}{dx} = 2x^2 + x$
First, factor the denominator: $x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1)$.
Rewrite the equation as:
$(x^2+1)(x+1) \frac{dy}{dx} = 2x^2 + x$
This is a separable differential equation. Separate the variables:
$dy = \frac{2x^2 + x}{(x^2+1)(x+1)} dx$
Integrate both sides:
$\int\limits dy = \int\limits \frac{2x^2 + x}{(x^2+1)(x+1)} dx$
The left side integral is $\int\limits dy = y$.
For the right side integral, we use partial fraction decomposition for the integrand $\frac{2x^2 + x}{(x^2+1)(x+1)}$.
$\frac{2x^2 + x}{(x^2+1)(x+1)} = \frac{Ax + B}{x^2+1} + \frac{C}{x+1}$
Multiplying by $(x^2+1)(x+1)$, we get:
$2x^2 + x = (Ax + B)(x+1) + C(x^2+1)$
Expanding the right side:
$2x^2 + x = Ax^2 + Ax + Bx + B + Cx^2 + C$
$2x^2 + x = (A+C)x^2 + (A+B)x + (B+C)$
Equating the coefficients of the powers of $x$:
Coefficient of $x^2$: $A + C = 2$
Coefficient of $x$: $A + B = 1$
Constant term: $B + C = 0 \implies B = -C$
Substitute $B = -C$ into $A + B = 1$:
$A - C = 1$
We have the system:
$A + C = 2$
$A - C = 1$
Adding these two equations: $2A = 3 \implies A = \frac{3}{2}$.
Substitute $A = \frac{3}{2}$ into $A + C = 2$: $\frac{3}{2} + C = 2 \implies C = 2 - \frac{3}{2} = \frac{1}{2}$.
Using $B = -C$: $B = -\frac{1}{2}$.
So, the partial fraction decomposition is:
$\frac{2x^2 + x}{(x^2+1)(x+1)} = \frac{\frac{3}{2}x - \frac{1}{2}}{x^2+1} + \frac{\frac{1}{2}}{x+1} = \frac{1}{2} \left( \frac{3x - 1}{x^2+1} + \frac{1}{x+1} \right)$
Now integrate:
$\int\limits \frac{1}{2} \left( \frac{3x - 1}{x^2+1} + \frac{1}{x+1} \right) dx = \frac{1}{2} \left( \int\limits \frac{3x}{x^2+1} dx - \int\limits \frac{1}{x^2+1} dx + \int\limits \frac{1}{x+1} dx \right)$
Evaluate each integral:
$\int\limits \frac{3x}{x^2+1} dx$: Let $u = x^2+1$, $du = 2x dx$. $\int\limits \frac{3}{u} \frac{du}{2} = \frac{3}{2} \int\limits \frac{1}{u} du = \frac{3}{2} \ln|u| = \frac{3}{2} \ln(x^2+1)$.
$\int\limits \frac{1}{x^2+1} dx = \arctan(x)$.
$\int\limits \frac{1}{x+1} dx = \ln|x+1|$.
Combining the integrals on the right side:
$\int\limits \frac{2x^2 + x}{(x^2+1)(x+1)} dx = \frac{1}{2} \left( \frac{3}{2} \ln(x^2+1) - \arctan(x) + \ln|x+1| \right) + C'$
Equating the left and right sides of the integrated equation:
$y = \frac{1}{2} \left( \frac{3}{2} \ln(x^2+1) - \arctan(x) + \ln|x+1| \right) + C$
$y = \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \arctan(x) + \frac{1}{2} \ln|x+1| + C$
Applying the initial condition:
We are given that $y = 1$ when $x = 0$. Substitute these values into the general solution:
$1 = \frac{3}{4} \ln(0^2+1) - \frac{1}{2} \arctan(0) + \frac{1}{2} \ln|0+1| + C$
$1 = \frac{3}{4} \ln(1) - \frac{1}{2} \arctan(0) + \frac{1}{2} \ln(1) + C$
We know that $\ln(1) = 0$ and $\arctan(0) = 0$.
$1 = \frac{3}{4} \cdot 0 - \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot 0 + C$
$1 = 0 - 0 + 0 + C$
$C = 1$
Particular Solution:
Substitute the value of $C$ back into the general solution:
$y = \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \arctan(x) + \frac{1}{2} \ln|x+1| + 1$
The particular solution satisfying the given condition is:
$\mathbf{y = \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \arctan(x) + \frac{1}{2} \ln|x+1| + 1}$
Question 12. $x(x^2 - 1) \frac{dy}{dx} = 1$ ; y = 0 when x = 2
Answer:
Given:
The differential equation is $x(x^2 - 1) \frac{dy}{dx} = 1$.
The initial condition is $y = 0$ when $x = 2$.
Solution:
The given differential equation is:
$x(x^2 - 1) \frac{dy}{dx} = 1$
First, factor the term $x^2 - 1 = (x-1)(x+1)$.
The equation is:
$x(x-1)(x+1) \frac{dy}{dx} = 1$
This is a separable differential equation. Separate the variables:
$dy = \frac{1}{x(x-1)(x+1)} dx$
Integrate both sides:
$\int\limits dy = \int\limits \frac{1}{x(x-1)(x+1)} dx$
The left side integral is $\int\limits dy = y$.
For the right side integral, we use partial fraction decomposition for the integrand $\frac{1}{x(x-1)(x+1)}$.
$\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$
Multiply by $x(x-1)(x+1)$:
$1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$
$1 = A(x^2 - 1) + B(x^2 + x) + C(x^2 - x)$
$1 = Ax^2 - A + Bx^2 + Bx + Cx^2 - Cx$
$1 = (A+B+C)x^2 + (B-C)x - A$
Equating coefficients of powers of $x$:
Coefficient of $x^2$: $A + B + C = 0$
Coefficient of $x$: $B - C = 0 \implies B = C$
Constant term: $-A = 1 \implies A = -1$
Substitute $A = -1$ into $A + B + C = 0$:
$-1 + B + C = 0 \implies B + C = 1$
Since $B = C$, substitute $B$ for $C$:
$B + B = 1 \implies 2B = 1 \implies B = \frac{1}{2}$
Since $B = C$, $C = \frac{1}{2}$.
So, the partial fraction decomposition is:
$\frac{1}{x(x-1)(x+1)} = \frac{-1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1}$
Now integrate:
$\int\limits \left( \frac{-1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)} \right) dx = - \int\limits \frac{1}{x} dx + \frac{1}{2} \int\limits \frac{1}{x-1} dx + \frac{1}{2} \int\limits \frac{1}{x+1} dx$
$= -\ln|x| + \frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| + C'$
Using logarithm properties:
$= -\ln|x| + \frac{1}{2} (\ln|x-1| + \ln|x+1|) + C'$
$= -\ln|x| + \frac{1}{2} \ln|(x-1)(x+1)| + C'$
$= -\ln|x| + \frac{1}{2} \ln|x^2-1| + C'$
Equating the left and right sides of the integrated equation:
$y = -\ln|x| + \frac{1}{2} \ln|x^2-1| + C$
Applying the initial condition:
We are given that $y = 0$ when $x = 2$. Substitute these values into the general solution:
$0 = -\ln|2| + \frac{1}{2} \ln|2^2-1| + C$
$0 = -\ln(2) + \frac{1}{2} \ln|4-1| + C$
$0 = -\ln(2) + \frac{1}{2} \ln(3) + C$
Solve for $C$:
$C = \ln(2) - \frac{1}{2} \ln(3)$
Using logarithm properties, $\frac{1}{2} \ln(3) = \ln(3^{1/2}) = \ln(\sqrt{3})$.
$C = \ln(2) - \ln(\sqrt{3})$
$C = \ln\left(\frac{2}{\sqrt{3}}\right)$
Particular Solution:
Substitute the value of $C$ back into the general solution:
$y = -\ln|x| + \frac{1}{2} \ln|x^2-1| + \ln\left(\frac{2}{\sqrt{3}}\right)$
Using logarithm properties, this can also be written as:
$y = \ln\left|\frac{\sqrt{x^2-1}}{x}\right| + \ln\left(\frac{2}{\sqrt{3}}\right)$
$y = \ln\left(\left|\frac{\sqrt{x^2-1}}{x}\right| \cdot \frac{2}{\sqrt{3}}\right)$
Since the initial condition is at $x=2$, where $x>0$ and $x^2-1 > 0$, we can remove the absolute values for the neighbourhood around $x=2$.
$y = \ln\left(\frac{\sqrt{x^2-1}}{x} \cdot \frac{2}{\sqrt{3}}\right)$
$y = \ln\left(\frac{2\sqrt{x^2-1}}{x\sqrt{3}}\right)$
The particular solution satisfying the given condition is:
$\mathbf{y = -\ln|x| + \frac{1}{2} \ln|x^2-1| + \ln\left(\frac{2}{\sqrt{3}}\right)}$
or equivalently
$\mathbf{y = \ln\left(\frac{2\sqrt{x^2-1}}{x\sqrt{3}}\right)}$ for $x>1$.
Question 13. $\cos \left( \frac{dy}{dx} \right) = a (a ∈ R)$ ; y = 1 when x = 0
Answer:
Given:
The differential equation is $\cos \left( \frac{dy}{dx} \right) = a$.
The initial condition is $y = 1$ when $x = 0$.
The parameter $a \in R$ (a is a real number).
Solution:
The given differential equation is:
$\cos \left( \frac{dy}{dx} \right) = a$
For the term $\frac{dy}{dx}$ to be a real value, the value of $\cos \left( \frac{dy}{dx} \right)$ must be within the range $[-1, 1]$. Therefore, a real solution exists only if the parameter $a$ satisfies the condition:
$-1 \leq a \leq 1$
If $|a| > 1$, there is no real solution for $\frac{dy}{dx}$.
Assuming $-1 \leq a \leq 1$, we can take the inverse cosine of both sides. The value of $\frac{dy}{dx}$ is given by the set of values $k$ such that $\cos(k) = a$. These values are $k = 2n\pi \pm \cos^{-1}(a)$ for any integer $n$, where $\cos^{-1}(a)$ denotes the principal value in $[0, \pi]$.
For simplicity, we can let $\frac{dy}{dx}$ be any constant value whose cosine is $a$. Let this constant be $k$.
$\frac{dy}{dx} = k$
where $\cos(k) = a$.
This is a separable differential equation. We can write it as:
$dy = k \, dx$
Now, integrate both sides of the equation:
$\int\limits dy = \int\limits k \, dx$
The integrals are:
$y = kx + C_{int}$
where $C_{int}$ is the constant of integration.
Substituting back the meaning of $k$, where $\cos(k)=a$, the general solution is $y = kx + C_{int}$ with the condition $\cos(k)=a$.
Applying the initial condition:
We are given that $y = 1$ when $x = 0$. Substitute these values into the general solution:
$1 = k(0) + C_{int}$
$1 = 0 + C_{int}$
$C_{int} = 1$
Particular Solution:
Substitute the value of $C_{int}$ back into the general solution:
$y = kx + 1$
where $k$ is any constant such that $\cos(k) = a$. This requires $-1 \leq a \leq 1$.
If we use the principal value $k = \cos^{-1}(a)$, the particular solution is:
$y = (\cos^{-1} a) x + 1$
Note that if $|a| > 1$, no real solution exists.
The particular solution satisfying the given condition, provided $|a| \leq 1$, is:
$\mathbf{y = kx + 1, \text{\$ where } \cos(k) = a}$
or using the principal value of inverse cosine:
$\mathbf{y = (\cos^{-1} a) x + 1, \text{\$ provided } -1 \leq a \leq 1}$
Question 14. $\frac{dy}{dx} y \tan x$ ; y = 1 when x = 0
Answer:
Given:
The differential equation is $\frac{dy}{dx} = y \tan x$.
The initial condition is $y = 1$ when $x = 0$.
Solution:
The given differential equation is:
$\frac{dy}{dx} = y \tan x$
This is a separable differential equation. We can separate the variables by dividing both sides by $y$ (assuming $y \neq 0$) and multiplying by $dx$:
$\frac{dy}{y} = \tan x \, dx$
Integrate both sides of the equation:
$\int\limits \frac{dy}{y} = \int\limits \tan x \, dx$
The integral on the left side is:
$\int\limits \frac{dy}{y} = \ln|y| + C_1$
The integral on the right side is:
$\int\limits \tan x \, dx = \int\limits \frac{\sin x}{\cos x} \, dx$
Let $u = \cos x$, then $du = -\sin x \, dx$, so $\sin x \, dx = -du$.
$\int\limits \frac{-du}{u} = - \int\limits \frac{du}{u} = -\ln|u| + C_2 = -\ln|\cos x| + C_2$
Using the property $-\ln A = \ln(1/A)$, we have $-\ln|\cos x| = \ln\left|\frac{1}{\cos x}\right| = \ln|\sec x|$.
So, $\int\limits \tan x \, dx = \ln|\sec x| + C_2$.
Equating the results from both sides:
$\ln|y| = \ln|\sec x| + C$
where $C = C_2 - C_1$ is the arbitrary constant of integration.
Exponentiate both sides with base $e$:
$|y| = e^{\ln|\sec x| + C} = e^{\ln|\sec x|} e^C = |\sec x| e^C$
Let $A = e^C$. Since $C$ is arbitrary, $A$ is an arbitrary positive constant ($A > 0$).
$|y| = A |\sec x|$
This implies $y = \pm A \sec x$. Let $K = \pm A$. Since $A$ is any positive constant, $K$ is any non-zero constant.
$y = K \sec x, \text{\$ } K \neq 0$
The case $y=0$ is also a solution ($0 = 0 \cdot \tan x$). This solution is included if we allow $K=0$.
The general solution is $y = K \sec x$, where $K$ is an arbitrary constant.
Applying the initial condition:
We are given that $y = 1$ when $x = 0$. Substitute these values into the general solution:
$1 = K \sec(0)$
Since $\sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1$, we have:
$1 = K \cdot 1$
$K = 1$
Particular Solution:
Substitute the value of $K$ back into the general solution:
$y = 1 \cdot \sec x$
$y = \sec x$
The particular solution satisfying the given condition is:
$\mathbf{y = \sec x}$
Question 15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is $y’ = e^x \sin x$.
Answer:
Given:
The differential equation is $y’ = e^x \sin x$.
This can be written as $\frac{dy}{dx} = e^x \sin x$.
The curve passes through the point $(0, 0)$, which gives the initial condition $y = 0$ when $x = 0$.
Solution:
The given differential equation is:
$\frac{dy}{dx} = e^x \sin x$
To find the equation of the curve (the general solution), we integrate both sides with respect to $x$:
$\int\limits dy = \int\limits e^x \sin x \, dx$
The left side integral is straightforward:
$\int\limits dy = y$
For the integral on the right side, $\int\limits e^x \sin x \, dx$, we use integration by parts twice. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.
Let $I = \int\limits e^x \sin x \, dx$.
Choose $u = \sin x$ and $dv = e^x dx$.
Then $du = \cos x \, dx$ and $v = \int e^x dx = e^x$.
$I = e^x \sin x - \int\limits e^x \cos x \, dx$
Now, evaluate the integral $\int\limits e^x \cos x \, dx$ using integration by parts again.
Choose $u' = \cos x$ and $dv' = e^x dx$.
Then $du' = -\sin x \, dx$ and $v' = \int e^x dx = e^x$.
$\int\limits e^x \cos x \, dx = e^x \cos x - \int\limits e^x (-\sin x) \, dx = e^x \cos x + \int\limits e^x \sin x \, dx$
Notice that the integral on the right is the original integral $I$.
Substitute this back into the expression for $I$:
$I = e^x \sin x - (e^x \cos x + I)$
$I = e^x \sin x - e^x \cos x - I$
$2I = e^x \sin x - e^x \cos x$
$I = \frac{1}{2} (e^x \sin x - e^x \cos x)$
Now, equate the results of the integration from both sides of the differential equation and add the constant of integration $C$:
$y = \frac{1}{2} (e^x \sin x - e^x \cos x) + C$
This is the general solution.
Applying the initial condition:
The curve passes through $(0, 0)$, so $y = 0$ when $x = 0$. Substitute these values into the general solution:
$0 = \frac{1}{2} (e^0 \sin 0 - e^0 \cos 0) + C$
We know that $e^0 = 1$, $\sin 0 = 0$, and $\cos 0 = 1$.
$0 = \frac{1}{2} (1 \cdot 0 - 1 \cdot 1) + C$
$0 = \frac{1}{2} (0 - 1) + C$
$0 = -\frac{1}{2} + C$
Solving for $C$:
$C = \frac{1}{2}$
Particular Solution:
Substitute the value of $C$ back into the general solution:
$y = \frac{1}{2} (e^x \sin x - e^x \cos x) + \frac{1}{2}$
This can also be written as:
$y = \frac{1}{2} e^x (\sin x - \cos x) + \frac{1}{2}$
or
$2y = e^x \sin x - e^x \cos x + 1$
The equation of the curve passing through the point $(0, 0)$ is:
$\mathbf{y = \frac{1}{2} (e^x \sin x - e^x \cos x) + \frac{1}{2}}$
Question 16. For the differential equation $xy \frac{dy}{dx} = (x + 2) (y + 2)$ , find the solution curve passing through the point (1, –1).
Answer:
Given:
The differential equation is $xy \frac{dy}{dx} = (x + 2) (y + 2)$.
The solution curve passes through the point $(1, –1)$, which gives the initial condition $y = -1$ when $x = 1$.
Solution:
The given differential equation is:
$xy \frac{dy}{dx} = (x + 2) (y + 2)$
This is a separable differential equation. Separate the variables by dividing by $xy$ and multiplying by $dx$ (assuming $x \neq 0$ and $y \neq 0$):
$\frac{y}{y + 2} dy = \frac{x + 2}{x} dx$
We can simplify the fractions on both sides:
$\frac{y + 2 - 2}{y + 2} dy = \frac{x}{x} + \frac{2}{x} dx$
$\left(1 - \frac{2}{y + 2}\right) dy = \left(1 + \frac{2}{x}\right) dx$
Integrate both sides of the equation:
$\int\limits \left(1 - \frac{2}{y + 2}\right) dy = \int\limits \left(1 + \frac{2}{x}\right) dx$
Integrate the left side:
$\int\limits 1 \, dy - \int\limits \frac{2}{y + 2} dy = y - 2 \ln|y + 2| + C_1$
Integrate the right side:
$\int\limits 1 \, dx + \int\limits \frac{2}{x} dx = x + 2 \ln|x| + C_2$
Equating the integrated results:
$y - 2 \ln|y + 2| = x + 2 \ln|x| + C$
where $C = C_2 - C_1$ is the arbitrary constant of integration.
Rearrange the terms:
$y - x = 2 \ln|x| + 2 \ln|y + 2| + C$
$y - x = 2 (\ln|x| + \ln|y + 2|) + C$
$y - x = 2 \ln|x(y + 2)| + C$
Applying the initial condition:
The curve passes through $(1, –1)$, so $y = -1$ when $x = 1$. Substitute these values into the general solution:
$-1 - 1 = 2 \ln|1(-1 + 2)| + C$
$-2 = 2 \ln|1(1)| + C$
$-2 = 2 \ln|1| + C$
We know that $\ln(1) = 0$.
$-2 = 2 \cdot 0 + C$
$-2 = 0 + C$
$C = -2$
Particular Solution:
Substitute the value of $C$ back into the general solution:
$y - x = 2 \ln|x(y + 2)| - 2$
Rearrange the equation:
$y - x + 2 = 2 \ln|x(y + 2)|$
Note that the point $(1, -1)$ lies in the region where $x > 0$ and $y+2 = -1+2 = 1 > 0$. In the neighbourhood of this point, $|x| = x$ and $|y+2| = y+2$. So, for the solution curve passing through this point, we can write:
$y - x + 2 = 2 \ln(x(y + 2))$
The equation of the solution curve passing through the point $(1, –1)$ is:
$\mathbf{y - x + 2 = 2 \ln|x(y + 2)|}$
or in the neighbourhood of $(1, -1)$:
$\mathbf{y - x + 2 = 2 \ln(x(y + 2))}$
Question 17. Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Answer:
Given:
The curve passes through the point $(0, –2)$. This means $y = -2$ when $x = 0$.
At any point $(x, y)$ on the curve, the product of the slope of its tangent ($\frac{dy}{dx}$) and the y coordinate ($y$) is equal to the x coordinate ($x$).
To Find:
The equation of the curve.
Solution:
From the given condition, the differential equation of the curve is:
$y \frac{dy}{dx} = x$
This is a separable differential equation. We can separate the variables by multiplying by $dx$:
$y \, dy = x \, dx$
Integrate both sides of the equation:
$\int\limits y \, dy = \int\limits x \, dx$
Integrating both sides gives:
$\frac{y^{1+1}}{1+1} = \frac{x^{1+1}}{1+1} + C_{int}$
$\frac{y^2}{2} = \frac{x^2}{2} + C_{int}$
where $C_{int}$ is the constant of integration.
Multiply the equation by 2 to simplify:
$y^2 = x^2 + 2 C_{int}$
Let $C = 2 C_{int}$. Since $C_{int}$ is an arbitrary constant, $C$ is also an arbitrary constant. The general solution is:
$y^2 = x^2 + C$
Applying the initial condition:
The curve passes through $(0, -2)$. Substitute $x = 0$ and $y = -2$ into the general solution:
$(-2)^2 = (0)^2 + C$
$4 = 0 + C$
$C = 4$
Particular Solution:
Substitute the value of $C$ back into the general solution $y^2 = x^2 + C$:
$y^2 = x^2 + 4$
We can also write this as $y^2 - x^2 = 4$. This represents a hyperbola.
The equation of the curve passing through the point $(0, –2)$ and satisfying the given condition is:
$\mathbf{y^2 = x^2 + 4}$
Question 18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).
Answer:
Given:
A curve passes through the point $(-2, 1)$.
At any point $(x, y)$ on the curve, the slope of the tangent is twice the slope of the line segment joining the point of contact $(x, y)$ to the point $(-4, -3)$.
To Find:
The equation of the curve.
Solution:
The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.
The slope of the line segment joining the point $(x, y)$ to the point $(-4, -3)$ is calculated using the formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Here, $(x_1, y_1) = (x, y)$ and $(x_2, y_2) = (-4, -3)$.
Slope of the line segment = $\frac{-3 - y}{-4 - x} = \frac{-(3+y)}{-(4+x)} = \frac{y+3}{x+4}$.
According to the problem statement, the slope of the tangent is twice the slope of this line segment:
$\frac{dy}{dx} = 2 \left(\frac{y+3}{x+4}\right)$
$\frac{dy}{dx} = 2 \frac{y+3}{x+4}$
(Differential Equation)
This is a separable differential equation. We separate the variables by moving terms involving $y$ to the left side and terms involving $x$ to the right side. Assume $y+3 \neq 0$ and $x+4 \neq 0$.
$\frac{dy}{y+3} = \frac{2}{x+4} dx$
Now, integrate both sides of the equation:
$\int\limits \frac{dy}{y+3} = \int\limits \frac{2}{x+4} dx$
$\int\limits \frac{dy}{y+3} = 2 \int\limits \frac{1}{x+4} dx$
Evaluating the integrals:
$\int\limits \frac{dy}{y+3} = \ln|y+3| + C_1$ (using substitution $u = y+3$)
$\int\limits \frac{1}{x+4} dx = \ln|x+4| + C_2$ (using substitution $v = x+4$)
So the integrated equation is:
$\ln|y+3| = 2 \ln|x+4| + C_{int}$
where $C_{int}$ is the constant of integration.
Using the logarithm property $k \ln A = \ln A^k$:
$\ln|y+3| = \ln|(x+4)^2| + C_{int}$
Exponentiate both sides with base $e$:
$|y+3| = e^{\ln|(x+4)^2| + C_{int}}$
$|y+3| = e^{\ln|(x+4)^2|} e^{C_{int}}$
$|y+3| = (x+4)^2 e^{C_{int}}$ (since $(x+4)^2 \geq 0$, $|(x+4)^2| = (x+4)^2$)
Let $A = e^{C_{int}}$. Since $C_{int}$ is an arbitrary constant, $A$ is an arbitrary positive constant ($A > 0$).
$|y+3| = A (x+4)^2$
This implies $y+3 = \pm A (x+4)^2$. Let $K = \pm A$. Since $A$ is any positive constant, $K$ is any non-zero constant.
$y+3 = K (x+4)^2, \text{\$ } K \neq 0$
Consider the case where $y+3=0$, i.e., $y=-3$. Substituting $y=-3$ into the original differential equation gives $x(-3)\frac{dy}{dx} = (x+2)(-3+2) = -(x+2)$. For $y=-3$ to be a solution curve, $\frac{dy}{dx}$ must be $0$ along this curve. This requires $-(x+2)=0$, so $x=-2$. Thus, the point $(-2, -3)$ is a potential singular solution point. The general solution $y+3 = K(x+4)^2$ includes the singular solution $y=-3$ if we allow $K=0$. Therefore, the constant $K$ can be any real number.
The general solution is:
$y+3 = K (x+4)^2$
Applying the initial condition:
The curve passes through the point $(-2, 1)$. Substitute $x = -2$ and $y = 1$ into the general solution:
$1 + 3 = K (-2 + 4)^2$
$4 = K (2)^2$
$4 = K \cdot 4$
Dividing by 4:
$K = 1$
Particular Solution:
Substitute the value of $K = 1$ back into the general solution $y+3 = K(x+4)^2$:
$y+3 = 1 \cdot (x+4)^2$
$y+3 = (x+4)^2$
Solving for $y$:
$y = (x+4)^2 - 3$
The equation of the curve passing through the point $(-2, 1)$ is:
$\mathbf{y = (x+4)^2 - 3}$
Question 19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Answer:
Given:
The volume of the spherical balloon changes at a constant rate.
Initial condition 1: Radius $r = 3$ units when time $t = 0$ seconds.
Initial condition 2: Radius $r = 6$ units when time $t = 3$ seconds.
To Find:
The radius of the balloon $r$ as a function of time $t$.
Solution:
Let $V$ be the volume of the spherical balloon and $r$ be its radius at time $t$.
The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$.
The problem states that the volume changes at a constant rate. Let this constant rate be $k$.
So, the differential equation is $\frac{dV}{dt} = k$.
We need to express $\frac{dV}{dt}$ in terms of $r$ and $\frac{dr}{dt}$. Differentiate the volume formula with respect to $t$:
$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi \frac{d}{dt}(r^3)$
Using the chain rule, $\frac{d}{dt}(r^3) = 3r^2 \frac{dr}{dt}$.
So, $\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Equating this to the constant rate $k$:
$4\pi r^2 \frac{dr}{dt} = k$
This is a separable differential equation. Separate the variables $r$ and $t$:
$4\pi r^2 \, dr = k \, dt$
Integrate both sides of the equation:
$\int\limits 4\pi r^2 \, dr = \int\limits k \, dt$
$4\pi \int\limits r^2 \, dr = k \int\limits 1 \, dt$
$4\pi \left( \frac{r^3}{3} \right) = kt + C$
$\frac{4}{3}\pi r^3 = kt + C$
This is the general solution relating $r$ and $t$. Note that the left side is the volume $V$. So $V = kt + C$, which means the volume is a linear function of time.
We need to find the values of the constants $k$ and $C$ using the given initial conditions.
Applying the initial conditions:
Condition 1: $r = 3$ when $t = 0$. Substitute these values into the general solution $\frac{4}{3}\pi r^3 = kt + C$:
$\frac{4}{3}\pi (3)^3 = k(0) + C$
[Using $r=3, t=0$] ... (i)
$\frac{4}{3}\pi (27) = 0 + C$
$36\pi = C$
Condition 2: $r = 6$ when $t = 3$. Substitute these values and the value of $C$ into the general solution:
$\frac{4}{3}\pi (6)^3 = k(3) + 36\pi$
[Using $r=6, t=3, C=36\pi$] ... (ii)
$\frac{4}{3}\pi (216) = 3k + 36\pi$
$4\pi (72) = 3k + 36\pi$
$288\pi = 3k + 36\pi$
$3k = 288\pi - 36\pi$
$3k = 252\pi$
$k = \frac{252\pi}{3} = 84\pi$
Equation for radius after t seconds:
Substitute the values of $k = 84\pi$ and $C = 36\pi$ back into the general solution $\frac{4}{3}\pi r^3 = kt + C$:
$\frac{4}{3}\pi r^3 = 84\pi t + 36\pi$
Divide the entire equation by $\frac{4}{3}\pi$:
$r^3 = \frac{3}{4\pi} (84\pi t + 36\pi)$
$r^3 = \frac{3}{4\pi} \cdot 84\pi t + \frac{3}{4\pi} \cdot 36\pi$
$r^3 = 3 \cdot \frac{84}{4} t + 3 \cdot \frac{36}{4}$
$r^3 = 3 \cdot 21 t + 3 \cdot 9$
$r^3 = 63t + 27$
To find the radius $r$, take the cube root of both sides:
$r = (63t + 27)^{1/3}$
The radius of the balloon after $t$ seconds is given by:
$\mathbf{r(t) = (63t + 27)^{1/3}}$
Question 20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).
Answer:
Given:
Principal increases continuously at the rate of $r\%$ per year.
Initial Principal $P_0 = \textsf{₹} 100$.
Principal doubles in 10 years, so $P = \textsf{₹} 200$ when $t = 10$ years.
$\log_e 2 = 0.6931$.
To Find:
The value of the rate $r$.
Solution:
Let $P$ be the principal at time $t$ years. The rate of increase of principal is proportional to the principal at that time.
The rate of increase is $r\%$ of $P$, which is $\frac{r}{100} P$.
The differential equation representing this situation is:
$\frac{dP}{dt} = \frac{r}{100} P$
This is a separable differential equation. Separate the variables $P$ and $t$:
$\frac{dP}{P} = \frac{r}{100} dt$
Integrate both sides of the equation:
$\int\limits \frac{dP}{P} = \int\limits \frac{r}{100} dt$
$\ln|P| = \frac{r}{100} t + C_{int}$
where $C_{int}$ is the constant of integration.
Exponentiate both sides:
$|P| = e^{\frac{r}{100} t + C_{int}} = e^{C_{int}} e^{\frac{r}{100} t}$
Since principal $P$ is always positive, $|P| = P$. Let $A = e^{C_{int}}$. Since $C_{int}$ is arbitrary, $A$ is an arbitrary positive constant ($A > 0$).
$P(t) = A e^{\frac{r}{100} t}$
Applying the initial conditions:
Condition 1: When $t = 0$, $P = 100$.
$100 = A e^{\frac{r}{100} \cdot 0} = A e^0 = A \cdot 1 = A$
So, $A = 100$. This constant $A$ represents the initial principal $P_0$.
The equation becomes:
$P(t) = 100 e^{\frac{r}{100} t}$
Condition 2: When $t = 10$, $P = 200$.
Substitute these values into the equation:
$200 = 100 e^{\frac{r}{100} \cdot 10}$
[Using $t=10, P=200$] ... (i)
$200 = 100 e^{\frac{r}{10}}$
Divide both sides by 100:
$2 = e^{\frac{r}{10}}$
Take the natural logarithm (logarithm to base $e$) of both sides:
$\ln(2) = \ln\left(e^{\frac{r}{10}}\right)$
Using the logarithm property $\ln(e^x) = x$:
$\ln(2) = \frac{r}{10}$
Solve for $r$:
$r = 10 \ln(2)$
Using the given value $\log_e 2 = 0.6931$:
$r = 10 \times 0.6931$
[Using $\ln(2) = 0.6931$] ... (ii)
$r = 6.931$
The value of $r$ is 6.931. This means the interest rate is 6.931% per year compounded continuously.
The value of r is:
$\mathbf{r = 6.931}$
Question 21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
Answer:
Given:
Principal increases continuously at the rate of $5\%$ per year. So, the rate $r = 5$.
Initial Principal $P_0 = \textsf{₹} 1000$.
Time period $t = 10$ years.
Value of $e^{0.5} = 1.648$.
To Find:
The value of the amount after 10 years ($P(10)$).
Solution:
Let $P$ be the principal at time $t$. The rate of increase of principal is proportional to the principal at that time, with the rate being $r\% = 5\% = \frac{5}{100}$.
The differential equation is:
$\frac{dP}{dt} = \frac{5}{100} P = 0.05 P$
This is a separable differential equation. Separate the variables:
$\frac{dP}{P} = 0.05 \, dt$
Integrate both sides:
$\int\limits \frac{dP}{P} = \int\limits 0.05 \, dt$
$\ln|P| = 0.05 t + C_{int}$
Exponentiate both sides:
$|P| = e^{0.05 t + C_{int}} = e^{C_{int}} e^{0.05 t}$
Since $P > 0$, $P = A e^{0.05 t}$, where $A = e^{C_{int}} > 0$.
Applying the initial condition:
When $t = 0$, $P = 1000$.
$1000 = A e^{0.05 \cdot 0} = A e^0 = A \cdot 1 = A$
So, $A = 1000$. The equation for the principal at time $t$ is:
$P(t) = 1000 e^{0.05 t}$
Finding the value after 10 years:
We need to find $P(10)$. Substitute $t = 10$ into the equation:
$P(10) = 1000 e^{0.05 \cdot 10}$
$P(10) = 1000 e^{0.5}$
Using the given value $e^{0.5} = 1.648$:
P(10) = 1000 $\times$ 1.648
[Using $e^{0.5} = 1.648$] ... (i)
P(10) = 1648
The amount will be $\textsf{₹} 1648$ after 10 years.
The value of the amount after 10 years is:
$\mathbf{\textsf{₹} 1648}$
Question 22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Answer:
Given:
Initial bacteria count $N_0 = 1,00,000$ at time $t = 0$.
The number of bacteria increases by 10% in 2 hours. This means the count at $t = 2$ hours is $N(2) = 1,00,000 + 10\% \text{ of } 1,00,000 = 1,00,000 + 0.10 \times 1,00,000 = 1,10,000$.
The rate of growth of bacteria is proportional to the number present.
$\log_e 2 = \ln 2 = 0.6931$.
To Find:
The time $t$ in hours when the bacteria count reaches $2,00,000$.
Solution:
Let $N(t)$ be the number of bacteria at time $t$. The statement "the rate of growth of bacteria is proportional to the number present" can be written as the differential equation:
$\frac{dN}{dt} = kN$
where $k$ is the constant of proportionality (growth rate).
This is a separable differential equation. Separate the variables:
$\frac{dN}{N} = k \, dt$
Integrate both sides:
$\int\limits \frac{dN}{N} = \int\limits k \, dt$
$\ln|N| = kt + C_{int}$
Since the number of bacteria $N$ is always positive, we can write $\ln N$. Exponentiate both sides:
$N(t) = e^{kt + C_{int}} = e^{C_{int}} e^{kt}$
Let $A = e^{C_{int}}$. Since $C_{int}$ is an arbitrary constant, $A$ is an arbitrary positive constant ($A > 0$).
$N(t) = A e^{kt}$
Apply the first initial condition: $N(0) = 1,00,000$ at $t = 0$.
$1,00,000 = A e^{k \cdot 0} = A e^0 = A \cdot 1$
So, $A = 1,00,000$. The equation representing the number of bacteria at time $t$ is:
$N(t) = 1,00,000 e^{kt}$
Apply the second initial condition: $N(2) = 1,10,000$ at $t = 2$.
$1,10,000 = 1,00,000 e^{k \cdot 2}$
Divide both sides by $1,00,000$:
$\frac{1,10,000}{1,00,000} = e^{2k}$
$1.1 = e^{2k}$
Take the natural logarithm ($\ln$) of both sides:
$\ln(1.1) = \ln(e^{2k})$
[Taking natural log] ... (i)
Using the property $\ln(e^x) = x$:
$\ln(1.1) = 2k$
$k = \frac{\ln(1.1)}{2}$
Now, we want to find the time $t$ when the count reaches $2,00,000$. So, set $N(t) = 2,00,000$:
$2,00,000 = 1,00,000 e^{kt}$
Divide both sides by $1,00,000$:
$\frac{2,00,000}{1,00,000} = e^{kt}$
$2 = e^{kt}$
Take the natural logarithm of both sides:
$\ln(2) = \ln(e^{kt})$
[Taking natural log] ... (ii)
$\ln(2) = kt$
Substitute the expression for $k = \frac{\ln(1.1)}{2}$ from equation (i) into this equation:
$\ln(2) = \left(\frac{\ln(1.1)}{2}\right) t$
Now, solve for $t$:
$t = \frac{2 \ln(2)}{\ln(1.1)}$
We are given the value $\ln 2 = 0.6931$. Substitute this value:
$t = \frac{2 \times 0.6931}{\ln(1.1)}$
$t = \frac{1.3862}{\ln(1.1)}$
Since the value of $\ln(1.1)$ is not provided, the time is expressed in terms of $\ln(1.1)$.
The count will reach 2,00,000 in $\mathbf{\frac{1.3862}{\ln(1.1)}}$ hours.
Question 23. The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is
(A) ex + e–y = C
(B) ex + ey = C
(C) e–x + ey = C
(D) e–x + e–y = C
Answer:
Given:
The differential equation is $\frac{dy}{dx} = e^{x+y}$.
Solution:
The given differential equation can be written as:
$\frac{dy}{dx} = e^x \cdot e^y$
This is a separable differential equation. Separate the variables by dividing both sides by $e^y$ and multiplying by $dx$:
$\frac{dy}{e^y} = e^x dx$
Rewrite the left side using a negative exponent:
$e^{-y} dy = e^x dx$
Now, integrate both sides of the equation:
$\int\limits e^{-y} dy = \int\limits e^x dx$
Evaluate the integrals. For the left side, use substitution $u = -y$, so $du = -dy$, which means $dy = -du$:
$\int\limits e^{-y} dy = \int\limits e^u (-du) = - \int\limits e^u du = -e^u + C_1$
Substitute back $u = -y$:
$\int\limits e^{-y} dy = -e^{-y} + C_1$
The right side integral is standard:
$\int\limits e^x dx = e^x + C_2$
Equating the integrated results:
$-e^{-y} + C_1 = e^x + C_2$
Rearrange the terms to isolate the exponential terms on one side and the constants on the other:
$C_1 - C_2 = e^x + e^{-y}$
Let $C = C_1 - C_2$ be the arbitrary constant of integration.
$e^x + e^{-y} = C$
Comparing this solution with the given options:
(A) $e^x + e^{–y} = C$
(B) $e^x + e^y = C$
(C) $e^{–x} + e^y = C$
(D) $e^{–x} + e^{–y} = C$
The general solution matches option (A).
The final answer is $\mathbf{e^x + e^{-y} = C}$.
The correct option is (A).
Example 15 to 18 (Before Exercise 9.5)
Example 15: Show that the differential equation $(x – y) \frac{dy}{dx} = x + 2y$ is homogeneous and solve it.
Answer:
Given:
The differential equation is $(x – y) \frac{dy}{dx} = x + 2y$.
To Show: The differential equation is homogeneous.
To Solve: Find the general solution of the differential equation.
Showing Homogeneity:
We can rewrite the differential equation as:
$\frac{dy}{dx} = \frac{x + 2y}{x - y}$
Let $f(x, y) = \frac{x + 2y}{x - y}$.
To check for homogeneity, consider $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda x + 2(\lambda y)}{\lambda x - \lambda y}$
Factor out $\lambda$ from the numerator and the denominator:
$f(\lambda x, \lambda y) = \frac{\lambda (x + 2y)}{\lambda (x - y)}$
Assuming $x \neq y$ and $\lambda \neq 0$, we can cancel $\lambda$:
$f(\lambda x, \lambda y) = \frac{x + 2y}{x - y}$
Thus, we have $f(\lambda x, \lambda y) = f(x, y)$.
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the function $f(x, y)$ is a homogeneous function of degree 0.
Therefore, the given differential equation is a homogeneous differential equation.
Solving the Homogeneous Equation:
The differential equation is $\frac{dy}{dx} = \frac{x + 2y}{x - y}$.
To solve a homogeneous differential equation, we use the substitution $y = vx$.
Differentiate $y = vx$ with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(vx) = v \frac{dx}{dx} + x \frac{dv}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:
$v + x \frac{dv}{dx} = \frac{x + 2(vx)}{x - vx}$
$v + x \frac{dv}{dx} = \frac{x(1 + 2v)}{x(1 - v)}$
Assuming $x \neq 0$, we can cancel $x$:
$v + x \frac{dv}{dx} = \frac{1 + 2v}{1 - v}$
Separate the variables $v$ and $x$. Move $v$ to the right side:
$x \frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v$
Find a common denominator on the right side:
$x \frac{dv}{dx} = \frac{1 + 2v - v(1 - v)}{1 - v} = \frac{1 + 2v - v + v^2}{1 - v} = \frac{v^2 + v + 1}{1 - v}$
Separate the variables $v$ and $x$:
$\frac{1 - v}{v^2 + v + 1} dv = \frac{dx}{x}$
... (i)
Integrate both sides of the equation (i):
$\int\limits \frac{1 - v}{v^2 + v + 1} dv = \int\limits \frac{1}{x} dx$
Let's evaluate the left side integral $\int\limits \frac{1 - v}{v^2 + v + 1} dv$. We can split it into two parts:
$\int\limits \frac{1}{v^2 + v + 1} dv - \int\limits \frac{v}{v^2 + v + 1} dv$
For the first integral, $\int\limits \frac{1}{v^2 + v + 1} dv$, complete the square in the denominator: $v^2 + v + 1 = \left(v + \frac{1}{2}\right)^2 + 1 - \left(\frac{1}{2}\right)^2 = \left(v + \frac{1}{2}\right)^2 + \frac{3}{4}$.
$\int\limits \frac{1}{\left(v + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dv$. This is of the form $\int \frac{1}{w^2 + a^2} dw = \frac{1}{a} \arctan\left(\frac{w}{a}\right)$.
Here $w = v + \frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$.
So, $\int\limits \frac{1}{v^2 + v + 1} dv = \frac{1}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right)$.
For the second integral, $\int\limits \frac{v}{v^2 + v + 1} dv$, we make the numerator proportional to the derivative of the denominator ($2v+1$).
$v = \frac{1}{2}(2v + 1) - \frac{1}{2}$. So the integral is:
$\int\limits \frac{\frac{1}{2}(2v + 1) - \frac{1}{2}}{v^2 + v + 1} dv = \frac{1}{2} \int\limits \frac{2v + 1}{v^2 + v + 1} dv - \frac{1}{2} \int\limits \frac{1}{v^2 + v + 1} dv$.
The first part of this is $\frac{1}{2} \ln|v^2 + v + 1|$. Since the discriminant of $v^2+v+1$ is negative and the coefficient of $v^2$ is positive, $v^2+v+1$ is always positive, so $\ln|v^2 + v + 1| = \ln(v^2 + v + 1)$.
The second part uses the result from the first integral calculation: $-\frac{1}{2} \left(\frac{2}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right)\right) = - \frac{1}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right)$.
So, $\int\limits \frac{v}{v^2 + v + 1} dv = \frac{1}{2} \ln(v^2 + v + 1) - \frac{1}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) + C'$.
Now, subtract the second integral result from the first one (ignoring constants for now):
$\int\limits \frac{1 - v}{v^2 + v + 1} dv = \left(\frac{2}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right)\right) - \left(\frac{1}{2} \ln(v^2 + v + 1) - \frac{1}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right)\right)$
$= \frac{2}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) - \frac{1}{2} \ln(v^2 + v + 1) + \frac{1}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right)$
$= \left(\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}\right) \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) - \frac{1}{2} \ln(v^2 + v + 1)$
$= \frac{3}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) - \frac{1}{2} \ln(v^2 + v + 1)$
$= \sqrt{3} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) - \frac{1}{2} \ln(v^2 + v + 1)$.
The integral on the right side of equation (i) is $\int\limits \frac{1}{x} dx = \ln|x|$.
Equating the results from both sides and adding the constant of integration $C$:
$\sqrt{3} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) - \frac{1}{2} \ln(v^2 + v + 1) = \ln|x| + C$
Substitute back $v = \frac{y}{x}$:
$\sqrt{3} \arctan\left(\frac{2\frac{y}{x} + 1}{\sqrt{3}}\right) - \frac{1}{2} \ln\left(\left(\frac{y}{x}\right)^2 + \frac{y}{x} + 1\right) = \ln|x| + C$
Simplify the terms:
$\sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) - \frac{1}{2} \ln\left(\frac{y^2 + xy + x^2}{x^2}\right) = \ln|x| + C$
Using the logarithm property $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$:
$\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) - \frac{1}{2} (\ln(x^2 + xy + y^2) - \ln(x^2)) = \ln|x| + C$
Using the property $\ln(x^2) = 2 \ln|x|$:
$\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) - \frac{1}{2} \ln(x^2 + xy + y^2) + \frac{1}{2} (2\ln|x|) = \ln|x| + C$
$\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) - \frac{1}{2} \ln(x^2 + xy + y^2) + \ln|x| = \ln|x| + C$
Cancel $\ln|x|$ from both sides:
$\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) - \frac{1}{2} \ln(x^2 + xy + y^2) = C$
Multiply by -2 (and let $-2C$ be the new constant $K$):
$\ln(x^2 + xy + y^2) - 2\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) = -2C$
Let $K = -2C$.
$\ln(x^2 + xy + y^2) - 2\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) = K$
The general solution of the given differential equation is:
$\mathbf{\ln(x^2 + xy + y^2) - 2\sqrt{3} \arctan\left(\frac{x + 2y}{\sqrt{3}x}\right) = K}$
where $K$ is an arbitrary constant.
Example 16: Show that the differential equation $x \cos x \left( \frac{y}{x} \right) \frac{dy}{dx} = y \cos \left( \frac{y}{x} \right) + x$ is homogeneous and solve it.
Answer:
Given:
The differential equation is $x \cos \left( \frac{y}{x} \right) \frac{dy}{dx} = y \cos \left( \frac{y}{x} \right) + x$.
To Show: The differential equation is homogeneous.
To Solve: Find the general solution of the differential equation.
Showing Homogeneity:
We can rewrite the given differential equation in the form $\frac{dy}{dx} = f(x, y)$:
$\frac{dy}{dx} = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}$
Let $f(x, y) = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}$.
To check for homogeneity, consider $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda y \cos \left( \frac{\lambda y}{\lambda x} \right) + \lambda x}{\lambda x \cos \left( \frac{\lambda y}{\lambda x} \right)}$
Assuming $\lambda \neq 0$, $\frac{\lambda y}{\lambda x} = \frac{y}{x}$. Substitute this:
$f(\lambda x, \lambda y) = \frac{\lambda y \cos \left( \frac{y}{x} \right) + \lambda x}{\lambda x \cos \left( \frac{y}{x} \right)}$
Factor out $\lambda$ from the numerator and denominator:
$f(\lambda x, \lambda y) = \frac{\lambda \left( y \cos \left( \frac{y}{x} \right) + x \right)}{\lambda \left( x \cos \left( \frac{y}{x} \right) \right)}$
Assuming $\lambda \neq 0$, cancel $\lambda$:
$f(\lambda x, \lambda y) = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}$
So, $f(\lambda x, \lambda y) = f(x, y)$.
This means $f(x, y)$ is a homogeneous function of degree 0.
Therefore, the given differential equation is a homogeneous differential equation.
Solving the Homogeneous Equation:
The differential equation is $\frac{dy}{dx} = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}$.
Use the substitution $y = vx$.
Differentiating $y = vx$ with respect to $x$, we get:
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:
$v + x \frac{dv}{dx} = \frac{vx \cos \left( \frac{vx}{x} \right) + x}{x \cos \left( \frac{vx}{x} \right)}$
Assuming $x \neq 0$, $\frac{vx}{x} = v$.
$v + x \frac{dv}{dx} = \frac{vx \cos(v) + x}{x \cos(v)}$
Factor out $x$ from the numerator:
$v + x \frac{dv}{dx} = \frac{x(v \cos(v) + 1)}{x \cos(v)}$
Cancel $x$ (assuming $x \neq 0$):
$v + x \frac{dv}{dx} = \frac{v \cos(v) + 1}{\cos(v)}$
Split the fraction on the right side:
$v + x \frac{dv}{dx} = \frac{v \cos(v)}{\cos(v)} + \frac{1}{\cos(v)}$
$v + x \frac{dv}{dx} = v + \frac{1}{\cos(v)}$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = \frac{1}{\cos(v)}$
This is a separable differential equation. Separate the variables $v$ and $x$:
$\cos(v) \, dv = \frac{dx}{x}$
Integrate both sides of the equation:
$\int\limits \cos(v) \, dv = \int\limits \frac{1}{x} \, dx$
Evaluating the integrals:
$\sin(v) = \ln|x| + C$
where $C$ is the arbitrary constant of integration.
Substitute back $v = \frac{y}{x}$:
$\sin\left(\frac{y}{x}\right) = \ln|x| + C$
The general solution of the given differential equation is:
$\mathbf{\sin\left(\frac{y}{x}\right) = \ln|x| + C}$
where $C$ is an arbitrary constant.
Example 17: Show that the differential equation $2y\; e^{\frac{x}{y}} dx + \left( y − 2x\; e^{\frac{x}{y}} \right) dy = 0$ is homogeneous and find its particular solution, given that, x = 0 when y = 1.
Answer:
Given:
The differential equation is $2y\; e^{\frac{x}{y}} dx + \left( y − 2x\; e^{\frac{x}{y}} \right) dy = 0$.
The initial condition is $x = 0$ when $y = 1$.
To Show: The differential equation is homogeneous.
To Find: The particular solution satisfying the given condition.
Showing Homogeneity:
The given differential equation is in the form $M(x, y) \, dx + N(x, y) \, dy = 0$, where $M(x, y) = 2y\; e^{\frac{x}{y}}$ and $N(x, y) = y − 2x\; e^{\frac{x}{y}}$.
Consider $M(\lambda x, \lambda y)$ and $N(\lambda x, \lambda y)$ for a non-zero constant $\lambda$:
$M(\lambda x, \lambda y) = 2(\lambda y)\; e^{\frac{\lambda x}{\lambda y}} = 2\lambda y\; e^{\frac{x}{y}} = \lambda (2y\; e^{\frac{x}{y}}) = \lambda M(x, y)$
$N(\lambda x, \lambda y) = (\lambda y) − 2(\lambda x)\; e^{\frac{\lambda x}{\lambda y}} = \lambda y − 2\lambda x\; e^{\frac{x}{y}} = \lambda (y − 2x\; e^{\frac{x}{y}}) = \lambda N(x, y)$
Since $M(\lambda x, \lambda y) = \lambda^1 M(x, y)$ and $N(\lambda x, \lambda y) = \lambda^1 N(x, y)$, both $M$ and $N$ are homogeneous functions of the same degree (degree 1).
Therefore, the given differential equation is a homogeneous differential equation.
Solving the Homogeneous Equation:
The differential equation is $2y\; e^{\frac{x}{y}} dx + \left( y − 2x\; e^{\frac{x}{y}} \right) dy = 0$.
Since the terms inside the exponential are of the form $\frac{x}{y}$, it is easier to use the substitution $x = vy$.
Differentiate $x = vy$ with respect to $y$:
$\frac{dx}{dy} = v + y \frac{dv}{dy}$
From the given differential equation, we can write $\frac{dx}{dy}$:
$2y\; e^{\frac{x}{y}} dx = - \left( y − 2x\; e^{\frac{x}{y}} \right) dy$
$\frac{dx}{dy} = \frac{- \left( y − 2x\; e^{\frac{x}{y}} \right)}{2y\; e^{\frac{x}{y}}} = \frac{2x\; e^{\frac{x}{y}} - y}{2y\; e^{\frac{x}{y}}}$
Now substitute $x = vy$ and $\frac{dx}{dy} = v + y \frac{dv}{dy}$:
$v + y \frac{dv}{dy} = \frac{2(vy)\; e^{\frac{vy}{y}} - y}{2y\; e^{\frac{vy}{y}}}$
Assuming $y \neq 0$, $\frac{vy}{y} = v$.
$v + y \frac{dv}{dy} = \frac{2vy\; e^v - y}{2y\; e^v}$
Factor out $y$ from the numerator:
$v + y \frac{dv}{dy} = \frac{y(2v\; e^v - 1)}{2y\; e^v}$
Assuming $y \neq 0$, cancel $y$:
$v + y \frac{dv}{dy} = \frac{2v\; e^v - 1}{2e^v}$
Split the fraction on the right side:
$v + y \frac{dv}{dy} = \frac{2v\; e^v}{2e^v} - \frac{1}{2e^v}$
$v + y \frac{dv}{dy} = v - \frac{1}{2} e^{-v}$
Subtract $v$ from both sides:
$y \frac{dv}{dy} = - \frac{1}{2} e^{-v}$
This is a separable differential equation. Separate the variables $v$ and $y$:
$\frac{dv}{e^{-v}} = - \frac{1}{2} \frac{dy}{y}$
$e^v \, dv = - \frac{1}{2} \frac{dy}{y}$
Integrate both sides:
$\int\limits e^v \, dv = \int\limits - \frac{1}{2} \frac{dy}{y}$
$\int\limits e^v \, dv = - \frac{1}{2} \int\limits \frac{1}{y} \, dy$
Evaluating the integrals:
$e^v = - \frac{1}{2} \ln|y| + C$
where $C$ is the arbitrary constant of integration.
Substitute back $v = \frac{x}{y}$:
$e^{\frac{x}{y}} = - \frac{1}{2} \ln|y| + C$
This is the general solution.
Applying the initial condition:
We are given that $x = 0$ when $y = 1$. Substitute these values into the general solution:
$e^{\frac{0}{1}} = - \frac{1}{2} \ln|1| + C$
$e^0 = - \frac{1}{2} \ln(1) + C$
We know that $e^0 = 1$ and $\ln(1) = 0$.
$1 = - \frac{1}{2} \cdot 0 + C$
$1 = 0 + C$
$C = 1$
Particular Solution:
Substitute the value of $C = 1$ back into the general solution:
$e^{\frac{x}{y}} = - \frac{1}{2} \ln|y| + 1$
Multiply by 2 to remove the fraction:
$2 e^{\frac{x}{y}} = - \ln|y| + 2$
Rearrange the terms:
$\ln|y| + 2 e^{\frac{x}{y}} = 2$
Since the initial condition is at $y=1 > 0$, we can write $\ln y$ instead of $\ln|y|$ for the particular solution in the neighborhood of this point.
$\ln y + 2 e^{\frac{x}{y}} = 2$
The particular solution satisfying the given condition is:
$\mathbf{\ln|y| + 2 e^{\frac{x}{y}} = 2}$
or for $y>0$:
$\mathbf{\ln y + 2 e^{\frac{x}{y}} = 2}$
Example 18: Show that the family of curves for which the slope of the tangent at any point (x, y) on it is $\frac{x^2 + y^2}{2xy}$ , is given by x2 – y2 = cx.
Answer:
Given:
The slope of the tangent at any point $(x, y)$ on a curve is given by $\frac{x^2 + y^2}{2xy}$.
To Show: The equation of the family of curves is $x^2 – y^2 = cx$.
Solution:
The slope of the tangent at any point $(x, y)$ on the curve is the derivative $\frac{dy}{dx}$.
So, the differential equation of the family of curves is:
$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$
We can check if this is a homogeneous differential equation. Let $f(x, y) = \frac{x^2 + y^2}{2xy}$.
Consider $f(\lambda x, \lambda y)$:
$f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{2(\lambda x)(\lambda y)} = \frac{\lambda^2 x^2 + \lambda^2 y^2}{2\lambda^2 xy} = \frac{\lambda^2 (x^2 + y^2)}{\lambda^2 (2xy)}$
Assuming $\lambda \neq 0$, cancel $\lambda^2$:
$f(\lambda x, \lambda y) = \frac{x^2 + y^2}{2xy} = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the function is homogeneous of degree 0. The differential equation is homogeneous.
To solve the homogeneous equation, use the substitution $y = vx$.
Differentiate $y = vx$ with respect to $x$:
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:
$v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)}$
$v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{2vx^2}$
Factor out $x^2$ from the numerator and denominator (assuming $x \neq 0$):
$v + x \frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2(2v)}$
$v + x \frac{dv}{dx} = \frac{1 + v^2}{2v}$
Separate the variables $v$ and $x$. Move $v$ to the right side:
$x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v$
Find a common denominator:
$x \frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$
Separate the variables $v$ and $x$:
$\frac{2v}{1 - v^2} dv = \frac{dx}{x}$
Integrate both sides:
$\int\limits \frac{2v}{1 - v^2} dv = \int\limits \frac{1}{x} dx$
For the left side integral, let $u = 1 - v^2$. Then $du = -2v \, dv$, so $2v \, dv = -du$.
$\int\limits \frac{-du}{u} = - \int\limits \frac{1}{u} du = -\ln|u| + C_1$
Substitute back $u = 1 - v^2$:
$\int\limits \frac{2v}{1 - v^2} dv = -\ln|1 - v^2| + C_1$
The right side integral is:
$\int\limits \frac{1}{x} dx = \ln|x| + C_2$
Equating the integrated results:
$-\ln|1 - v^2| + C_1 = \ln|x| + C_2$
Rearrange the terms:
$C_1 - C_2 = \ln|x| + \ln|1 - v^2|$
Let $C_{int} = C_1 - C_2$ be the arbitrary constant. Using logarithm properties:
$C_{int} = \ln|x(1 - v^2)|$
Exponentiate both sides:
$e^{C_{int}} = |x(1 - v^2)|$
Let $K = \pm e^{C_{int}}$. $K$ is a non-zero constant.
$K = x(1 - v^2)$
Substitute back $v = \frac{y}{x}$:
$K = x \left(1 - \left(\frac{y}{x}\right)^2\right)$
$K = x \left(1 - \frac{y^2}{x^2}\right)$
$K = x \left(\frac{x^2 - y^2}{x^2}\right)$
$K = \frac{x^2 - y^2}{x}$
Multiply by $x$:
$Kx = x^2 - y^2$
$x^2 - y^2 = Kx$
Since $K$ is an arbitrary non-zero constant, let's rename it $c$.
$x^2 - y^2 = cx$
Note that if $x=0$, the original differential equation is undefined. If $y=0$ but $x \neq 0$, $\frac{dy}{dx} = \frac{x^2}{0}$ which is undefined unless $x=0$. The equation $x^2 - y^2 = cx$ represents a family of curves. If $c=0$, $x^2-y^2=0 \implies y = \pm x$. These lines might need to be checked separately depending on the context, but the derivation leads to $x^2 - y^2 = cx$ where $c$ can be any real number to include potential singular solutions $y=\pm x$ if applicable (which would require careful analysis of the original equation at $y=\pm x$). However, the question asks to show that the family is given by $x^2 - y^2 = cx$, implying $c$ is the arbitrary constant from integration.
We have shown that the general solution of the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$ is $x^2 - y^2 = cx$, where $c$ is an arbitrary constant. This represents the family of curves satisfying the given condition.
Exercise 9.5
In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.
Question 1. $(x^2 + xy) dy = (x^2 + y^2) dx$
Answer:
Given:
The differential equation is $(x^2 + xy) dy = (x^2 + y^2) dx$.
To Show:
The given differential equation is homogeneous and to solve it.
Solution:
Rewrite the given differential equation in the form $\frac{dy}{dx} = f(x, y)$.
$\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$
Let $f(x, y) = \frac{x^2 + y^2}{x^2 + xy}$.
To check if it is a homogeneous differential equation, replace $x$ with $\lambda x$ and $y$ with $\lambda y$ in $f(x, y)$.
$f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{(\lambda x)^2 + (\lambda x)(\lambda y)}$
$f(\lambda x, \lambda y) = \frac{\lambda^2 x^2 + \lambda^2 y^2}{\lambda^2 x^2 + \lambda^2 xy}$
$f(\lambda x, \lambda y) = \frac{\lambda^2 (x^2 + y^2)}{\lambda^2 (x^2 + xy)}$
$f(\lambda x, \lambda y) = \frac{x^2 + y^2}{x^2 + xy}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
To solve the homogeneous equation, substitute $y = vx$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot \frac{d}{dx}(x) + x \cdot \frac{dv}{dx}$
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$.
$v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{x^2 + x(vx)}$
$v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{x^2 + v x^2}$
$v + x \frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2(1 + v)}$
$v + x \frac{dv}{dx} = \frac{1 + v^2}{1 + v}$
Now, separate the variables $v$ and $x$.
$x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v$
$x \frac{dv}{dx} = \frac{1 + v^2 - v(1 + v)}{1 + v}$
$x \frac{dv}{dx} = \frac{1 + v^2 - v - v^2}{1 + v}$
$x \frac{dv}{dx} = \frac{1 - v}{1 + v}$
$\frac{1 + v}{1 - v} dv = \frac{dx}{x}$
Integrate both sides:
$\int \frac{1 + v}{1 - v} dv = \int \frac{dx}{x}$
Consider the left integral. Rewrite the integrand $\frac{1 + v}{1 - v}$:
$\frac{1 + v}{1 - v} = \frac{-(1 - v) + 2}{1 - v} = -1 + \frac{2}{1 - v}$
So the integral becomes:
$\int \left(-1 + \frac{2}{1 - v}\right) dv = \int \frac{dx}{x}$
Integrating, we get:
$-v + 2 \int \frac{1}{1 - v} dv = \int \frac{dx}{x}$
$-v + 2 (-\ln|1 - v|) = \ln|x| + C$
$-v - 2\ln|1 - v| = \ln|x| + C$
Now, substitute back $v = \frac{y}{x}$.
$-\frac{y}{x} - 2\ln\left|1 - \frac{y}{x}\right| = \ln|x| + C$
$-\frac{y}{x} - 2\ln\left|\frac{x - y}{x}\right| = \ln|x| + C$
$-\frac{y}{x} - 2(\ln|x - y| - \ln|x|) = \ln|x| + C$
$-\frac{y}{x} - 2\ln|x - y| + 2\ln|x| = \ln|x| + C$
Combine the $\ln|x|$ terms:
$-\frac{y}{x} - 2\ln|x - y| + \ln|x| = C$
Rearrange the terms:
$\ln|x| - 2\ln|x - y| - \frac{y}{x} = C$
$\ln|x| - \ln((x - y)^2) = C + \frac{y}{x}$
$\ln\left|\frac{x}{(x - y)^2}\right| = C + \frac{y}{x}$
Exponentiate both sides:
$\left|\frac{x}{(x - y)^2}\right| = e^{C + y/x}$
$\frac{x}{(x - y)^2} = \pm e^C e^{y/x}$
Let $A = \pm e^C$, where $A$ is an arbitrary non-zero constant.
$\frac{x}{(x - y)^2} = A e^{y/x}$
The general solution is $(x - y)^2 = \frac{x}{A} e^{-y/x}$. Let $B = 1/A$, where $B$ is an arbitrary non-zero constant.
The general solution can be written as $(x - y)^2 = B x e^{-y/x}$.
Question 2. $y’ = \frac{x + y}{x}$
Answer:
Given:
The differential equation is $y' = \frac{x + y}{x}$.
This can be written as $\frac{dy}{dx} = \frac{x + y}{x}$.
To Show:
The given differential equation is homogeneous and to solve it.
Solution:
Let $f(x, y) = \frac{x + y}{x}$.
To check if it is a homogeneous differential equation, replace $x$ with $\lambda x$ and $y$ with $\lambda y$ in $f(x, y)$.
$f(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x}$
$f(\lambda x, \lambda y) = \frac{\lambda (x + y)}{\lambda x}$
$f(\lambda x, \lambda y) = \frac{x + y}{x}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
To solve the homogeneous equation, substitute $y = vx$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot \frac{d}{dx}(x) + x \cdot \frac{dv}{dx}$
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$.
$v + x \frac{dv}{dx} = \frac{x + vx}{x}$
$v + x \frac{dv}{dx} = \frac{x(1 + v)}{x}$
$v + x \frac{dv}{dx} = 1 + v$
Now, separate the variables $v$ and $x$.
$x \frac{dv}{dx} = 1 + v - v$
$x \frac{dv}{dx} = 1$
$dv = \frac{1}{x} dx$
Integrate both sides:
$\int dv = \int \frac{1}{x} dx$
Integrating, we get:
$v = \ln|x| + C$
Now, substitute back $v = \frac{y}{x}$.
$\frac{y}{x} = \ln|x| + C$
Multiply both sides by $x$ to solve for $y$:
$y = x (\ln|x| + C)$
The general solution is $y = x \ln|x| + Cx$.
Question 3. (x – y) dy – (x + y) dx = 0
Answer:
Given:
The differential equation is $(x – y) dy – (x + y) dx = 0$.
To Show:
The given differential equation is homogeneous and to solve it.
Solution:
Rewrite the given differential equation in the form $\frac{dy}{dx} = f(x, y)$.
$(x – y) dy = (x + y) dx$
$\frac{dy}{dx} = \frac{x + y}{x - y}$
Let $f(x, y) = \frac{x + y}{x - y}$.
To check if it is a homogeneous differential equation, replace $x$ with $\lambda x$ and $y$ with $\lambda y$ in $f(x, y)$.
$f(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y}$
$f(\lambda x, \lambda y) = \frac{\lambda (x + y)}{\lambda (x - y)}$
$f(\lambda x, \lambda y) = \frac{x + y}{x - y}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
To solve the homogeneous equation, substitute $y = vx$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot \frac{d}{dx}(x) + x \cdot \frac{dv}{dx}$
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$.
$v + x \frac{dv}{dx} = \frac{x + vx}{x - vx}$
$v + x \frac{dv}{dx} = \frac{x(1 + v)}{x(1 - v)}$
$v + x \frac{dv}{dx} = \frac{1 + v}{1 - v}$
Now, separate the variables $v$ and $x$.
$x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v$
$x \frac{dv}{dx} = \frac{1 + v - v(1 - v)}{1 - v}$
$x \frac{dv}{dx} = \frac{1 + v - v + v^2}{1 - v}$
$x \frac{dv}{dx} = \frac{1 + v^2}{1 - v}$
$\frac{1 - v}{1 + v^2} dv = \frac{dx}{x}$
Integrate both sides:
$\int \frac{1 - v}{1 + v^2} dv = \int \frac{dx}{x}$
$\int \left(\frac{1}{1 + v^2} - \frac{v}{1 + v^2}\right) dv = \int \frac{dx}{x}$
$\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \int \frac{dx}{x}$
The integrals are:
$\int \frac{1}{1 + v^2} dv = \arctan(v)$
For $\int \frac{v}{1 + v^2} dv$, let $u = 1 + v^2$, then $du = 2v dv$. So, $v dv = \frac{1}{2} du$.
$\int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(1 + v^2)$ (since $1+v^2 > 0$).
$\int \frac{dx}{x} = \ln|x| + C$
Substituting these back into the integrated equation:
$\arctan(v) - \frac{1}{2} \ln(1 + v^2) = \ln|x| + C$
Now, substitute back $v = \frac{y}{x}$.
$\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C$
$\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(\frac{x^2 + y^2}{x^2}\right) = \ln|x| + C$
$\arctan\left(\frac{y}{x}\right) - \frac{1}{2} (\ln(x^2 + y^2) - \ln(x^2)) = \ln|x| + C$
$\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2 + y^2) + \frac{1}{2} (2\ln|x|) = \ln|x| + C$
$\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2 + y^2) + \ln|x| = \ln|x| + C$
Cancel $\ln|x|$ from both sides:
$\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2 + y^2) = C$
Multiply by 2:
$2\arctan\left(\frac{y}{x}\right) - \ln(x^2 + y^2) = 2C$
Let $C_1 = 2C$ be a new arbitrary constant.
$2\arctan\left(\frac{y}{x}\right) - \ln(x^2 + y^2) = C_1$
The general solution is $2\arctan\left(\frac{y}{x}\right) - \ln(x^2 + y^2) = C_1$.
Question 4. $(x^2 – y^2) dx + 2xy\; dy = 0$
Answer:
Given:
The differential equation is $(x^2 – y^2) dx + 2xy\; dy = 0$.
To Show:
The given differential equation is homogeneous and to solve it.
Solution:
Rewrite the given differential equation in the form $\frac{dy}{dx} = f(x, y)$.
$(x^2 – y^2) dx + 2xy\; dy = 0$
$2xy\; dy = -(x^2 – y^2) dx$
$2xy\; dy = (y^2 - x^2) dx$
$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$
Let $f(x, y) = \frac{y^2 - x^2}{2xy}$.
To check if it is a homogeneous differential equation, replace $x$ with $\lambda x$ and $y$ with $\lambda y$ in $f(x, y)$.
$f(\lambda x, \lambda y) = \frac{(\lambda y)^2 - (\lambda x)^2}{2(\lambda x)(\lambda y)}$
$f(\lambda x, \lambda y) = \frac{\lambda^2 y^2 - \lambda^2 x^2}{2\lambda^2 xy}$
$f(\lambda x, \lambda y) = \frac{\lambda^2 (y^2 - x^2)}{\lambda^2 (2xy)}$
$f(\lambda x, \lambda y) = \frac{y^2 - x^2}{2xy}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
To solve the homogeneous equation, substitute $y = vx$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot \frac{d}{dx}(x) + x \cdot \frac{dv}{dx}$
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.
$v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)}$
$v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2vx^2}$
$v + x \frac{dv}{dx} = \frac{x^2(v^2 - 1)}{x^2(2v)}$
$v + x \frac{dv}{dx} = \frac{v^2 - 1}{2v}$
Now, separate the variables $v$ and $x$.
$x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v$
$x \frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v}$
$x \frac{dv}{dx} = \frac{-v^2 - 1}{2v}$
$x \frac{dv}{dx} = -\frac{v^2 + 1}{2v}$
$\frac{2v}{v^2 + 1} dv = -\frac{1}{x} dx$
Integrate both sides:
$\int \frac{2v}{v^2 + 1} dv = \int -\frac{1}{x} dx$
Integrating the left side:
Let $u = v^2 + 1$, so $du = 2v dv$.
$\int \frac{1}{u} du = \ln|u| = \ln(v^2 + 1)$ (since $v^2+1 > 0$).
Integrating the right side:
$\int -\frac{1}{x} dx = -\ln|x| + C$, where $C$ is the constant of integration.
So, we have:
$\ln(v^2 + 1) = -\ln|x| + C$
Rearrange the terms:
$\ln(v^2 + 1) + \ln|x| = C$
Using logarithm properties $\ln a + \ln b = \ln(ab)$:
$\ln((v^2 + 1)|x|) = C$
Exponentiate both sides:
$(v^2 + 1)|x| = e^C$
Substitute back $v = \frac{y}{x}$.
$\left(\left(\frac{y}{x}\right)^2 + 1\right)|x| = e^C$
$\left(\frac{y^2}{x^2} + 1\right)|x| = e^C$
$\left(\frac{y^2 + x^2}{x^2}\right)|x| = e^C$
Since $x^2 = |x|^2$, we have:
$\frac{y^2 + x^2}{|x|^2}|x| = e^C$
$\frac{y^2 + x^2}{|x|} = e^C$
Multiply both sides by $|x|$:
$y^2 + x^2 = e^C |x|$
Let $K = e^C$. Since $e^C$ is always positive, $K > 0$.
$x^2 + y^2 = K |x|$
This is a form of the general solution. However, the constant can absorb the sign of $x$. Consider the cases $x > 0$ and $x < 0$ separately during integration, or by allowing the constant to be any non-zero value.
The general solution is commonly written in the form:
$x^2 + y^2 = Cx$
where $C$ is an arbitrary constant. This form covers the case where $x > 0$ (with $C > 0$) and the case where $x < 0$ (with $C < 0$), as well as the trivial solution $(0,0)$ when $C=0$ (although the derivation using $y=vx$ does not strictly apply at $(0,0)$ where $\frac{dy}{dx}$ is undefined, the original equation is satisfied). To show this equivalence from $x^2+y^2 = K|x|$, if $x>0$, $x^2+y^2 = Kx$, letting $C=K$ ($C>0$). If $x<0$, $x^2+y^2 = K(-x)$, letting $C=-K$ ($C<0$).
Thus, the general solution is $x^2 + y^2 = Cx$, where $C$ is an arbitrary constant.
Question 5. $x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy$
Answer:
Given:
The differential equation is $x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy$.
To Show:
The given differential equation is homogeneous and to solve it.
Solution:
Rewrite the given differential equation in the form $\frac{dy}{dx} = f(x, y)$.
$\frac{dy}{dx} = \frac{x^2 - 2y^2 + xy}{x^2}$
$\frac{dy}{dx} = \frac{x^2}{x^2} - \frac{2y^2}{x^2} + \frac{xy}{x^2}$
$\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$
Let $f(x, y) = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$.
To check if it is a homogeneous differential equation, replace $x$ with $\lambda x$ and $y$ with $\lambda y$ in $f(x, y)$.
$f(\lambda x, \lambda y) = 1 - 2\left(\frac{\lambda y}{\lambda x}\right)^2 + \frac{\lambda y}{\lambda x}$
$f(\lambda x, \lambda y) = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
To solve the homogeneous equation, substitute $y = vx$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot \frac{d}{dx}(x) + x \cdot \frac{dv}{dx}$
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$.
$v + x \frac{dv}{dx} = 1 - 2v^2 + v$
Now, separate the variables $v$ and $x$.
$x \frac{dv}{dx} = 1 - 2v^2$
$\frac{dv}{1 - 2v^2} = \frac{dx}{x}$
Integrate both sides:
$\int \frac{dv}{1 - 2v^2} = \int \frac{dx}{x}$
Consider the integral on the left side: $\int \frac{dv}{1 - 2v^2} = \int \frac{dv}{1^2 - (\sqrt{2}v)^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right|$. Here $a=1$ and the variable is $\sqrt{2}v$. Let $u = \sqrt{2}v$, so $du = \sqrt{2} dv$, which means $dv = \frac{du}{\sqrt{2}}$.
$\int \frac{1}{1 - u^2} \frac{du}{\sqrt{2}} = \frac{1}{\sqrt{2}} \int \frac{du}{1 - u^2} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2(1)} \ln\left|\frac{1+u}{1-u}\right| + C_1$
$= \frac{1}{2\sqrt{2}} \ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| + C_1$
The integral on the right side is $\int \frac{dx}{x} = \ln|x| + C_2$.
Equating the results of the integration:
$\frac{1}{2\sqrt{2}} \ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \ln|x| + C$, where $C = C_2 - C_1$.
Multiply by $2\sqrt{2}$:
$\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = 2\sqrt{2} (\ln|x| + C)$
$\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = 2\sqrt{2} \ln|x| + 2\sqrt{2}C$
Let $C_1 = 2\sqrt{2}C$ be a new arbitrary constant.
$\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \ln(|x|^{2\sqrt{2}}) + C_1$
$\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| - \ln(|x|^{2\sqrt{2}}) = C_1$
Using logarithm property $\ln a - \ln b = \ln(a/b)$:
$\ln\left|\frac{1+\sqrt{2}v}{(1-\sqrt{2}v)|x|^{2\sqrt{2}}}\right| = C_1$
Exponentiate both sides:
$\left|\frac{1+\sqrt{2}v}{(1-\sqrt{2}v)|x|^{2\sqrt{2}}}\right| = e^{C_1}$
$\frac{1+\sqrt{2}v}{(1-\sqrt{2}v)|x|^{2\sqrt{2}}} = \pm e^{C_1}$
Let $K = \pm e^{C_1}$, where $K$ is an arbitrary non-zero constant.
$\frac{1+\sqrt{2}v}{(1-\sqrt{2}v)|x|^{2\sqrt{2}}} = K$
Substitute back $v = \frac{y}{x}$.
$\frac{1+\sqrt{2}(y/x)}{(1-\sqrt{2}(y/x))|x|^{2\sqrt{2}}} = K$
$\frac{(x+\sqrt{2}y)/x}{(x-\sqrt{2}y)/x} \cdot \frac{1}{|x|^{2\sqrt{2}}} = K$
$\frac{x+\sqrt{2}y}{x-\sqrt{2}y} \cdot \frac{1}{|x|^{2\sqrt{2}}} = K$
$\frac{x+\sqrt{2}y}{(x-\sqrt{2}y)|x|^{2\sqrt{2}}} = K$
This can be rearranged to express the general solution:
$x+\sqrt{2}y = K |x|^{2\sqrt{2}} (x-\sqrt{2}y)$
The constant $K$ can absorb the absolute value sign, provided we specify that $K$ is non-zero. So the solution can also be written as:
$x+\sqrt{2}y = C x^{2\sqrt{2}} (x-\sqrt{2}y)$, where $C$ is an arbitrary constant (assuming $x > 0$ or incorporating the sign into $C$). A more rigorous form keeping absolute values is preferred unless specified otherwise.
The general solution is $(x+\sqrt{2}y) = C (x-\sqrt{2}y) |x|^{2\sqrt{2}}$, where $C$ is an arbitrary constant.
Question 6. $x \;dy \;–\; y \;dx = \sqrt{x^2 + y^2}\; dx$
Answer:
Given Differential Equation:
$x \;dy \;–\; y \;dx = \sqrt{x^2 + y^2}\; dx$
Rearrange the equation to find $\frac{dy}{dx}$:
$x \;dy = (y + \sqrt{x^2 + y^2})\; dx$
$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$
Let $f(x, y) = \frac{y + \sqrt{x^2 + y^2}}{x}$. We check if it is a homogeneous equation by evaluating $f(\lambda x, \lambda y)$:
$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^2 + (\lambda y)^2}}{\lambda x}$
$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^2 x^2 + \lambda^2 y^2}}{\lambda x}$
$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^2(x^2 + y^2)}}{\lambda x}$
$f(\lambda x, \lambda y) = \frac{\lambda y + |\lambda|\sqrt{x^2 + y^2}}{\lambda x}$
For $\lambda > 0$, $|\lambda| = \lambda$.
$f(\lambda x, \lambda y) = \frac{\lambda y + \lambda\sqrt{x^2 + y^2}}{\lambda x} = \frac{\lambda (y + \sqrt{x^2 + y^2})}{\lambda x} = \frac{y + \sqrt{x^2 + y^2}}{x} = f(x, y)$
Thus, the given differential equation is homogeneous.
Solution Method:
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} = v + x \frac{dv}{dx}$
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$:
$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + (vx)^2}}{x}$
$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2 x^2}}{x}$
$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2(1 + v^2)}}{x}$
$v + x \frac{dv}{dx} = \frac{vx + |x|\sqrt{1 + v^2}}{x}$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = \frac{vx + |x|\sqrt{1 + v^2}}{x} - v$
$x \frac{dv}{dx} = v + \frac{|x|\sqrt{1 + v^2}}{x} - v$
$x \frac{dv}{dx} = \frac{|x|}{x}\sqrt{1 + v^2}$
This is a separable differential equation. We separate the variables $v$ and $x$:
$\frac{dv}{\sqrt{1 + v^2}} = \frac{|x|}{x} \frac{dx}{x} = \frac{|x|}{x^2} dx$
Integrate both sides:
$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{|x|}{x^2} dx$
The integral on the left is a standard form: $\int \frac{du}{\sqrt{a^2+u^2}} = \ln|u + \sqrt{a^2+u^2}|$. Here $a=1, u=v$.
The integral on the right depends on the sign of $x$. Note that $\frac{|x|}{x^2} = \frac{|x|}{|x|^2} = \frac{1}{|x|}$.
So, $\int \frac{|x|}{x^2} dx = \int \frac{1}{|x|} dx$. This integral is $\ln|x|$.
Thus, the integrated equation is:
$\ln|v + \sqrt{1 + v^2}| = \ln|x| + C_1$
where $C_1$ is the constant of integration.
$\ln|v + \sqrt{1 + v^2}| - \ln|x| = C_1$
$\ln\left|\frac{v + \sqrt{1 + v^2}}{x}\right| = C_1$
Exponentiating both sides:
$\left|\frac{v + \sqrt{1 + v^2}}{x}\right| = e^{C_1}$
Since $v + \sqrt{1 + v^2} > 0$ (because $\sqrt{1+v^2} > |v|$), the absolute value on the left is $\frac{v + \sqrt{1 + v^2}}{|x|}$.
So, $\frac{v + \sqrt{1 + v^2}}{|x|} = e^{C_1}$
Let $A = e^{C_1}$, where $A > 0$.
$v + \sqrt{1 + v^2} = A|x|$
Substitute back $v = \frac{y}{x}$:
$\frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = A|x|$
$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = A|x|$
$\frac{y}{x} + \sqrt{\frac{x^2 + y^2}{x^2}} = A|x|$
$\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{|x|} = A|x|$
Multiply both sides by $|x|$:
$\frac{y|x|}{x} + \sqrt{x^2 + y^2} = A|x|^2$
$\frac{y|x|}{x} + \sqrt{x^2 + y^2} = Ax^2$
We consider the cases based on the sign of $x$:
Case 1: $x > 0$. Then $|x|=x$.
The equation becomes $\frac{yx}{x} + \sqrt{x^2 + y^2} = Ax^2$:
$y + \sqrt{x^2 + y^2} = Ax^2$
Here $A > 0$. Rearrange to isolate the square root:
$\sqrt{x^2 + y^2} = Ax^2 - y$
For this equation to hold, we must have $Ax^2 - y \geq 0$. Squaring both sides:
$x^2 + y^2 = (Ax^2 - y)^2 = (Ax^2)^2 - 2(Ax^2)y + y^2$
$x^2 + y^2 = A^2x^4 - 2Ax^2y + y^2$
$x^2 = A^2x^4 - 2Ax^2y$
Since $x > 0$, $x \neq 0$. We can divide by $x^2$:
$1 = A^2x^2 - 2Ay$
Rearrange to solve for $y$:
$2Ay = A^2x^2 - 1$
Since $A > 0$, $A \neq 0$. Divide by $2A$:
$y = \frac{A^2x^2}{2A} - \frac{1}{2A} = \frac{A}{2}x^2 - \frac{1}{2A}$
Let $C = \frac{A}{2}$. Since $A > 0$, $C > 0$. The solution for $x > 0$ is of the form:
$y = Cx^2 - \frac{1}{4C}$
where $C$ is a positive constant.
We verify the condition $Ax^2 - y \geq 0$: $2Cx^2 - (Cx^2 - \frac{1}{4C}) = Cx^2 + \frac{1}{4C}$. Since $C>0$, this is $\geq 0$ for all $x$. The solution is valid for $x>0$.
Case 2: $x < 0$. Then $|x|=-x$.
The equation becomes $\frac{y(-x)}{x} + \sqrt{x^2 + y^2} = Ax^2$:
$-y + \sqrt{x^2 + y^2} = Ax^2$
Here $A > 0$. Rearrange to isolate the square root:
$\sqrt{x^2 + y^2} = Ax^2 + y$
For this equation to hold, we must have $Ax^2 + y \geq 0$. Squaring both sides:
$x^2 + y^2 = (Ax^2 + y)^2 = (Ax^2)^2 + 2(Ax^2)y + y^2$
$x^2 + y^2 = A^2x^4 + 2Ax^2y + y^2$
$x^2 = A^2x^4 + 2Ax^2y$
Since $x < 0$, $x \neq 0$. We can divide by $x^2$:
$1 = A^2x^2 + 2Ay$
Rearrange to solve for $y$:
$2Ay = 1 - A^2x^2$
Since $A > 0$, $A \neq 0$. Divide by $2A$:
$y = \frac{1}{2A} - \frac{A^2x^2}{2A} = \frac{1}{2A} - \frac{A}{2}x^2$
Let $C = -\frac{A}{2}$. Since $A > 0$, $C < 0$. The solution for $x < 0$ is of the form:
$y = -Cx^2 - \frac{1}{4(-C)} = -Cx^2 + \frac{1}{4C}$
Let's use the same form as in Case 1. If $y = Cx^2 - \frac{1}{4C}$, then $\frac{1}{2A} - \frac{A}{2}x^2 = Cx^2 - \frac{1}{4C}$. Since $C = -A/2$, $A = -2C$.
$y = \frac{1}{2(-2C)} - \frac{-2C}{2}x^2 = -\frac{1}{4C} + Cx^2 = Cx^2 - \frac{1}{4C}$
where $C$ is a negative constant.
We verify the condition $Ax^2 + y \geq 0$: $(-2C)x^2 + (Cx^2 - \frac{1}{4C}) = -Cx^2 - \frac{1}{4C} = (-C)x^2 - \frac{1}{4C}$. Since $C<0$, $-C>0$. This is $\geq 0$ for all $x$. The solution is valid for $x<0$.
General Solution:
Combining Case 1 and Case 2, the general solution to the differential equation is the family of parabolas given by:
$y = Cx^2 - \frac{1}{4C}$
where $C$ is a non-zero constant ($C > 0$ for the solution to be valid for $x > 0$, and $C < 0$ for the solution to be valid for $x < 0$).
Singular Solutions:
We examine points or curves where steps in the derivation (like division by $x$) might not be valid.
Consider the original equation at $x=0$: $0 \cdot dy - y \cdot dx = \sqrt{0^2 + y^2} \, dx$
$-y \, dx = \sqrt{y^2} \, dx$
$-y \, dx = |y| \, dx$
This equation holds if and only if $-y = |y|$. This condition is true when $y \leq 0$.
Therefore, the line segment on the y-axis defined by $x=0$ and $y \leq 0$ is a solution to the differential equation.
Consider $y=0$. Substituting $y=0$ and $dy=0$ into the original equation:
$x \cdot 0 - 0 \cdot dx = \sqrt{x^2 + 0^2} \, dx$
$0 = \sqrt{x^2} \, dx$
$0 = |x| \, dx$
This holds if and only if $|x|=0$, which means $x=0$. So, the point $(0,0)$ satisfies the equation. The point $(0,0)$ is included in the singular solution set $x=0, y \leq 0$.
The general solution $y = Cx^2 - \frac{1}{4C}$ does not include the point $(0,0)$ as $y(0) = -\frac{1}{4C} \neq 0$.
Summary of Solutions:
The solution consists of a family of parabolas given by $y = Cx^2 - \frac{1}{4C}$ where $C$ is a non-zero constant.
For $C > 0$, the solution curve is valid for $x > 0$.
For $C < 0$, the solution curve is valid for $x < 0$.
Additionally, there is a singular solution given by the line segment $x = 0$, $y \leq 0$.
Question 7. $\left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y \;dx = \left\{ y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right\} x \;dy$
Answer:
Given Differential Equation:
$\left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y \;dx = \left\{ y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right\} x \;dy$
Rearrange and Check Homogeneity:
We first rearrange the equation to express $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y}{\left\{ y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right\} x}$
$\frac{dy}{dx} = \frac{y}{x} \cdot \frac{x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right)}{y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right)}$
Let $F(x, y) = \frac{y}{x} \cdot \frac{x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right)}{y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right)}$.
To check for homogeneity, we evaluate $F(\lambda x, \lambda y)$ for $\lambda \neq 0$:
$F(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} \cdot \frac{\lambda x \cos \left( \frac{\lambda y}{\lambda x} \right) + \lambda y \sin \left( \frac{\lambda y}{\lambda x} \right)}{\lambda y \sin \left( \frac{\lambda y}{\lambda x} \right) − \lambda x \cos \left( \frac{\lambda y}{\lambda x} \right)}$
$F(\lambda x, \lambda y) = \frac{y}{x} \cdot \frac{\lambda \left( x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right)}{\lambda \left( y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right)}$
$F(\lambda x, \lambda y) = \frac{y}{x} \cdot \frac{x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right)}{y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right)} = F(x, y)$
Since $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$, the given differential equation is homogeneous.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:
$v + x \frac{dv}{dx} = \frac{vx}{x} \cdot \frac{x \cos \left( \frac{vx}{x} \right) + vx \sin \left( \frac{vx}{x} \right)}{vx \sin \left( \frac{vx}{x} \right) − x \cos \left( \frac{vx}{x} \right)}$
$v + x \frac{dv}{dx} = v \cdot \frac{x \cos(v) + vx \sin(v)}{vx \sin(v) − x \cos(v)}$
$v + x \frac{dv}{dx} = v \cdot \frac{x(\cos(v) + v \sin(v))}{x(v \sin(v) − \cos(v))}$
For $x \neq 0$, this simplifies to:
$v + x \frac{dv}{dx} = v \frac{\cos(v) + v \sin(v)}{v \sin(v) − \cos(v)}$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = v \frac{\cos(v) + v \sin(v)}{v \sin(v) − \cos(v)} - v$
$x \frac{dv}{dx} = v \left( \frac{\cos(v) + v \sin(v)}{v \sin(v) − \cos(v)} - 1 \right)$
$x \frac{dv}{dx} = v \left( \frac{\cos(v) + v \sin(v) - (v \sin(v) − \cos(v))}{v \sin(v) − \cos(v)} \right)$
$x \frac{dv}{dx} = v \left( \frac{\cos(v) + v \sin(v) - v \sin(v) + \cos(v)}{v \sin(v) − \cos(v)} \right)$
$x \frac{dv}{dx} = v \left( \frac{2 \cos(v)}{v \sin(v) − \cos(v)} \right)$
Separation of Variables and Integration:
Assuming $x \neq 0$ and $v \cos(v) \neq 0$ and $v \sin(v) - \cos(v) \neq 0$, we separate the variables:
$\frac{v \sin(v) − \cos(v)}{v \cos(v)} \;dv = \frac{2}{x} \;dx$
$\left( \frac{v \sin(v)}{v \cos(v)} - \frac{\cos(v)}{v \cos(v)} \right) \;dv = \frac{2}{x} \;dx$
$\left( \frac{\sin(v)}{\cos(v)} - \frac{1}{v} \right) \;dv = \frac{2}{x} \;dx$
$\left( \tan(v) - \frac{1}{v} \right) \;dv = \frac{2}{x} \;dx$
Integrate both sides:
$\int \left( \tan(v) - \frac{1}{v} \right) \;dv = \int \frac{2}{x} \;dx$
$\int \tan(v) \;dv - \int \frac{1}{v} \;dv = 2 \int \frac{1}{x} \;dx$
$-\ln|\cos(v)| - \ln|v| = 2 \ln|x| + C_1$
where $C_1$ is the constant of integration.
$-(\ln|\cos(v)| + \ln|v|) = \ln(x^2) + C_1$
$-\ln|v \cos(v)| = \ln(x^2) + C_1$
$\ln|v \cos(v)| = -\ln(x^2) - C_1$
$\ln|v \cos(v)| = \ln(x^{-2}) - C_1$
$\ln|v \cos(v)| - \ln(x^{-2}) = -C_1$
$\ln\left|\frac{v \cos(v)}{x^{-2}}\right| = -C_1$
$\ln|v x^2 \cos(v)| = -C_1$
Exponentiating both sides:
$|v x^2 \cos(v)| = e^{-C_1}$
$v x^2 \cos(v) = \pm e^{-C_1}$
Let $K = \pm e^{-C_1}$. Since $e^{-C_1}$ is a positive constant, $K$ can be any non-zero constant.
$v x^2 \cos(v) = K$
Back-substitution and General Solution:
Substitute back $v = \frac{y}{x}$:
$\frac{y}{x} x^2 \cos\left(\frac{y}{x}\right) = K$
$y x \cos\left(\frac{y}{x}\right) = K$
This is the general solution, where $K$ is a non-zero constant.
Consideration of Singular Cases:
During the separation of variables, we assumed $x \neq 0$, $v \cos(v) \neq 0$, and $v \sin(v) - \cos(v) \neq 0$.
1. $x=0$: The original equation is not defined for $x=0$ due to the terms $\frac{y}{x}$.
2. $y=0$: Substituting $y=0$ and $dy=0$ into the original equation gives:
$\left\{ x \cos \left( 0 \right) + 0 \sin \left( 0 \right) \right\} \cdot 0 \;dx = \left\{ 0 \sin \left( 0 \right) − x \cos \left( 0 \right) \right\} x \cdot 0$
$\{x \cdot 1 + 0\} \cdot 0 = \{0 - x \cdot 1\} \cdot 0$
$0 = 0$
Thus, $y=0$ is a solution.
Let's see if $y=0$ is included in the general solution $y x \cos\left(\frac{y}{x}\right) = K$. If $y=0$, the equation becomes $0 \cdot x \cos(0/x) = K$, which is $0 = K$. This implies that $y=0$ is the particular solution corresponding to $K=0$. So the general solution form includes $y=0$ if we allow $K=0$.
3. $v \cos(v) = 0$: This corresponds to $y/x \cdot \cos(y/x) = 0$. This means either $y=0$ (already covered) or $\cos(y/x) = 0$. If $\cos(y/x) = 0$, then $y/x = \frac{\pi}{2} + n\pi$ for integer $n$. This gives $y = (\frac{\pi}{2} + n\pi)x$. Let $m = \frac{\pi}{2} + n\pi$. So $y=mx$. For these lines, $y/x = m$, so $\cos(y/x)=0$. Substituting into the general solution $y x \cos(y/x) = K$ gives $y x \cdot 0 = K$, so $0 = K$. These lines are also part of the solution when $K=0$.
4. $v \sin(v) - \cos(v) = 0$: This is equivalent to $\tan(v) = 1/v$. These correspond to constant values of $v$ where the denominator in the separated equation for $v$ is zero. If $v$ is a constant $v_0$ such that $v_0 \sin(v_0) - \cos(v_0) = 0$, then $x \frac{dv_0}{dx} = x \cdot 0 = 0$. The right side is $v_0 \left( \frac{2 \cos(v_0)}{v_0 \sin(v_0) − \cos(v_0)} \right)$, which is undefined. So, solutions cannot cross curves where $v \sin(v) - \cos(v) = 0$. However, solutions might exist *on* such curves if they are also singular points of the original ODE. Let's check if $y=v_0x$ where $v_0 \tan(v_0)=1$ is a solution to the original ODE by direct substitution. $\frac{dy}{dx}=v_0$. We tested $y=mx$ and found it is a solution iff $\cos(m)=0$. Since $v_0 \tan(v_0) = 1$ implies $v_0 \neq 0$ and $\tan(v_0)$ is defined, $\cos(v_0) \neq 0$. Thus, lines $y=v_0 x$ with $v_0 \tan(v_0)=1$ are not solutions.
Final Answer:
The general solution to the given differential equation is:
$y x \cos\left(\frac{y}{x}\right) = C$
where $C$ is an arbitrary constant (including zero).
The case $C=0$ covers the solutions $y=0$ and the family of lines $y = (\frac{\pi}{2} + n\pi)x$ for integer $n$, which are potential singular solutions identified during the process.
Question 8. $x \frac{dy}{dx} - y + x \sin \left( \frac{y}{x} \right) = 0$
Answer:
Given Differential Equation:
$x \frac{dy}{dx} - y + x \sin \left( \frac{y}{x} \right) = 0$
Rearrange and Check Homogeneity:
Rearrange the equation to express $\frac{dy}{dx}$:
$x \frac{dy}{dx} = y - x \sin \left( \frac{y}{x} \right)$
Divide by $x$ (assuming $x \neq 0$):
$\frac{dy}{dx} = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$
Let $f(x, y) = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \sin \left( \frac{\lambda y}{\lambda x} \right)$
$f(\lambda x, \lambda y) = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous of degree 0.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} = v + x \frac{dv}{dx}$
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$:
$v + x \frac{dv}{dx} = \frac{vx}{x} - \sin \left( \frac{vx}{x} \right)$
$v + x \frac{dv}{dx} = v - \sin(v)$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = - \sin(v)$
Separation of Variables and Integration:
Assuming $\sin(v) \neq 0$ and $x \neq 0$, we can separate the variables $v$ and $x$:
$\frac{dv}{\sin(v)} = - \frac{dx}{x}$
We can rewrite $\frac{1}{\sin(v)}$ as $\text{cosec}(v)$:
$\text{cosec}(v) \; dv = - \frac{1}{x} \; dx$
Integrate both sides:
$\int \text{cosec}(v) \; dv = \int - \frac{1}{x} \; dx$
The integral of $\text{cosec}(v)$ is $\ln|\text{cosec}(v) - \cot(v)|$. The integral of $- \frac{1}{x}$ is $-\ln|x|$.
$\ln|\text{cosec}(v) - \cot(v)| = -\ln|x| + \ln|C|$
where $\ln|C|$ is the constant of integration, and $C$ is a non-zero constant from the property of logarithms.
Rearrange the terms:
$\ln|\text{cosec}(v) - \cot(v)| + \ln|x| = \ln|C|$
Using the logarithm property $\ln A + \ln B = \ln(AB)$:
$\ln\left|x (\text{cosec}(v) - \cot(v))\right| = \ln|C|$
Exponentiate both sides:
$\left|x (\text{cosec}(v) - \cot(v))\right| = |C|$
$x (\text{cosec}(v) - \cot(v)) = \pm C$
Let the arbitrary constant be $K = \pm C$. Since $C$ was non-zero, $K$ is initially a non-zero constant. We will check the $K=0$ case later.
$x \left( \frac{1}{\sin(v)} - \frac{\cos(v)}{\sin(v)} \right) = K$
$x \frac{1 - \cos(v)}{\sin(v)} = K$
Using Trigonometric Identity:
We use the half-angle identities: $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$.
So, $\frac{1 - \cos(v)}{\sin(v)} = \frac{2 \sin^2(v/2)}{2 \sin(v/2) \cos(v/2)}$. Assuming $\sin(v/2) \neq 0$ and $\cos(v/2) \neq 0$, this simplifies to $\frac{\sin(v/2)}{\cos(v/2)} = \tan(v/2)$.
The equation becomes:
$x \tan\left(\frac{v}{2}\right) = K$
Back-substitution:
Substitute back $v = \frac{y}{x}$ into the equation:
$x \tan\left(\frac{y}{2x}\right) = K$
This is the general solution, where $K$ is an arbitrary constant.
Singular Solutions:
In our derivation, we made assumptions $x \neq 0$ and $\sin(v) = \sin(y/x) \neq 0$. Let's examine these cases.
The original differential equation is not defined at $x=0$ due to the terms $\frac{dy}{dx}$ and $\frac{y}{x}$.
Consider the case where $\sin(y/x) = 0$. This happens when $\frac{y}{x} = n\pi$ for any integer $n$ ($n = 0, \pm 1, \pm 2, ...$). This implies $y = n\pi x$.
Let's check if $y = n\pi x$ is a solution to the original equation $x \frac{dy}{dx} - y + x \sin \left( \frac{y}{x} \right) = 0$.
If $y = n\pi x$, then the derivative $\frac{dy}{dx} = n\pi$.
Substituting these into the equation:
$x (n\pi) - (n\pi x) + x \sin(n\pi) = 0$
$n\pi x - n\pi x + x \cdot 0 = 0$
$0 = 0$
This identity holds for all $x$. Therefore, the lines $y = n\pi x$ for any integer $n$ are solutions to the differential equation.
Now let's see if these lines are included in the general solution $x \tan\left(\frac{y}{2x}\right) = K$.
Substitute $y = n\pi x$ into the general solution:
$x \tan\left(\frac{n\pi x}{2x}\right) = K$
$x \tan\left(\frac{n\pi}{2}\right) = K$
If $n$ is an even integer, say $n=2m$ for some integer $m$, then $\frac{n\pi}{2} = \frac{2m\pi}{2} = m\pi$. $\tan(m\pi) = 0$ for any integer $m$.
So, $x \cdot 0 = K$, which implies $K = 0$. This means the solutions $y = 2m\pi x$ (even multiples of $\pi x$) are included in the general solution family when the constant $K=0$.
If $n$ is an odd integer, say $n=2m+1$ for some integer $m$, then $\frac{n\pi}{2} = \frac{(2m+1)\pi}{2}$. $\tan\left(\frac{(2m+1)\pi}{2}\right)$ is undefined (approaches $\pm \infty$). For the product $x \tan\left(\frac{(2m+1)\pi}{2}\right)$ to be a finite constant $K$, it must be that the expression is undefined, which doesn't lead to a finite $K$ unless $x=0$ (which is not in the domain of $\frac{dy}{dx}$). Therefore, the lines $y = (2m+1)\pi x$ (odd multiples of $\pi x$) are not included in the general solution $x \tan\left(\frac{y}{2x}\right) = K$ for $x \neq 0$. These are singular solutions.
Final Answer:
The general solution to the given differential equation is the family of curves given by:
$x \tan\left(\frac{y}{2x}\right) = C$
where $C$ is an arbitrary constant. This family includes the solutions $y = 2k\pi x$ for any integer $k$, which correspond to $C=0$.
Additionally, there are singular solutions corresponding to the lines $y = (2k+1)\pi x$ for any integer $k$, which are not contained within the general solution family obtained by the method of separation of variables.
Question 9. $y \;dx + x \log \left( \frac{y}{x} \right) dy - 2x \;dy = 0$
Answer:
Given Differential Equation:
$y \;dx + x \log \left( \frac{y}{x} \right) dy - 2x \;dy = 0$
Rearrange and Check Homogeneity:
We can rewrite the equation as:
$y \;dx + \left( x \log \left( \frac{y}{x} \right) - 2x \right) dy = 0$
$y \;dx = - \left( x \log \left( \frac{y}{x} \right) - 2x \right) dy$
$y \;dx = \left( 2x - x \log \left( \frac{y}{x} \right) \right) dy$
$y \;dx = x \left( 2 - \log \left( \frac{y}{x} \right) \right) dy$
For $x \neq 0$ and $2 - \log \left( \frac{y}{x} \right) \neq 0$, we can write $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{y}{x \left( 2 - \log \left( \frac{y}{x} \right) \right)} = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$
Let $f(x, y) = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$. For the term $\log \left( \frac{y}{x} \right)$ to be defined, we must have $\frac{y}{x} > 0$. Thus, $x$ and $y$ must have the same sign.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$ (such that $\frac{\lambda y}{\lambda x} = \frac{y}{x} > 0$):
$f(\lambda x, \lambda y) = \frac{\frac{\lambda y}{\lambda x}}{2 - \log \left( \frac{\lambda y}{\lambda x} \right)} = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous of degree 0.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$. Since $y/x > 0$, we have $v > 0$.
Differentiating $y = vx$ with respect to $x$ using the product rule, we get:
$\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} = v + x \frac{dv}{dx}$
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{v}{2 - \log(v)}$:
$v + x \frac{dv}{dx} = \frac{v}{2 - \log(v)}$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = \frac{v}{2 - \log(v)} - v$
$x \frac{dv}{dx} = v \left( \frac{1}{2 - \log(v)} - 1 \right)$
$x \frac{dv}{dx} = v \left( \frac{1 - (2 - \log(v))}{2 - \log(v)} \right)$
$x \frac{dv}{dx} = v \left( \frac{1 - 2 + \log(v)}{2 - \log(v)} \right)$
$x \frac{dv}{dx} = v \frac{\log(v) - 1}{2 - \log(v)}$
Separation of Variables and Integration:
Assuming $x \neq 0$, $v \neq 0$, $\log(v) - 1 \neq 0$, and $2 - \log(v) \neq 0$, we separate the variables $v$ and $x$:
$\frac{2 - \log(v)}{v (\log(v) - 1)} \; dv = \frac{dx}{x}$
Integrate both sides:
$\int \frac{2 - \log(v)}{v (\log(v) - 1)} \; dv = \int \frac{dx}{x}$
For the integral on the left side, let $u = \log(v) - 1$. Then $du = \frac{1}{v} dv$. Also, $\log(v) = u + 1$. The numerator $2 - \log(v) = 2 - (u+1) = 1 - u$.
The left integral becomes:
$\int \frac{1 - u}{u} \; du = \int \left( \frac{1}{u} - 1 \right) \; du$
$= \int \frac{1}{u} \; du - \int 1 \; du$
$= \ln|u| - u + C_1$
Substitute back $u = \log(v) - 1$:
$= \ln|\log(v) - 1| - (\log(v) - 1) + C_1$
$= \ln|\log(v) - 1| - \log(v) + 1 + C_1$
Let $C_2 = 1 + C_1$. So, the left integral is $\ln|\log(v) - 1| - \log(v) + C_2$.
The integral on the right side is $\int \frac{dx}{x} = \ln|x| + C_3$.
Equating the integrals:
$\ln|\log(v) - 1| - \log(v) + C_2 = \ln|x| + C_3$
$\ln|\log(v) - 1| - \log(v) - \ln|x| = C_3 - C_2$
Using logarithm properties, $\ln A - \ln B = \ln(A/B)$: ($v > 0, x \neq 0$)
$\ln\left|\frac{\log(v) - 1}{v}\right| - \ln|x| = C_4$ (where $C_4 = C_3 - C_2$)
$\ln\left|\frac{\log(v) - 1}{vx}\right| = C_4$
Exponentiating both sides:
$\left|\frac{\log(v) - 1}{vx}\right| = e^{C_4}$
$\frac{\log(v) - 1}{vx} = \pm e^{C_4}$
Let $C = \pm e^{C_4}$. $C$ is a non-zero constant initially from the separation step. However, we will check the case $C=0$ later.
$\log(v) - 1 = Cvx$
Back-substitution and General Solution:
Substitute back $v = \frac{y}{x}$ (remembering $y/x > 0$):
$\log\left(\frac{y}{x}\right) - 1 = C \left(\frac{y}{x}\right) x$
$\log\left(\frac{y}{x}\right) - 1 = Cy$
This is the general solution, where $C$ is an arbitrary constant.
Consideration of Singular Cases:
During the separation, we assumed $v(\log(v) - 1) \neq 0$. This means $v \neq 0$ and $\log(v) \neq 1$.
1. $v=0$: $y/x = 0 \implies y=0$. As noted earlier, the term $\log(y/x)$ in the original equation is only defined for $y/x > 0$, so $y=0$ (for $x \neq 0$) is outside the domain of the original equation.
2. $\log(v) = 1$: $v = e$. This means $y/x = e$, or $y = ex$. This case corresponds to the denominator $v(\log(v)-1)$ in the separated equation being zero.
Let's check if $y=ex$ is a solution to the original differential equation $y \;dx + x \log \left( \frac{y}{x} \right) dy - 2x \;dy = 0$.
If $y=ex$, then $dy = e \;dx$. Substituting these into the original equation:
$ex \;dx + x \log\left(\frac{ex}{x}\right) (e \;dx) - 2x (e \;dx) = 0$
$ex \;dx + x \log(e) (e \;dx) - 2ex \;dx = 0$
$ex \;dx + x \cdot 1 \cdot e \;dx - 2ex \;dx = 0$
$ex \;dx + ex \;dx - 2ex \;dx = 0$
$0 = 0$
This holds true for all $x$. Thus, $y=ex$ is a solution.
Let's check if $y=ex$ is included in the general solution $\log\left(\frac{y}{x}\right) - 1 = Cy$.
Substitute $y=ex$ into the general solution:
$\log\left(\frac{ex}{x}\right) - 1 = C(ex)$
$\log(e) - 1 = Cex$
$1 - 1 = Cex$
$0 = Cex$
Since this must hold for any $x$ in the domain ($x \neq 0$ and $y/x > 0$, so $ex/x > 0 \implies e > 0$, which is true), we must have $C = 0$.
Therefore, the solution $y=ex$ is included in the general solution when the constant $C$ is zero.
We also assumed $2 - \log(v) \neq 0$, which means $\log(v) \neq 2$, so $v \neq e^2$. This corresponds to $y/x \neq e^2$, or $y \neq e^2 x$. We checked this case when setting up the $\frac{dy}{dx}$ form. If $y = e^2 x$, then $2 - \log(y/x) = 2 - \log(e^2) = 2 - 2 = 0$. The original equation is $y \;dx + x(\log(y/x) - 2) dy = 0$. If $y=e^2x$, this becomes $e^2 x \;dx + x(\log(e^2) - 2) dy = 0 \implies e^2 x \;dx + x(2-2) dy = 0 \implies e^2 x \;dx = 0$. This implies $x=0$ (since $e^2 \neq 0$), but the equation is not defined for $x=0$. So $y=e^2x$ is not a solution for $x \neq 0$.
Final Answer:
The general solution to the given differential equation, defined for regions where $\frac{y}{x} > 0$, is given by:
$\log\left(\frac{y}{x}\right) - 1 = Cy$
where $C$ is an arbitrary real constant. This includes the solution $y=ex$ when $C=0$.
Question 10. $\left( 1 + e^{\frac{x}{y}} \right) dx + e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right) dy = 0$
Answer:
Given Differential Equation:
$\left( 1 + e^{\frac{x}{y}} \right) dx + e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right) dy = 0$
Check for Homogeneity:
The equation is of the form $M(x, y) dx + N(x, y) dy = 0$, where $M(x, y) = 1 + e^{\frac{x}{y}}$ and $N(x, y) = e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right)$.
We check if $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.
$M(\lambda x, \lambda y) = 1 + e^{\frac{\lambda x}{\lambda y}} = 1 + e^{\frac{x}{y}} = M(x, y) = \lambda^0 M(x, y)$
$N(\lambda x, \lambda y) = e^{\frac{\lambda x}{\lambda y}} \left( 1 − \frac{\lambda x}{\lambda y} \right) = e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right) = N(x, y) = \lambda^0 N(x, y)$
Both functions are homogeneous of degree 0. Thus, the differential equation is homogeneous.
Solution Method: Substitution
Since the equation involves the term $\frac{x}{y}$, we use the substitution $x = vy$, where $v$ is a function of $y$.
Differentiating $x = vy$ with respect to $y$, we get:
$dx = v \;dy + y \frac{dv}{dy} \;dy$
Substitute $x = vy$ and $dx = v \;dy + y \frac{dv}{dy} \;dy$ into the original differential equation:
$\left( 1 + e^{\frac{vy}{y}} \right) \left( v \;dy + y \frac{dv}{dy} \;dy \right) + e^{\frac{vy}{y}} \left( 1 − \frac{vy}{y} \right) dy = 0$
$\left( 1 + e^{v} \right) (v \;dy + y \;dv) + e^{v} (1 − v) dy = 0$
Expand the terms:
$(1 + e^{v}) v \;dy + (1 + e^{v}) y \;dv + e^{v} (1 − v) dy = 0$
Group the terms with $dy$ and $dv$:
$\left[ (1 + e^{v}) v + e^{v} (1 − v) \right] dy + (1 + e^{v}) y \;dv = 0$
Simplify the coefficient of $dy$:
$[ v + v e^{v} + e^{v} - v e^{v} ] dy + (1 + e^{v}) y \;dv = 0$
$[ v + e^{v} ] dy + (1 + e^{v}) y \;dv = 0$
Separation of Variables and Integration:
We separate the variables $v$ and $y$, assuming $y \neq 0$ and $v + e^v \neq 0$. Note that $1 + e^v > 0$ always.
$(1 + e^{v}) y \;dv = - (v + e^{v}) dy$
$\frac{1 + e^{v}}{v + e^{v}} \;dv = - \frac{dy}{y}$
Integrate both sides:
$\int \frac{1 + e^{v}}{v + e^{v}} \;dv = \int - \frac{1}{y} \;dy$
For the integral on the left side, let $u = v + e^v$. Then the differential $du = (1 + e^v) dv$. The integral becomes $\int \frac{du}{u}$.
$\int \frac{du}{u} = \int - \frac{1}{y} \;dy$
$\ln|u| = -\ln|y| + C_1$
where $C_1$ is the constant of integration.
Substitute back $u = v + e^v$:
$\ln|v + e^v| = -\ln|y| + C_1$
Rearrange the terms using logarithm properties:
$\ln|v + e^v| + \ln|y| = C_1$
$\ln|y(v + e^v)| = C_1$
Exponentiate both sides:
$|y(v + e^v)| = e^{C_1}$
$y(v + e^v) = \pm e^{C_1}$
Let $C = \pm e^{C_1}$. Since $e^{C_1} > 0$, $C$ is a non-zero constant.
Back-substitution and General Solution:
Substitute back $v = \frac{x}{y}$:
$y \left( \frac{x}{y} + e^{\frac{x}{y}} \right) = C$
Distribute $y$ on the left side:
$y \cdot \frac{x}{y} + y e^{\frac{x}{y}} = C$
$x + y e^{\frac{x}{y}} = C$
This is the general solution. The constant $C$ can be any real number, as we will see below.
Consideration of Singular Case:
We assumed $y \neq 0$ (implied by the $\frac{x}{y}$ terms) and $v + e^v \neq 0$, which is $\frac{x}{y} + e^{\frac{x}{y}} \neq 0$.
Let $z = x/y$. We check if $z + e^z = 0$ has a solution. Let $g(z) = z + e^z$. $g'(z) = 1 + e^z > 1$ for all real $z$. $g(z)$ is strictly increasing. $g(z) \to -\infty$ as $z \to -\infty$ and $g(z) \to \infty$ as $z \to \infty$. By the Intermediate Value Theorem, there is a unique root $z_0$ such that $z_0 + e^{z_0} = 0$. ($z_0 \approx -0.567$).
If the solution curve passes through points where $\frac{x}{y} = z_0$, then $v + e^v = 0$. The step involving dividing by $v+e^v$ is not valid along the curve $x/y = z_0$, which is the line $x = z_0 y$.
Let's check if $x = z_0 y$ is a solution to the original differential equation $\left( 1 + e^{\frac{x}{y}} \right) dx + e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right) dy = 0$.
If $x = z_0 y$, then $dx = z_0 dy$. Substituting $x = z_0 y$ and $dx = z_0 dy$ (assuming $y \neq 0$, so $x/y = z_0$) into the equation:
$\left( 1 + e^{\frac{z_0 y}{y}} \right) (z_0 dy) + e^{\frac{z_0 y}{y}} \left( 1 − \frac{z_0 y}{y} \right) dy = 0$
$(1 + e^{z_0}) z_0 \;dy + e^{z_0} (1 − z_0) dy = 0$
$[z_0 (1 + e^{z_0}) + e^{z_0} (1 − z_0)] dy = 0$
$[z_0 + z_0 e^{z_0} + e^{z_0} - z_0 e^{z_0}] dy = 0$
$[z_0 + e^{z_0}] dy = 0$
Since $z_0 + e^{z_0} = 0$ by definition of $z_0$, this equation becomes $0 \cdot dy = 0$. This is true for all $y$ in the domain. Thus, the line $x = z_0 y$ (where $z_0$ is the root of $z+e^z=0$) is a solution.
Let's see if this singular solution is included in the general solution $x + y e^{x/y} = C$.
Substitute $x = z_0 y$ into the general solution:
$z_0 y + y e^{z_0} = C$
$y(z_0 + e^{z_0}) = C$
Since $z_0 + e^{z_0} = 0$, this gives $y \cdot 0 = C$, which means $0 = C$.
Therefore, the singular solution $x = z_0 y$ is included in the general solution when the constant $C=0$. The initial assumption that $C$ must be non-zero was based on the separation step; the value $C=0$ corresponds to the case where $y(v+e^v)=0$, which leads to $v+e^v=0$ (since $y \neq 0$), which is the singular case.
Final Answer:
The general solution to the given differential equation is:
$x + y e^{\frac{x}{y}} = C$
where $C$ is an arbitrary constant.
For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:
Question 11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1
Answer:
Given Differential Equation:
$(x + y) dy + (x – y) dx = 0$
Given Condition:
$y = 1$ when $x = 1$
Rearrange and Check Homogeneity:
Rearrange the equation to express $\frac{dy}{dx}$:
$(x + y) dy = - (x – y) dx$
$(x + y) dy = (y – x) dx$
$\frac{dy}{dx} = \frac{y - x}{x + y}$
Let $f(x, y) = \frac{y - x}{x + y}$.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda y - \lambda x}{\lambda x + \lambda y} = \frac{\lambda(y - x)}{\lambda(x + y)} = \frac{y - x}{x + y}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y - x}{x + y}$:
$v + x \frac{dv}{dx} = \frac{vx - x}{x + vx}$
$v + x \frac{dv}{dx} = \frac{x(v - 1)}{x(1 + v)}$
For $x \neq 0$, this simplifies to:
$v + x \frac{dv}{dx} = \frac{v - 1}{1 + v}$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = \frac{v - 1}{1 + v} - v$
$x \frac{dv}{dx} = \frac{v - 1 - v(1 + v)}{1 + v}$
$x \frac{dv}{dx} = \frac{v - 1 - v - v^2}{1 + v}$
$x \frac{dv}{dx} = \frac{-1 - v^2}{1 + v}$
$x \frac{dv}{dx} = - \frac{1 + v^2}{1 + v}$
Separation of Variables and Integration:
Assuming $x \neq 0$ and $1+v^2 \neq 0$ (which is always true), and $1+v \neq 0$, we separate the variables:
$\frac{1 + v}{1 + v^2} \; dv = - \frac{dx}{x}$
Integrate both sides:
$\int \frac{1 + v}{1 + v^2} \; dv = \int - \frac{1}{x} \; dx$
Split the integral on the left side:
$\int \left( \frac{1}{1 + v^2} + \frac{v}{1 + v^2} \right) \; dv = \int - \frac{1}{x} \; dx$
$\int \frac{1}{1 + v^2} \; dv + \int \frac{v}{1 + v^2} \; dv = \int - \frac{1}{x} \; dx$
The first integral on the left is $\arctan(v)$. For the second integral on the left, use substitution $u = 1 + v^2$, so $du = 2v \; dv$. $\int \frac{v}{1 + v^2} \; dv = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(1 + v^2)$ (since $1+v^2$ is always positive). The integral on the right is $-\ln|x|$.
$\arctan(v) + \frac{1}{2} \ln(1 + v^2) = -\ln|x| + C$
where $C$ is the constant of integration.
Back-substitution:
Substitute back $v = \frac{y}{x}$:
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln\left(1 + \left(\frac{y}{x}\right)^2\right) = -\ln|x| + C$
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln\left(1 + \frac{y^2}{x^2}\right) = -\ln|x| + C$
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln\left(\frac{x^2 + y^2}{x^2}\right) = -\ln|x| + C$
Using logarithm properties, $\ln(A/B) = \ln A - \ln B$:
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \left( \ln(x^2 + y^2) - \ln(x^2) \right) = -\ln|x| + C$
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln(x^2 + y^2) - \frac{1}{2} \ln(x^2) = -\ln|x| + C$
Since $\ln(x^2) = 2\ln|x|$:
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln(x^2 + y^2) - \frac{1}{2} (2\ln|x|) = -\ln|x| + C$
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln(x^2 + y^2) - \ln|x| = -\ln|x| + C$
Add $\ln|x|$ to both sides:
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln(x^2 + y^2) = C$
This is the general solution.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 1$ when $x = 1$. Substitute these values into the general solution:
$\arctan\left(\frac{1}{1}\right) + \frac{1}{2} \ln(1^2 + 1^2) = C$
$\arctan(1) + \frac{1}{2} \ln(1 + 1) = C$
$\frac{\pi}{4} + \frac{1}{2} \ln(2) = C$
$C = \frac{\pi}{4} + \frac{1}{2} \ln(2)$
... (i)
Substitute the value of $C$ back into the general solution:
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln(x^2 + y^2) = \frac{\pi}{4} + \frac{1}{2} \ln(2)$
This is the particular solution satisfying the given condition.
Final Answer:
The particular solution is:
$\arctan\left(\frac{y}{x}\right) + \frac{1}{2} \ln(x^2 + y^2) = \frac{\pi}{4} + \frac{1}{2} \ln(2)$
Question 12. x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
Answer:
Given Differential Equation:
$x^2 dy + (xy + y^2) dx = 0$
Given Condition:
$y = 1$ when $x = 1$
Rearrange and Check Homogeneity:
Rearrange the equation to express $\frac{dy}{dx}$:
$x^2 dy = - (xy + y^2) dx$
$\frac{dy}{dx} = \frac{-(xy + y^2)}{x^2} = \frac{-xy - y^2}{x^2}$
Let $f(x, y) = \frac{-xy - y^2}{x^2}$.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{-\lambda x \lambda y - (\lambda y)^2}{(\lambda x)^2} = \frac{-\lambda^2 xy - \lambda^2 y^2}{\lambda^2 x^2} = \frac{\lambda^2(-xy - y^2)}{\lambda^2 x^2} = \frac{-xy - y^2}{x^2}$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{-xy - y^2}{x^2}$:
$v + x \frac{dv}{dx} = \frac{-x(vx) - (vx)^2}{x^2}$
$v + x \frac{dv}{dx} = \frac{-vx^2 - v^2x^2}{x^2}$
For $x \neq 0$, this simplifies to:
$v + x \frac{dv}{dx} = -v - v^2$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = -v - v^2 - v$
$x \frac{dv}{dx} = -2v - v^2$
$x \frac{dv}{dx} = -v(v + 2)$
Separation of Variables and Integration:
Assuming $x \neq 0$ and $v(v+2) \neq 0$, we separate the variables:
$\frac{dv}{v(v + 2)} = - \frac{dx}{x}$
Integrate both sides:
$\int \frac{dv}{v(v + 2)} = \int - \frac{1}{x} \; dx$
We use partial fraction decomposition for the left side integrand: $\frac{1}{v(v+2)} = \frac{1/2}{v} - \frac{1/2}{v+2}$.
$\int \left( \frac{1/2}{v} - \frac{1/2}{v+2} \right) dv = \int - \frac{1}{x} \; dx$
$\frac{1}{2} \ln|v| - \frac{1}{2} \ln|v+2| = -\ln|x| + C'$
where $C'$ is the constant of integration.
$\frac{1}{2} \left( \ln|v| - \ln|v+2| \right) = -\ln|x| + C'$
$\frac{1}{2} \ln\left|\frac{v}{v+2}\right| = -\ln|x| + C'$
Multiply by 2:
$\ln\left|\frac{v}{v+2}\right| = -2\ln|x| + 2C'$
$\ln\left|\frac{v}{v+2}\right| = \ln(x^{-2}) + C''$ (where $C'' = 2C'$)
Exponentiate both sides:
$\left|\frac{v}{v+2}\right| = e^{\ln(x^{-2}) + C''} = e^{\ln(x^{-2})} e^{C''} = x^{-2} e^{C''}$
$\frac{v}{v+2} = \pm e^{C''} \frac{1}{x^2}$
Let $C = \pm e^{C''}$. $C$ is an arbitrary non-zero constant initially. We can allow $C=0$ later as shown in the singular solution analysis.
$\frac{v}{v+2} = \frac{C}{x^2}$
Back-substitution:
Substitute back $v = \frac{y}{x}$:
$\frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C}{x^2}$
Simplify the left side:
$\frac{\frac{y}{x}}{\frac{y + 2x}{x}} = \frac{C}{x^2}$
$\frac{y}{y + 2x} = \frac{C}{x^2}$
This is the general solution form. Note that $C=0$ gives $y=0$, which is a solution. The case $y+2x=0$ (or $v=-2$) leads to division by zero here and corresponds to the singular solution $y=-2x$.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 1$ when $x = 1$. Substitute these values into the general solution:
$\frac{1}{1 + 2(1)} = \frac{C}{1^2}$
$\frac{1}{3} = C$
The constant is $C = \frac{1}{3}$.
Particular Solution:
Substitute the value of $C$ back into the general solution $\frac{y}{y + 2x} = \frac{C}{x^2}$:
$\frac{y}{y + 2x} = \frac{1/3}{x^2}$
$\frac{y}{y + 2x} = \frac{1}{3x^2}$
Cross-multiply:
$y (3x^2) = 1 (y + 2x)$
$3x^2 y = y + 2x$
Rearrange to solve for $y$:
$3x^2 y - y = 2x$
$y(3x^2 - 1) = 2x$
$y = \frac{2x}{3x^2 - 1}$
This is the particular solution satisfying the given condition. Note that the initial condition $(1,1)$ is not on the singular solution line $y=-2x$. The particular solution is valid where $3x^2 - 1 \neq 0$.
Final Answer:
The particular solution is:
$y = \frac{2x}{3x^2 - 1}$
Question 13. $\left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx + x \;dy = 0 \; ; \; y = \frac{π}{4} \;when \;x = 1$
Answer:
Given Differential Equation:
$\left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx + x \;dy = 0$
Given Condition:
$y = \frac{π}{4}$ when $x = 1$
Rearrange and Check Homogeneity:
Rearrange the equation to express $\frac{dy}{dx}$:
$x \;dy = - \left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx$
$x \;dy = \left[ y - x \sin^2 \left( \frac{y}{x} \right) \right] dx$
$\frac{dy}{dx} = \frac{y - x \sin^2 \left( \frac{y}{x} \right)}{x}$
$\frac{dy}{dx} = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$
Let $f(x, y) = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \sin^2 \left( \frac{\lambda y}{\lambda x} \right) = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$:
$v + x \frac{dv}{dx} = v - \sin^2(v)$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = - \sin^2(v)$
Separation of Variables and Integration:
Assuming $x \neq 0$ and $\sin^2(v) \neq 0$, we separate the variables:
$\frac{dv}{\sin^2(v)} = - \frac{dx}{x}$
We can rewrite $\frac{1}{\sin^2(v)}$ as $\text{cosec}^2(v)$:
$\text{cosec}^2(v) \; dv = - \frac{1}{x} \; dx$
Integrate both sides:
$\int \text{cosec}^2(v) \; dv = \int - \frac{1}{x} \; dx$
The integral of $\text{cosec}^2(v)$ is $-\cot(v)$. The integral of $-\frac{1}{x}$ is $-\ln|x|$.
$-\cot(v) = -\ln|x| + C$
where $C$ is the constant of integration.
Multiply by -1:
$\cot(v) = \ln|x| - C$
Let the arbitrary constant be $K = -C$. Then:
$\cot(v) = \ln|x| + K$
Back-substitution:
Substitute back $v = \frac{y}{x}$:
$\cot\left(\frac{y}{x}\right) = \ln|x| + K$
This is the general solution.
Apply Given Condition to Find Particular Solution:
Given condition: $y = \frac{\pi}{4}$ when $x = 1$. Substitute these values into the general solution:
$\cot\left(\frac{\pi/4}{1}\right) = \ln|1| + K$
$\cot\left(\frac{\pi}{4}\right) = 0 + K$
$1 = K$
The constant is $K = 1$.
Particular Solution:
Substitute the value of $K$ back into the general solution $\cot\left(\frac{y}{x}\right) = \ln|x| + K$:
$\cot\left(\frac{y}{x}\right) = \ln|x| + 1$
This is the particular solution satisfying the given condition.
Consideration of Singular Solutions:
We assumed $\sin(v) \neq 0$, which means $\sin(y/x) \neq 0$. This occurs when $y/x \neq n\pi$ for any integer $n$. So, $y \neq n\pi x$.
Let's check if $y = n\pi x$ is a solution to the original equation $\left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx + x \;dy = 0$.
If $y = n\pi x$, then $\sin(y/x) = \sin(n\pi) = 0$. Also, $dy = n\pi \;dx$.
Substituting these into the equation:
$[ x \cdot 0^2 - n\pi x ] dx + x (n\pi \;dx) = 0$
$[-n\pi x] dx + n\pi x \;dx = 0$
$-n\pi x \;dx + n\pi x \;dx = 0$
$0 = 0$
This identity holds for all $x$ where the equation is defined ($x \neq 0$). Thus, the lines $y = n\pi x$ for any integer $n$ are solutions to the differential equation.
Let's check if the particular solution $\cot(y/x) = \ln|x| + 1$ includes any of these singular solutions. For $y=n\pi x$, $\cot(y/x) = \cot(n\pi)$ is undefined for integers $n$. Therefore, the particular solution found does not include these singular solutions.
The given initial condition $(1, \pi/4)$ has $y/x = \pi/4$. Since $\sin(\pi/4) = 1/\sqrt{2} \neq 0$, the initial condition is not on a singular solution line. The particular solution found is valid around the initial point.
Final Answer:
The particular solution is:
$\cot\left(\frac{y}{x}\right) = \ln|x| + 1$
Question 14. $\frac{dy}{dx} - \frac{y}{x} + cosec \left( \frac{y}{x} \right) = 0 \; ; \;y = 0 \;when \;x = 1$
Answer:
Given Differential Equation:
$\frac{dy}{dx} - \frac{y}{x} + \text{cosec} \left( \frac{y}{x} \right) = 0$
Given Condition:
$y = 0$ when $x = 1$
Rearrange and Check Homogeneity:
Rearrange the equation to express $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$
Let $f(x, y) = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$.
The term $\text{cosec} \left( \frac{y}{x} \right) = \frac{1}{\sin(y/x)}$ requires $x \neq 0$ and $\sin(y/x) \neq 0$, i.e., $\frac{y}{x} \neq n\pi$ for any integer $n$.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \text{cosec} \left( \frac{\lambda y}{\lambda x} \right) = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$:
$v + x \frac{dv}{dx} = v - \text{cosec}(v)$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = - \text{cosec}(v)$
Separation of Variables and Integration:
Assuming $x \neq 0$ and $\text{cosec}(v) \neq 0$ (which is always true as $|\text{cosec}(v)| \geq 1$), we separate the variables:
$\frac{dv}{\text{cosec}(v)} = - \frac{dx}{x}$
$\sin(v) \; dv = - \frac{1}{x} \; dx$
Integrate both sides:
$\int \sin(v) \; dv = \int - \frac{1}{x} \; dx$
$-\cos(v) = -\ln|x| + C$
where $C$ is the constant of integration.
Multiply by -1:
$\cos(v) = \ln|x| - C$
Let $K = -C$. Then:
$\cos(v) = \ln|x| + K$
Back-substitution:
Substitute back $v = \frac{y}{x}$:
$\cos\left(\frac{y}{x}\right) = \ln|x| + K$
This is the general solution. Note that this solution is typically obtained under the assumption $\sin(y/x) \neq 0$, which is required by the original differential equation.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 0$ when $x = 1$. Substitute these values into the general solution:
$\cos\left(\frac{0}{1}\right) = \ln|1| + K$
$\cos(0) = 0 + K$
$1 = K$
The constant is $K = 1$.
Particular Solution:
Substitute the value of $K$ back into the general solution $\cos\left(\frac{y}{x}\right) = \ln|x| + K$:
$\cos\left(\frac{y}{x}\right) = \ln|x| + 1$
This is the particular solution satisfying the given condition.
Note on the initial condition:
The original differential equation involves the term $\text{cosec}(y/x) = 1/\sin(y/x)$, which is undefined when $\sin(y/x) = 0$. This occurs when $y/x = n\pi$ for any integer $n$, i.e., $y = n\pi x$. The given initial condition $(1, 0)$ lies on the line $y=0$, which is $y=0 \cdot x$, corresponding to $n=0$. Thus, the initial condition point is a point where the original differential equation is not defined. The derived particular solution $\cos(y/x) = \ln|x| + 1$ is a curve that passes through the point $(1, 0)$, and it satisfies the differential equation for points $(x, y)$ on the curve where $x \neq 0$ and $\sin(y/x) \neq 0$.
Final Answer:
The particular solution is:
$\cos\left(\frac{y}{x}\right) = \ln|x| + 1$
Question 15. $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0 \; ; \;y = 2 \;when \;x = 1$
Answer:
Given Differential Equation:
$2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$
Given Condition:
$y = 2$ when $x = 1$
Rearrange and Check Homogeneity:
Rearrange the equation to express $\frac{dy}{dx}$:
$2x^2 \frac{dy}{dx} = 2xy + y^2$
$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}$
$\frac{dy}{dx} = \frac{2xy}{2x^2} + \frac{y^2}{2x^2}$
$\frac{dy}{dx} = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$
Let $f(x, y) = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$.
To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:
$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} + \frac{1}{2} \left(\frac{\lambda y}{\lambda x}\right)^2 = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$
$f(\lambda x, \lambda y) = f(x, y)$
Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$, the given differential equation is homogeneous.
Solution Method: Substitution
We use the substitution $y = vx$, where $v$ is a function of $x$.
Differentiating $y = vx$ with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$:
$v + x \frac{dv}{dx} = v + \frac{1}{2} v^2$
Subtract $v$ from both sides:
$x \frac{dv}{dx} = \frac{1}{2} v^2$
Separation of Variables and Integration:
Assuming $x \neq 0$ and $v^2 \neq 0$, we separate the variables:
$\frac{dv}{v^2} = \frac{1}{2x} \; dx$
Integrate both sides:
$\int v^{-2} \; dv = \frac{1}{2} \int \frac{1}{x} \; dx$
$\frac{v^{-1}}{-1} = \frac{1}{2} \ln|x| + C$
$-\frac{1}{v} = \frac{1}{2} \ln|x| + C$
where $C$ is the constant of integration.
Back-substitution:
Substitute back $v = \frac{y}{x}$:
$-\frac{1}{\frac{y}{x}} = \frac{1}{2} \ln|x| + C$
$-\frac{x}{y} = \frac{1}{2} \ln|x| + C$
This is the general solution. Note the assumption $v^2 \neq 0$, which means $y/x \neq 0$, i.e., $y \neq 0$. If $y=0$, substituting into the original ODE $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$ gives $0 + 0 - 2x^2 \cdot 0 = 0$, which is $0=0$. So $y=0$ is a solution. This corresponds to the case where $v=0$, where we divided by $v^2$. Let's see if $y=0$ is part of the general solution form. $-\frac{x}{0}$ is undefined. So the general solution does not include $y=0$. $y=0$ is a singular solution.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 2$ when $x = 1$. Substitute these values into the general solution:
$-\frac{1}{2} = \frac{1}{2} \ln|1| + C$
$-\frac{1}{2} = \frac{1}{2} \cdot 0 + C$
$-\frac{1}{2} = C$
The constant is $C = -\frac{1}{2}$.
Particular Solution:
Substitute the value of $C$ back into the general solution $-\frac{x}{y} = \frac{1}{2} \ln|x| + C$:
$-\frac{x}{y} = \frac{1}{2} \ln|x| - \frac{1}{2}$
Multiply by -2:
$\frac{2x}{y} = -\ln|x| + 1$
Rearrange to solve for $y$:
$y = \frac{2x}{1 - \ln|x|}$
This is the particular solution satisfying the given condition. The initial condition $(1,2)$ does not lie on the singular solution line $y=0$. The particular solution is valid where $1 - \ln|x| \neq 0$, i.e., $\ln|x| \neq 1$, which means $|x| \neq e$.
Final Answer:
The particular solution is:
$y = \frac{2x}{1 - \ln|x|}$
Question 16. A homogeneous differential equation of the from $\frac{dy}{dx} = h \left( \frac{x}{y} \right)$ can be solved by making the substitution.
(A) y = vx
(B) v = yx
(C) x = vy
(D) x = v
Answer:
Given Differential Equation Form:
$\frac{dy}{dx} = h \left( \frac{x}{y} \right)$
Analysis of Homogeneous Equation Forms:
A homogeneous differential equation $\frac{dy}{dx} = f(x, y)$ can be solved by a substitution if $f(x, y)$ is a homogeneous function of degree zero, i.e., $f(\lambda x, \lambda y) = f(x, y)$ for a non-zero constant $\lambda$.
If the equation is in the form $\frac{dy}{dx} = F \left( \frac{y}{x} \right)$, the standard substitution is $y = vx$.
If the equation is in the form $\frac{dx}{dy} = G \left( \frac{x}{y} \right)$, the standard substitution is $x = vy$.
Applying to the Given Form:
The given equation is $\frac{dy}{dx} = h \left( \frac{x}{y} \right)$. The function on the right side, $h \left( \frac{x}{y} \right)$, depends on the ratio $\frac{x}{y}$.
To transform this equation into a separable form, we can rewrite it by taking the reciprocal of both sides (assuming $h \left( \frac{x}{y} \right) \neq 0$):
$\frac{1}{\frac{dy}{dx}} = \frac{1}{h \left( \frac{x}{y} \right)}$
$\frac{dx}{dy} = \frac{1}{h \left( \frac{x}{y} \right)}$
Let $g \left( \frac{x}{y} \right) = \frac{1}{h \left( \frac{x}{y} \right)}$. The equation becomes:
$\frac{dx}{dy} = g \left( \frac{x}{y} \right)$
This equation is of the form $\frac{dx}{dy} = G \left( \frac{x}{y} \right)$.
The appropriate substitution for this form is $x = vy$, where $v$ is considered a function of $y$. Differentiating $x = vy$ with respect to $y$ yields $\frac{dx}{dy} = v + y \frac{dv}{dy}$. Substituting this into the ODE separates the variables $v$ and $y$.
Evaluating the Options:
(A) $y = vx$: This substitution is suitable for equations of the form $\frac{dy}{dx} = F \left( \frac{y}{x} \right)$.
(B) $v = yx$: This is not a standard substitution for this type of ODE.
(C) $x = vy$: This substitution is suitable for equations of the form $\frac{dx}{dy} = G \left( \frac{x}{y} \right)$, which the given equation can be rewritten as.
(D) $x = v$: This is a simple change of variable, not a substitution designed to solve homogeneous ODEs by reducing them to separable form.
Conclusion:
A homogeneous differential equation of the form $\frac{dy}{dx} = h \left( \frac{x}{y} \right)$ is best solved by rewriting it as $\frac{dx}{dy} = \frac{1}{h \left( \frac{x}{y} \right)}$ and then using the substitution $x = vy$.
The correct option is (C).
Question 17. Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(B) (xy) dx – (x3 + y3 ) dy = 0
(C) (x3 + 2y2) dx + 2xy dy = 0
(D) y2 dx + (x2 – xy – y2 ) dy = 0
Answer:
A differential equation of the form $M(x, y) dx + N(x, y) dy = 0$ is called a homogeneous differential equation if both $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.
A function $f(x, y)$ is a homogeneous function of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$ for some constant $n$ and for every non-zero $\lambda$ where $(x,y)$ and $(\lambda x, \lambda y)$ are in the domain of $f$.
Let's examine each option:
(A) $(4x + 6y + 5) dy – (3y + 2x + 4) dx = 0$
Here, $M(x, y) = -(3y + 2x + 4) = -3y - 2x - 4$ and $N(x, y) = 4x + 6y + 5$.
$M(\lambda x, \lambda y) = -3(\lambda y) - 2(\lambda x) - 4 = \lambda(-3y - 2x) - 4$. Due to the constant term $-4$, $M(x, y)$ is not a homogeneous function.
Therefore, the differential equation is not homogeneous.
(B) $(xy) dx – (x^3 + y^3) dy = 0$
Here, $M(x, y) = xy$ and $N(x, y) = -(x^3 + y^3) = -x^3 - y^3$.
Check $M(x, y)$: $M(\lambda x, \lambda y) = (\lambda x)(\lambda y) = \lambda^2 xy = \lambda^2 M(x, y)$. $M(x, y)$ is homogeneous of degree 2.
Check $N(x, y)$: $N(\lambda x, \lambda y) = -(\lambda x)^3 - (\lambda y)^3 = -\lambda^3 x^3 - \lambda^3 y^3 = \lambda^3(-x^3 - y^3) = \lambda^3 N(x, y)$. $N(x, y)$ is homogeneous of degree 3.
Since $M(x, y)$ and $N(x, y)$ are homogeneous functions of different degrees (2 and 3), the differential equation is not homogeneous.
(C) $(x^3 + 2y^2) dx + 2xy dy = 0$
Here, $M(x, y) = x^3 + 2y^2$ and $N(x, y) = 2xy$.
Check $M(x, y)$: $M(\lambda x, \lambda y) = (\lambda x)^3 + 2(\lambda y)^2 = \lambda^3 x^3 + 2\lambda^2 y^2$. This cannot be written in the form $\lambda^n (x^3 + 2y^2)$ for a single value of $n$. For example, $\lambda^2 M(x, y) = \lambda^2 x^3 + 2\lambda^2 y^2 \neq M(\lambda x, \lambda y)$. $\lambda^3 M(x, y) = \lambda^3 x^3 + 2\lambda^3 y^2 \neq M(\lambda x, \lambda y)$. Thus, $M(x, y)$ is not a homogeneous function.
Therefore, the differential equation is not homogeneous.
(D) $y^2 dx + (x^2 – xy – y^2 ) dy = 0$
Here, $M(x, y) = y^2$ and $N(x, y) = x^2 – xy – y^2$.
Check $M(x, y)$: $M(\lambda x, \lambda y) = (\lambda y)^2 = \lambda^2 y^2 = \lambda^2 M(x, y)$. $M(x, y)$ is homogeneous of degree 2.
Check $N(x, y)$: $N(\lambda x, \lambda y) = (\lambda x)^2 – (\lambda x)(\lambda y) – (\lambda y)^2 = \lambda^2 x^2 – \lambda^2 xy – \lambda^2 y^2 = \lambda^2 (x^2 – xy – y^2) = \lambda^2 N(x, y)$. $N(x, y)$ is homogeneous of degree 2.
Since both $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree (degree 2), the differential equation is homogeneous.
Alternatively, we can write the equation in the form $\frac{dy}{dx} = f(x,y)$ and check if $f(x,y)$ can be expressed as a function of $\frac{y}{x}$ only.
From option (D): $y^2 dx = -(x^2 – xy – y^2 ) dy = (xy + y^2 - x^2) dy$.
$\frac{dy}{dx} = \frac{y^2}{xy + y^2 - x^2}$
Divide the numerator and denominator by $x^2$ (assuming $x \neq 0$):
$\frac{dy}{dx} = \frac{\frac{y^2}{x^2}}{\frac{xy}{x^2} + \frac{y^2}{x^2} - \frac{x^2}{x^2}} = \frac{(\frac{y}{x})^2}{\frac{y}{x} + (\frac{y}{x})^2 - 1}$
This expression is a function of $\frac{y}{x}$ only. Therefore, the differential equation is homogeneous.
The correct option is (D).
Example 19 to 23 (Before Exercise 9.6)
Example 19: Find the general solution of the differential equation $\frac{dy}{dx} - y = \cos x$ .
Answer:
Given Differential Equation:
$\frac{dy}{dx} - y = \cos x$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = -1$
$Q(x) = \cos x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int -1 dx}$
IF = $e^{-x}$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot e^{-x} = \int \cos x \cdot e^{-x} dx + C$
Evaluate the Integral:
We need to evaluate the integral $I = \int e^{-x} \cos x dx$. We use integration by parts twice.
Let $I = \int e^{-x} \cos x dx$.
Apply integration by parts $\int u dv = uv - \int v du$.
Choose $u = \cos x$ and $dv = e^{-x} dx$.
Then $du = -\sin x dx$ and $v = \int e^{-x} dx = -e^{-x}$.
$I = (\cos x)(-e^{-x}) - \int (-e^{-x}) (-\sin x) dx$
$I = -e^{-x} \cos x - \int e^{-x} \sin x dx$
Now, apply integration by parts to the new integral $\int e^{-x} \sin x dx$.
Choose $u = \sin x$ and $dv = e^{-x} dx$.
Then $du = \cos x dx$ and $v = \int e^{-x} dx = -e^{-x}$.
$\int e^{-x} \sin x dx = (\sin x)(-e^{-x}) - \int (-e^{-x}) (\cos x) dx$
$\int e^{-x} \sin x dx = -e^{-x} \sin x + \int e^{-x} \cos x dx$
Notice that the integral on the right is the original integral $I$.
Substitute this back into the expression for $I$:
$I = -e^{-x} \cos x - [-e^{-x} \sin x + I]$
$I = -e^{-x} \cos x + e^{-x} \sin x - I$
Move the $I$ term to the left side:
$I + I = e^{-x} \sin x - e^{-x} \cos x$
$2I = e^{-x}(\sin x - \cos x)$
$I = \frac{1}{2} e^{-x}(\sin x - \cos x)$
Final General Solution:
Substitute the result of the integral back into the general solution equation:
$y \cdot e^{-x} = \frac{1}{2} e^{-x}(\sin x - \cos x) + C$
Divide both sides by $e^{-x}$ to solve for $y$:
$y = \frac{1}{2} e^{-x}(\sin x - \cos x) e^{x} + C e^{x}$
$y = \frac{1}{2} (\sin x - \cos x) + C e^{x}$
The general solution is:
$y = \frac{1}{2} (\sin x - \cos x) + C e^{x}$
Example 20: Find the general solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ (x ≠ 0).
Answer:
Given Differential Equation:
$x \frac{dy}{dx} + 2y = x^2$ (x ≠ 0)
This is a first-order linear differential equation. To put it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $x$ (since $x \neq 0$ is given):
$\frac{1}{x} \left( x \frac{dy}{dx} + 2y \right) = \frac{1}{x} (x^2)$
$\frac{dy}{dx} + \frac{2}{x} y = x$
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
$P(x) = \frac{2}{x}$
$Q(x) = x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \frac{2}{x} dx}$
IF = $e^{2 \int \frac{1}{x} dx}$
IF = $e^{2 \ln|x|}$
Using the logarithm property $a \ln b = \ln b^a$:
IF = $e^{\ln(x^2)}$
Using the property $e^{\ln A} = A$:
IF = $x^2$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot x^2 = \int x \cdot x^2 dx + C$
$y x^2 = \int x^3 dx + C$
Evaluate the integral $\int x^3 dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$.
$y x^2 = \frac{x^4}{4} + C$
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $x^2$ (since $x \neq 0$):
$y = \frac{1}{x^2} \left( \frac{x^4}{4} + C \right)$
$y = \frac{x^4}{4x^2} + \frac{C}{x^2}$
$y = \frac{x^2}{4} + \frac{C}{x^2}$
The general solution is:
$y = \frac{x^2}{4} + \frac{C}{x^2}$
Example 21: Find the general solution of the differential equation y dx – (x + 2y2 ) dy = 0.
Answer:
Given Differential Equation:
$y \; dx – (x + 2y^2 ) dy = 0$
This equation is not in the standard form for a linear differential equation in $y$ as a function of $x$ ($\frac{dy}{dx} + P(x)y = Q(x)$) due to the presence of $y^2$.
However, we can rearrange it to be a linear differential equation in $x$ as a function of $y$ ($\frac{dx}{dy} + P(y)x = Q(y)$).
Rearrange the equation to express $\frac{dx}{dy}$:
$y \; dx = (x + 2y^2 ) dy$
Divide by $dy$ (assuming $dy \neq 0$):
$y \frac{dx}{dy} = x + 2y^2$
Divide by $y$ (assuming $y \neq 0$):
$\frac{dx}{dy} = \frac{x}{y} + \frac{2y^2}{y}$
$\frac{dx}{dy} = \frac{1}{y} x + 2y$
Rearrange into the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$:
$\frac{dx}{dy} - \frac{1}{y} x = 2y$
Comparing with the standard form, we identify:
$P(y) = -\frac{1}{y}$
$Q(y) = 2y$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(y) dy}$.
IF = $e^{\int -\frac{1}{y} dy}$
IF = $e^{- \int \frac{1}{y} dy}$
IF = $e^{-\ln|y|}$
Using the logarithm property $a \ln b = \ln b^a$:
IF = $e^{\ln(|y|^{-1})}$
Using the property $e^{\ln A} = A$:
IF = $|y|^{-1} = \frac{1}{|y|}$
We can use $\frac{1}{y}$ as the integrating factor (assuming $y \neq 0$).
IF = $\frac{1}{y}$
General Solution:
The general solution of a linear first-order differential equation in $x(y)$ is given by:
$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) dy + C$
Substitute the values of IF and $Q(y)$:
$x \cdot \frac{1}{y} = \int 2y \cdot \frac{1}{y} dy + C$
$\frac{x}{y} = \int 2 \; dy + C$
Evaluate the integral $\int 2 \; dy = 2y + C$.
$\frac{x}{y} = 2y + C$
Solve for x (optional, depending on desired form):
Multiply both sides by $y$ to solve for $x$:
$x = y(2y + C)$
$x = 2y^2 + Cy$
The general solution is:
$\frac{x}{y} = 2y + C$ or $x = 2y^2 + Cy$
Example 22: Find the particular solution of the differential equation
$\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$ (x ≠ 0)
Given that y = 0 when $x = \frac{\pi}{2}$.
Answer:
Given Differential Equation:
$\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$ (x ≠ 0)
Given Condition:
$y = 0$ when $x = \frac{\pi}{2}$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = \cot x$
$Q(x) = 2x + x^2 \cot x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \cot x dx}$
IF = $e^{\int \frac{\cos x}{\sin x} dx}$
Let $u = \sin x$, then $du = \cos x dx$. The integral is $\int \frac{du}{u} = \ln|u| = \ln|\sin x|$.
IF = $e^{\ln|\sin x|}$
IF = $|\sin x|$
Since the initial condition is given at $x = \frac{\pi}{2}$, which is in the interval where $\sin x > 0$ (e.g., $0 < x < \pi$), we can use $\sin x$ as the integrating factor.
IF = $\sin x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot \sin x = \int (2x + x^2 \cot x) \cdot \sin x dx + C$
$y \sin x = \int (2x \sin x + x^2 \cot x \sin x) dx + C$
$y \sin x = \int \left( 2x \sin x + x^2 \frac{\cos x}{\sin x} \sin x \right) dx + C$
$y \sin x = \int (2x \sin x + x^2 \cos x) dx + C$
Evaluate the Integral:
We need to evaluate $\int (2x \sin x + x^2 \cos x) dx$. Observe that this integral has the form of the result from the product rule for integration, $\int (u'v + uv') dx = uv$.
Let $u = x^2$ and $v = \sin x$. Then $u' = 2x$ and $v' = \cos x$.
The integrand is $2x \sin x + x^2 \cos x = u'v + uv'$.
Therefore, $\int (2x \sin x + x^2 \cos x) dx = \int \frac{d}{dx}(x^2 \sin x) dx = x^2 \sin x$.
Substitute this back into the general solution equation:
$y \sin x = x^2 \sin x + C$
Apply Given Condition to Find Particular Solution:
Given condition: $y = 0$ when $x = \frac{\pi}{2}$. Substitute these values into the general solution:
$0 \cdot \sin\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + C$
$0 \cdot 1 = \left(\frac{\pi^2}{4}\right) \cdot 1 + C$
$0 = \frac{\pi^2}{4} + C$
$C = - \frac{\pi^2}{4}$
Particular Solution:
Substitute the value of $C$ back into the general solution $y \sin x = x^2 \sin x + C$:
$y \sin x = x^2 \sin x - \frac{\pi^2}{4}$
To express $y$ explicitly (assuming $\sin x \neq 0$):
$y = \frac{x^2 \sin x - \frac{\pi^2}{4}}{\sin x}$
$y = x^2 - \frac{\pi^2}{4 \sin x}$
Final Answer:
The particular solution is:
$y \sin x = x^2 \sin x - \frac{\pi^2}{4}$ or $y = x^2 - \frac{\pi^2}{4} \text{cosec} x$
Example 23: Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point.
Answer:
Problem Statement:
We are given that the curve passes through the point (0, 1).
The slope of the tangent to the curve at any point (x, y) is given by $\frac{dy}{dx}$.
According to the problem, the slope is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate.
Slope = x-coordinate + (x-coordinate $\times$ y-coordinate)
$\frac{dy}{dx} = x + xy$
Differential Equation:
The differential equation representing the curve is:
$\frac{dy}{dx} = x(1 + y)$
This is a first-order differential equation.
Solution Method: Separation of Variables
We can separate the variables $y$ and $x$, assuming $1+y \neq 0$:
$\frac{dy}{1 + y} = x \; dx$
Integrate both sides:
$\int \frac{dy}{1 + y} = \int x \; dx$
The integral on the left side is $\ln|1 + y|$. The integral on the right side is $\frac{x^2}{2}$.
$\ln|1 + y| = \frac{x^2}{2} + C$
where $C$ is the constant of integration.
Apply Given Condition to Find Particular Solution:
The curve passes through the point (0, 1). Substitute $x=0$ and $y=1$ into the general solution:
$\ln|1 + 1| = \frac{0^2}{2} + C$
$\ln|2| = 0 + C$
$C = \ln 2$
Particular Solution:
Substitute the value of $C$ back into the general solution $\ln|1 + y| = \frac{x^2}{2} + C$:
$\ln|1 + y| = \frac{x^2}{2} + \ln 2$}
Rearrange the terms:
$\ln|1 + y| - \ln 2 = \frac{x^2}{2}$
Using the logarithm property $\ln A - \ln B = \ln(A/B)$:
$\ln\left|\frac{1 + y}{2}\right| = \frac{x^2}{2}$
Exponentiate both sides:
$\left|\frac{1 + y}{2}\right| = e^{\frac{x^2}{2}}$
$\frac{1 + y}{2} = \pm e^{\frac{x^2}{2}}$
Since the curve passes through $(0, 1)$, $1+y = 1+1 = 2$, and $e^{x^2/2} = e^0 = 1$. For the equation to hold at $(0,1)$, $\frac{1+1}{2} = \pm e^0 \implies 1 = \pm 1$. We must choose the positive sign.
So, $\frac{1 + y}{2} = e^{\frac{x^2}{2}}$
Multiply by 2:
$1 + y = 2 e^{\frac{x^2}{2}}$
Solve for $y$:
$y = 2 e^{\frac{x^2}{2}} - 1$
Consideration of Singular Solution:
We assumed $1+y \neq 0$ when separating variables. If $1+y=0$, then $y=-1$. Substituting $y=-1$ and $\frac{dy}{dx}=0$ into the differential equation $\frac{dy}{dx} = x(1 + y)$: $0 = x(1 + (-1)) = x \cdot 0 = 0$. This holds for all $x$. So, the line $y=-1$ is a singular solution.
Let's check if the particular solution $y = 2 e^{\frac{x^2}{2}} - 1$ includes $y=-1$. $2 e^{\frac{x^2}{2}} - 1 = -1 \implies 2 e^{\frac{x^2}{2}} = 0 \implies e^{\frac{x^2}{2}} = 0$, which has no real solution for $x$. Thus, the particular solution found does not include the singular solution $y=-1$. The initial condition $(0, 1)$ is not on the line $y=-1$.
Final Answer:
The equation of the curve passing through the point (0, 1) is:
$y = 2 e^{\frac{x^2}{2}} - 1$
Exercise 9.6
For each of the differential equations given in Exercises 1 to 12, find the general solution:
Question 1. $\frac{dy}{dx} + 2y = \sin x$
Answer:
Given Differential Equation:
$\frac{dy}{dx} + 2y = \sin x$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = 2$
$Q(x) = \sin x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int 2 dx}$
IF = $e^{2x}$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y e^{2x} = \int \sin x \cdot e^{2x} dx + C$
Evaluate the Integral:
We need to evaluate the integral $I = \int e^{2x} \sin x dx$. We use integration by parts twice.
Let $I = \int e^{2x} \sin x dx$.
Using integration by parts ($\int u dv = uv - \int v du$):
Let $u = \sin x$ and $dv = e^{2x} dx$. Then $du = \cos x dx$ and $v = \frac{1}{2} e^{2x}$.
$I = \frac{1}{2} e^{2x} \sin x - \int \frac{1}{2} e^{2x} \cos x dx$
$I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \int e^{2x} \cos x dx$
Apply integration by parts again to $\int e^{2x} \cos x dx$:
Let $u = \cos x$ and $dv = e^{2x} dx$. Then $du = -\sin x dx$ and $v = \frac{1}{2} e^{2x}$.
$\int e^{2x} \cos x dx = \frac{1}{2} e^{2x} \cos x - \int \frac{1}{2} e^{2x} (-\sin x) dx$
$\int e^{2x} \cos x dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \int e^{2x} \sin x dx$
Substitute this result back into the expression for $I$:
$I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left( \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I \right)$
$I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} I$
Add $\frac{1}{4} I$ to both sides:
$I + \frac{1}{4} I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x$
$\frac{5}{4} I = \frac{1}{4} e^{2x} (2 \sin x - \cos x)$
Multiply by $\frac{4}{5}$:
$I = \frac{1}{5} e^{2x} (2 \sin x - \cos x)$
Substitute the Integral Result back into the General Solution:
$y e^{2x} = \frac{1}{5} e^{2x} (2 \sin x - \cos x) + C$
Divide both sides by $e^{2x}$:
$y = \frac{1}{e^{2x}} \left( \frac{1}{5} e^{2x} (2 \sin x - \cos x) + C \right)$
$y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x}$
The general solution is:
$y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x}$
Question 2. $\frac{dy}{dx} + 3y = e^{−2x}$
Answer:
Given Differential Equation:
$\frac{dy}{dx} + 3y = e^{-2x}$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = 3$
$Q(x) = e^{-2x}$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int 3 dx}$
IF = $e^{3x}$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y e^{3x} = \int e^{-2x} \cdot e^{3x} dx + C$
$y e^{3x} = \int e^{-2x + 3x} dx + C$
$y e^{3x} = \int e^{x} dx + C$
Evaluate the integral $\int e^{x} dx = e^{x} + C$.
$y e^{3x} = e^{x} + C$
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $e^{3x}$:
$y = \frac{1}{e^{3x}} (e^{x} + C)$
$y = \frac{e^{x}}{e^{3x}} + \frac{C}{e^{3x}}$
$y = e^{x - 3x} + C e^{-3x}$
$y = e^{-2x} + C e^{-3x}$
The general solution is:
$y = e^{-2x} + C e^{-3x}$
Question 3. $\frac{dy}{dx} + \frac{y}{x} = x^2$
Answer:
Given Differential Equation:
$\frac{dy}{dx} + \frac{y}{x} = x^2$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = \frac{1}{x}$
$Q(x) = x^2$
The equation is defined for $x \neq 0$.
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \frac{1}{x} dx}$
IF = $e^{\ln|x|}$
IF = $|x|$
We can use $x$ as the integrating factor for $x > 0$, or $-x$ for $x < 0$. However, using $x$ covers both cases with the understanding that $|x|$ is involved in the strict derivation involving logarithms. A simpler approach is to use $x$ directly as the IF, as multiplying by the IF yields $(xy)'$ on the left side, whether IF is $x$ or $|x|$. Let's use IF = $x$.
IF = $x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot x = \int x^2 \cdot x dx + C$
$xy = \int x^3 dx + C$
Evaluate the integral $\int x^3 dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$.
$xy = \frac{x^4}{4} + C$
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $x$ (since $x \neq 0$):
$y = \frac{1}{x} \left( \frac{x^4}{4} + C \right)$
$y = \frac{x^4}{4x} + \frac{C}{x}$
$y = \frac{x^3}{4} + \frac{C}{x}$
The general solution is:
$y = \frac{x^3}{4} + \frac{C}{x}$
Question 4. $\frac{dy}{dx} + (\sec x) y = \tan x \left( 0 ≤ x < \frac{π}{2} \right)$
Answer:
Given Differential Equation:
$\frac{dy}{dx} + (\sec x) y = \tan x$
The equation is given for the interval $0 \leq x < \frac{\pi}{2}$.
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = \sec x$
$Q(x) = \tan x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \sec x dx}$
The integral of $\sec x$ is $\ln|\sec x + \tan x|$.
IF = $e^{\ln|\sec x + \tan x|}$
Since the interval is $0 \leq x < \frac{\pi}{2}$, $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$ are defined. For $0 < x < \frac{\pi}{2}$, $\cos x > 0$, so $\sec x > 0$ and $\tan x > 0$. At $x=0$, $\sec 0 = 1$ and $\tan 0 = 0$. Thus, $\sec x + \tan x \geq 1 > 0$ on the given interval. Therefore, $|\sec x + \tan x| = \sec x + \tan x$.
IF = $e^{\ln(\sec x + \tan x)}$
IF = $\sec x + \tan x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$}
Substitute the values of IF and $Q(x)$:
$y (\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C$
$y (\sec x + \tan x) = \int (\tan x \sec x + \tan^2 x) dx + C$
Evaluate the Integral:
We need to evaluate the integral $\int (\tan x \sec x + \tan^2 x) dx$. We can split this into two standard integrals:
$\int (\tan x \sec x + \tan^2 x) dx = \int \tan x \sec x dx + \int \tan^2 x dx$
The integral of $\tan x \sec x$ is $\sec x$.
For the integral of $\tan^2 x$, we use the trigonometric identity $\tan^2 x = \sec^2 x - 1$:
$\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \int \sec^2 x dx - \int 1 dx = \tan x - x$
Combining these results, the integral is:
$\int (\tan x \sec x + \tan^2 x) dx = \sec x + (\tan x - x) = \sec x + \tan x - x$
Substitute the Integral Result back into the General Solution:
$y (\sec x + \tan x) = \sec x + \tan x - x + C$
where $C$ is the constant of integration.
Solve for y:
Divide both sides by $(\sec x + \tan x)$ (which is non-zero on the given interval):
$y = \frac{\sec x + \tan x - x + C}{\sec x + \tan x}$
$y = \frac{\sec x + \tan x}{\sec x + \tan x} - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x}$
$y = 1 - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x}$
The general solution is:
$y = 1 - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x}$
Question 5. $\cos^2 x \frac{dy}{dx} + y = \tan x \left( 0 ≤ x < \frac{π}{2} \right)$
Answer:
Given Differential Equation:
$\cos^2 x \frac{dy}{dx} + y = \tan x$
The equation is given for the interval $0 \leq x < \frac{\pi}{2}$.
This is a first-order differential equation. To put it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $\cos^2 x$. Note that for $0 \leq x < \frac{\pi}{2}$, $\cos x \neq 0$ except possibly at $x=\pi/2$. For $0 \leq x < \pi/2$, $\cos^2 x > 0$.
Divide by $\cos^2 x$:
$\frac{\cos^2 x}{\cos^2 x} \frac{dy}{dx} + \frac{y}{\cos^2 x} = \frac{\tan x}{\cos^2 x}$
$\frac{dy}{dx} + \frac{1}{\cos^2 x} y = \tan x \cdot \frac{1}{\cos^2 x}$
$\frac{dy}{dx} + \sec^2 x \cdot y = \tan x \sec^2 x$
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
$P(x) = \sec^2 x$
$Q(x) = \tan x \sec^2 x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \sec^2 x dx}$
The integral of $\sec^2 x$ is $\tan x$.
IF = $e^{\tan x}$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y e^{\tan x} = \int (\tan x \sec^2 x) e^{\tan x} dx + C$}
Evaluate the Integral:
We need to evaluate the integral $\int \tan x \sec^2 x e^{\tan x} dx$. We can use substitution.
Let $u = \tan x$. Then the differential $du = \sec^2 x dx$.
The integral becomes $\int u e^{u} du$.
We evaluate this integral using integration by parts ($\int w dz = wz - \int z dw$).
Let $w = u$ and $dz = e^u du$. Then $dw = du$ and $z = \int e^u du = e^u$.
$\int u e^{u} du = u e^u - \int e^u du = u e^u - e^u + C'$
Substitute back $u = \tan x$:
$\int \tan x \sec^2 x e^{\tan x} dx = \tan x e^{\tan x} - e^{\tan x}$
Substitute the Integral Result back into the General Solution:
$y e^{\tan x} = \tan x e^{\tan x} - e^{\tan x} + C$
where $C$ is the constant of integration.
Solve for y:
Divide both sides by $e^{\tan x}$ (which is always positive):
$y = \frac{\tan x e^{\tan x} - e^{\tan x} + C}{e^{\tan x}}$
$y = \frac{\tan x e^{\tan x}}{e^{\tan x}} - \frac{e^{\tan x}}{e^{\tan x}} + \frac{C}{e^{\tan x}}$
$y = \tan x - 1 + C e^{-\tan x}$
The general solution is:
$y = \tan x - 1 + C e^{-\tan x}$
Question 6. $x \frac{dy}{dx} + 2y = x^2 \log x$
Answer:
Given Differential Equation:
$x \frac{dy}{dx} + 2y = x^2 \log x$
This is a first-order linear differential equation. To put it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $x$. Note that for $\log x$ to be defined, we must have $x > 0$. In this domain, $x \neq 0$, so we can divide by $x$:
$\frac{1}{x} \left( x \frac{dy}{dx} + 2y \right) = \frac{1}{x} (x^2 \log x)$
$\frac{dy}{dx} + \frac{2}{x} y = x \log x$
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
$P(x) = \frac{2}{x}$
$Q(x) = x \log x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \frac{2}{x} dx}$
IF = $e^{2 \int \frac{1}{x} dx}$
For $x > 0$, $\int \frac{1}{x} dx = \ln x$.
IF = $e^{2 \ln x}$
Using the logarithm property $a \ln b = \ln b^a$:
IF = $e^{\ln(x^2)}$
Using the property $e^{\ln A} = A$:
IF = $x^2$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot x^2 = \int (x \log x) \cdot x^2 dx + C$
$y x^2 = \int x^3 \log x dx + C$
Evaluate the Integral:
We need to evaluate the integral $\int x^3 \log x dx$. We use integration by parts, $\int u dv = uv - \int v du$.
Let $u = \log x$ and $dv = x^3 dx$.
Then $du = \frac{1}{x} dx$ and $v = \int x^3 dx = \frac{x^4}{4}$.
$\int x^3 \log x dx = (\log x) \left(\frac{x^4}{4}\right) - \int \left(\frac{x^4}{4}\right) \left(\frac{1}{x}\right) dx$
$= \frac{x^4}{4} \log x - \int \frac{x^3}{4} dx$
$= \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx$
$= \frac{x^4}{4} \log x - \frac{1}{4} \left(\frac{x^4}{4}\right)$
$= \frac{x^4}{4} \log x - \frac{x^4}{16}$
Substitute the Integral Result back into the General Solution:
$y x^2 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C$
where $C$ is the constant of integration.
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $x^2$ (since $x > 0$):
$y = \frac{1}{x^2} \left( \frac{x^4}{4} \log x - \frac{x^4}{16} + C \right)$
$y = \frac{x^4}{4x^2} \log x - \frac{x^4}{16x^2} + \frac{C}{x^2}$
$y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2}$
The general solution is:
$y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2}$
Question 7. $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$
Answer:
Given Differential Equation:
$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$}
This is a first-order differential equation. To put it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $x \log x$. Note that for $\log x$ to be defined, we must have $x > 0$. Also, we must have $x \log x \neq 0$, which implies $x \neq 1$ and $x \neq 0$. So, the domain for this equation is $x \in (0, 1) \cup (1, \infty)$.
Divide by $x \log x$:
$\frac{x \log x}{x \log x} \frac{dy}{dx} + \frac{y}{x \log x} = \frac{\frac{2}{x} \log x}{x \log x}$}
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x} \log x \cdot \frac{1}{x \log x}$
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2}$
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
$P(x) = \frac{1}{x \log x}$
$Q(x) = \frac{2}{x^2}$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \frac{1}{x \log x} dx}$}
For the integral $\int \frac{1}{x \log x} dx$, let $u = \log x$. Then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1}{u} du = \ln|u| = \ln|\log x|$.
IF = $e^{\ln|\log x|}$
IF = $|\log x|$
For simplicity in the following steps, we can use $\log x$ as the IF, understanding that the absolute value will be handled correctly in the product $y \cdot (\text{IF})$. Let's use IF = $\log x$.
IF = $\log x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot \log x = \int \frac{2}{x^2} \cdot \log x dx + C$
$y \log x = 2 \int \frac{\log x}{x^2} dx + C$
Evaluate the Integral:
We need to evaluate the integral $\int \frac{\log x}{x^2} dx = \int x^{-2} \log x dx$. We use integration by parts, $\int u dv = uv - \int v du$.
Let $u = \log x$ and $dv = x^{-2} dx$.
Then $du = \frac{1}{x} dx$ and $v = \int x^{-2} dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$.
$\int x^{-2} \log x dx = (\log x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) dx$
$= -\frac{\log x}{x} - \int -\frac{1}{x^2} dx$
$= -\frac{\log x}{x} + \int x^{-2} dx$
$= -\frac{\log x}{x} + \left(-\frac{1}{x}\right)$
$= -\frac{\log x}{x} - \frac{1}{x}$
$= -\frac{1 + \log x}{x}$
Substitute the Integral Result back into the General Solution:
$y \log x = 2 \left( -\frac{1 + \log x}{x} \right) + C$
$y \log x = - \frac{2(1 + \log x)}{x} + C$
where $C$ is the constant of integration.
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $\log x$ (assuming $\log x \neq 0$, i.e., $x \neq 1$):
$y = \frac{1}{\log x} \left( - \frac{2(1 + \log x)}{x} + C \right)$
$y = - \frac{2(1 + \log x)}{x \log x} + \frac{C}{\log x}$
$y = - \frac{2}{x \log x} - \frac{2 \log x}{x \log x} + \frac{C}{\log x}$}
$y = - \frac{2}{x \log x} - \frac{2}{x} + \frac{C}{\log x}$}
The general solution is:
$y = \frac{C}{\log x} - \frac{2}{x} - \frac{2}{x \log x}$
Question 8. (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
Answer:
Given Differential Equation:
$(1 + x^2) dy + 2xy dx = \cot x dx$ (x ≠ 0)
This is a first-order differential equation. To put it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, we first rearrange the terms to group $dx$ and $dy$ and then divide by the coefficient of $dy$ (which is $1+x^2$) and $dx$ (to get $\frac{dy}{dx}$).
$(1 + x^2) dy = \cot x dx - 2xy dx$
$(1 + x^2) dy = (\cot x - 2xy) dx$
Divide by $(1 + x^2) dx$ (assuming $1+x^2 \neq 0$, which is always true for real $x$, and $dx \neq 0$):
$\frac{(1 + x^2) dy}{(1 + x^2) dx} = \frac{\cot x - 2xy}{(1 + x^2) dx} dx$
$\frac{dy}{dx} = \frac{\cot x}{1 + x^2} - \frac{2xy}{1 + x^2}$
Rearrange into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2}$
Comparing this with the standard form, we identify:
$P(x) = \frac{2x}{1 + x^2}$}
$Q(x) = \frac{\cot x}{1 + x^2}$}
The equation is defined where $\cot x$ is defined, i.e., $x \neq n\pi$ for any integer $n$. Given $x \neq 0$, we consider intervals like $(0, \pi)$, $(\pi, 2\pi)$, etc.
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \frac{2x}{1 + x^2} dx}$}
For the integral $\int \frac{2x}{1 + x^2} dx$, let $u = 1 + x^2$. Then $du = 2x dx$. The integral becomes $\int \frac{du}{u} = \ln|u| = \ln|1 + x^2|$. Since $1+x^2 > 0$ for real $x$, we have $\ln(1 + x^2)$.
IF = $e^{\ln(1 + x^2)}$
IF = $1 + x^2$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y (1 + x^2) = \int \frac{\cot x}{1 + x^2} \cdot (1 + x^2) dx + C$
$y (1 + x^2) = \int \cot x dx + C$
The integral of $\cot x$ is $\ln|\sin x|$.
$y (1 + x^2) = \ln|\sin x| + C$
where $C$ is the constant of integration.
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $(1 + x^2)$:
$y = \frac{\ln|\sin x| + C}{1 + x^2}$
The general solution is:
$y = \frac{\ln|\sin x| + C}{1 + x^2}$
Question 9. $x \frac{dy}{dx} + y - x + xy\; \cot x = 0$ (x ≠ 0)
Answer:
Given Differential Equation:
$x \frac{dy}{dx} + y - x + xy\; \cot x = 0$ (x ≠ 0)
This is a first-order differential equation. We need to rearrange it into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Group terms containing $y$:
$x \frac{dy}{dx} + y + xy\; \cot x = x$
$x \frac{dy}{dx} + y (1 + x \cot x) = x$
Divide the entire equation by $x$ (since $x \neq 0$ is given):
$\frac{x}{x} \frac{dy}{dx} + \frac{y (1 + x \cot x)}{x} = \frac{x}{x}$
$\frac{dy}{dx} + \left( \frac{1}{x} + \cot x \right) y = 1$
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
$P(x) = \frac{1}{x} + \cot x$
$Q(x) = 1$
The equation is defined where $\cot x$ is defined, i.e., $x \neq n\pi$ for any integer $n$.
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \left( \frac{1}{x} + \cot x \right) dx}$}
IF = $e^{\int \frac{1}{x} dx + \int \cot x dx}$
IF = $e^{\ln|x| + \ln|\sin x|}$
Using the logarithm property $\ln A + \ln B = \ln(AB)$:
IF = $e^{\ln|x \sin x|}$
IF = $|x \sin x|$
We can use $x \sin x$ as the integrating factor, considering the sign of $x \sin x$ within specific intervals. Let's use IF = $x \sin x$.
IF = $x \sin x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y (x \sin x) = \int 1 \cdot (x \sin x) dx + C$
$y x \sin x = \int x \sin x dx + C$
Evaluate the Integral:
We need to evaluate the integral $\int x \sin x dx$. We use integration by parts, $\int u dv = uv - \int v du$.
Let $u = x$ and $dv = \sin x dx$.
Then $du = dx$ and $v = \int \sin x dx = -\cos x$.
$\int x \sin x dx = x(-\cos x) - \int (-\cos x) dx$
$= -x \cos x + \int \cos x dx$
$= -x \cos x + \sin x$
Substitute the Integral Result back into the General Solution:
$y x \sin x = -x \cos x + \sin x + C$
where $C$ is the constant of integration.
Solve for y:
To find the general solution explicitly in terms of $y$, divide both sides by $x \sin x$ (assuming $x \sin x \neq 0$, which means $x \neq n\pi$):
$y = \frac{-x \cos x + \sin x + C}{x \sin x}$
$y = \frac{-x \cos x}{x \sin x} + \frac{\sin x}{x \sin x} + \frac{C}{x \sin x}$}
$y = - \frac{\cos x}{\sin x} + \frac{1}{x} + \frac{C}{x \sin x}$}
$y = -\cot x + \frac{1}{x} + C \text{cosec} x \cdot \frac{1}{x}$}
$y = \frac{1}{x} - \cot x + \frac{C}{x \sin x}$}
The general solution is:
$y = \frac{1}{x} - \cot x + \frac{C}{x \sin x}$
Question 10. $(x + y) \frac{dy}{dx} = 1$
Answer:
Given Differential Equation:
$(x + y) \frac{dy}{dx} = 1$
This equation is not linear in $y$ as a function of $x$. However, we can rewrite it by considering $x$ as a function of $y$.
Take the reciprocal of both sides (assuming $\frac{dy}{dx} \neq 0$, which implies $1 \neq 0$, always true):
$\frac{1}{\frac{dy}{dx}} = x + y$
$\frac{dx}{dy} = x + y$
Rearrange this into the standard linear form for $x$ as a function of $y$: $\frac{dx}{dy} + P(y)x = Q(y)$.
$\frac{dx}{dy} - x = y$
Comparing this with the standard form $\frac{dx}{dy} + P(y)x = Q(y)$, we identify:
$P(y) = -1$}
$Q(y) = y$}
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(y) dy}$.
IF = $e^{\int -1 dy}$
IF = $e^{-y}$
General Solution:
The general solution of a linear first-order differential equation in $x(y)$ is given by:
$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) dy + C$
Substitute the values of IF and $Q(y)$:
$x e^{-y} = \int y \cdot e^{-y} dy + C$
Evaluate the Integral:
We need to evaluate the integral $\int y e^{-y} dy$. We use integration by parts, $\int u dv = uv - \int v du$.
Let $u = y$ and $dv = e^{-y} dy$.
Then $du = dy$ and $v = \int e^{-y} dy = -e^{-y}$.
$\int y e^{-y} dy = y(-e^{-y}) - \int (-e^{-y}) dy$
$= -y e^{-y} + \int e^{-y} dy$
$= -y e^{-y} + (-e^{-y})$
$= -y e^{-y} - e^{-y}$
$= -e^{-y}(y + 1)$
Substitute the Integral Result back into the General Solution:
$x e^{-y} = -e^{-y}(y + 1) + C$
where $C$ is the constant of integration.
General Solution Form:
We can express the solution in terms of $x$ and $y$. We can multiply by $e^y$:
$x = -(y + 1) + C e^y$
$x = -y - 1 + C e^y$
Or, rearrange to isolate the constant term:
$x e^{-y} + e^{-y}(y + 1) = C$
$e^{-y}(x + y + 1) = C$
The general solution can be written as:
$e^{-y}(x + y + 1) = C$ or $x + y + 1 = C e^y$
Question 11. $y \;dx + (x – y^2) \;dy = 0$
Answer:
Given Differential Equation:
$y \; dx + (x – y^2) \; dy = 0$
This equation is not in the standard form for a linear differential equation in $y$ as a function of $x$. However, we can rearrange it to be a linear differential equation in $x$ as a function of $y$ ($\frac{dx}{dy} + P(y)x = Q(y)$).
Rearrange the equation to express $\frac{dx}{dy}$:
$y \; dx = - (x – y^2) \; dy$
$y \; dx = (y^2 - x) \; dy$
Divide by $dy$ (assuming $dy \neq 0$):
$y \frac{dx}{dy} = y^2 - x$
Divide by $y$ (assuming $y \neq 0$):
$\frac{dx}{dy} = \frac{y^2 - x}{y} = \frac{y^2}{y} - \frac{x}{y}$
$\frac{dx}{dy} = y - \frac{1}{y} x$
Rearrange into the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$:
$\frac{dx}{dy} + \frac{1}{y} x = y$
Comparing with the standard form, we identify:
$P(y) = \frac{1}{y}$}
$Q(y) = y$}
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(y) dy}$.
IF = $e^{\int \frac{1}{y} dy}$
IF = $e^{\ln|y|}$
IF = $|y|$
We can use $y$ as the integrating factor (assuming $y \neq 0$).
IF = $y$
General Solution:
The general solution of a linear first-order differential equation in $x(y)$ is given by:
$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) dy + C$
Substitute the values of IF and $Q(y)$:
$x \cdot y = \int y \cdot y dy + C$
$xy = \int y^2 dy + C$
Evaluate the integral $\int y^2 dy = \frac{y^{2+1}}{2+1} + C = \frac{y^3}{3} + C$.
$xy = \frac{y^3}{3} + C$
General Solution Form:
The general solution can be written as:
$xy = \frac{y^3}{3} + C$
We can also solve for $x$ explicitly (assuming $y \neq 0$):
$x = \frac{1}{y} \left( \frac{y^3}{3} + C \right)$
$x = \frac{y^2}{3} + \frac{C}{y}$
The general solution is:
$xy = \frac{y^3}{3} + C$ or $x = \frac{y^2}{3} + \frac{C}{y}$
Question 12. $(x + 3y^2) \frac{dy}{dx} = y$ (y > 0)
Answer:
Given Differential Equation:
$(x + 3y^2) \frac{dy}{dx} = y$ (y > 0)
This equation is not linear in $y$ as a function of $x$. However, we can rewrite it by considering $x$ as a function of $y$.
Take the reciprocal of both sides (assuming $\frac{dy}{dx} \neq 0$):
$\frac{1}{\frac{dy}{dx}} = \frac{x + 3y^2}{y}$
$\frac{dx}{dy} = \frac{x}{y} + \frac{3y^2}{y}$
$\frac{dx}{dy} = \frac{1}{y} x + 3y$
Rearrange this into the standard linear form for $x$ as a function of $y$: $\frac{dx}{dy} + P(y)x = Q(y)$.
$\frac{dx}{dy} - \frac{1}{y} x = 3y$
Comparing this with the standard form $\frac{dx}{dy} + P(y)x = Q(y)$, we identify:
$P(y) = -\frac{1}{y}$}
$Q(y) = 3y$}
The equation is defined for $y > 0$. In this domain, $y \neq 0$.
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(y) dy}$.
IF = $e^{\int -\frac{1}{y} dy}$
IF = $e^{- \int \frac{1}{y} dy}$
For $y > 0$, $\int \frac{1}{y} dy = \ln y$.
IF = $e^{-\ln y}$
Using the logarithm property $a \ln b = \ln b^a$:
IF = $e^{\ln(y^{-1})}$
Using the property $e^{\ln A} = A$:
IF = $y^{-1} = \frac{1}{y}$
General Solution:
The general solution of a linear first-order differential equation in $x(y)$ is given by:
$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) dy + C$
Substitute the values of IF and $Q(y)$:
$x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} dy + C$
$\frac{x}{y} = \int 3 \; dy + C$
Evaluate the integral $\int 3 \; dy = 3y + C$.
$\frac{x}{y} = 3y + C$
where $C$ is the constant of integration.
General Solution Form:
Multiply both sides by $y$ (since $y > 0$, $y \neq 0$) to solve for $x$:
$x = y(3y + C)$
$x = 3y^2 + Cy$
The general solution for $y > 0$ is:
$x = 3y^2 + Cy$
For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition:
Question 13. $\frac{dy}{dx} + 2y \tan x = \sin x \;; \;y = 0 \;when \;x = \frac{π}{3}$
Answer:
Given Differential Equation:
$\frac{dy}{dx} + 2y \tan x = \sin x$}
Given Condition:
$y = 0$ when $x = \frac{\pi}{3}$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = 2 \tan x$
$Q(x) = \sin x$
The term $\tan x$ is defined for $x \neq \frac{\pi}{2} + n\pi$. The initial condition $x=\pi/3$ is in the interval where $\tan x$ is defined, e.g., $(-\pi/2, \pi/2)$.
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int 2 \tan x dx}$
IF = $e^{2 \int \tan x dx}$
The integral of $\tan x$ is $\ln|\sec x|$.
IF = $e^{2 \ln|\sec x|}$
Using the logarithm property $a \ln b = \ln b^a$:
IF = $e^{\ln|\sec x|^2}$
IF = $|\sec x|^2 = \sec^2 x$
Since the initial condition is at $x = \frac{\pi}{3}$, which is in the interval $(-\pi/2, \pi/2)$ where $\sec x$ is defined and non-zero, $\sec^2 x > 0$.
IF = $\sec^2 x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \sec^2 x = \int \sin x \cdot \sec^2 x dx + C$
$y \sec^2 x = \int \sin x \cdot \frac{1}{\cos^2 x} dx + C$
$y \sec^2 x = \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} dx + C$
$y \sec^2 x = \int \tan x \sec x dx + C$
The integral of $\tan x \sec x$ is $\sec x$.
$y \sec^2 x = \sec x + C$
where $C$ is the constant of integration.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 0$ when $x = \frac{\pi}{3}$. Substitute these values into the general solution:
$0 \cdot \sec^2 \left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) + C$
Recall $\sec(\pi/3) = \frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2$.
$0 \cdot (2)^2 = 2 + C$
$0 = 2 + C$
$C = -2$
Particular Solution:
Substitute the value of $C$ back into the general solution $y \sec^2 x = \sec x + C$:
$y \sec^2 x = \sec x - 2$
To express $y$ explicitly (assuming $\sec x \neq 0$, which is true on the relevant interval):
$y = \frac{\sec x - 2}{\sec^2 x}$
$y = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x}$
$y = \frac{1}{\sec x} - 2 \frac{1}{\sec^2 x}$
$y = \cos x - 2 \cos^2 x$
Final Answer:
The particular solution is:
$y = \cos x - 2 \cos^2 x$
Question 14. $(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2} \;;\; y = 0 \;when \;x = 1$
Answer:
Given Differential Equation:
$(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$}
Given Condition:
$y = 0$ when $x = 1$
This is a first-order differential equation. To put it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $(1 + x^2)$. Note that $1 + x^2 > 0$ for all real $x$, so we can always divide by it.
Divide by $(1 + x^2)$:
$\frac{(1 + x^2)}{(1 + x^2)} \frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{\frac{1}{1 + x^2}}{1 + x^2}$}
$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2}$}
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
$P(x) = \frac{2x}{1 + x^2}$}
$Q(x) = \frac{1}{(1 + x^2)^2}$}
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int \frac{2x}{1 + x^2} dx}$}
For the integral $\int \frac{2x}{1 + x^2} dx$, let $u = 1 + x^2$. Then $du = 2x dx$. The integral becomes $\int \frac{du}{u} = \ln|u| = \ln|1 + x^2|$. Since $1+x^2 > 0$, we have $\ln(1 + x^2)$.
IF = $e^{\ln(1 + x^2)}$
IF = $1 + x^2$}
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y (1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) dx + C$
$y (1 + x^2) = \int \frac{1}{1 + x^2} dx + C$
The integral of $\frac{1}{1 + x^2}$ is $\arctan(x)$.
$y (1 + x^2) = \arctan(x) + C$
where $C$ is the constant of integration.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 0$ when $x = 1$. Substitute these values into the general solution:
$0 \cdot (1 + 1^2) = \arctan(1) + C$
$0 \cdot (1 + 1) = \frac{\pi}{4} + C$
$0 \cdot 2 = \frac{\pi}{4} + C$
$0 = \frac{\pi}{4} + C$
$C = - \frac{\pi}{4}$
Particular Solution:
Substitute the value of $C$ back into the general solution $y (1 + x^2) = \arctan(x) + C$:
$y (1 + x^2) = \arctan(x) - \frac{\pi}{4}$
To express $y$ explicitly (divide by $1+x^2$):
$y = \frac{\arctan(x) - \frac{\pi}{4}}{1 + x^2}$
Final Answer:
The particular solution is:
$y = \frac{\arctan(x) - \frac{\pi}{4}}{1 + x^2}$
Question 15. $\frac{dy}{dx} - 3y \;\cot x = \sin 2x \;;\; y = 2 \;when \;x = \frac{π}{2}$
Answer:
Given Differential Equation:
$\frac{dy}{dx} - 3y \cot x = \sin 2x$}
Given Condition:
$y = 2$ when $x = \frac{\pi}{2}$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Comparing the given equation with the standard form, we identify:
$P(x) = -3 \cot x$
$Q(x) = \sin 2x$
The term $\cot x$ is defined for $x \neq n\pi$. The initial condition $x=\pi/2$ is in the interval where $\cot x$ is defined, e.g., $(0, \pi)$.
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int -3 \cot x dx}$
IF = $e^{-3 \int \cot x dx}$
The integral of $\cot x$ is $\ln|\sin x|$.
IF = $e^{-3 \ln|\sin x|}$
Using the logarithm property $a \ln b = \ln b^a$:
IF = $e^{\ln|\sin x|^{-3}}$
IF = $|\sin x|^{-3} = \frac{1}{|\sin x|^3}$}
Since the initial condition is at $x = \frac{\pi}{2}$, which is in the interval $(0, \pi)$ where $\sin x > 0$, $|\sin x| = \sin x$.
IF = $\frac{1}{\sin^3 x} = \text{cosec}^3 x$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y \cdot \frac{1}{\sin^3 x} = \int \sin 2x \cdot \frac{1}{\sin^3 x} dx + C$
Use the identity $\sin 2x = 2 \sin x \cos x$:
$\frac{y}{\sin^3 x} = \int (2 \sin x \cos x) \cdot \frac{1}{\sin^3 x} dx + C$
$\frac{y}{\sin^3 x} = \int \frac{2 \sin x \cos x}{\sin^3 x} dx + C$
$\frac{y}{\sin^3 x} = \int \frac{2 \cos x}{\sin^2 x} dx + C$
$\frac{y}{\sin^3 x} = 2 \int \frac{\cos x}{\sin^2 x} dx + C$
Evaluate the Integral:
We need to evaluate the integral $\int \frac{\cos x}{\sin^2 x} dx$. We can use substitution.
Let $u = \sin x$. Then the differential $du = \cos x dx$.
The integral becomes $\int \frac{du}{u^2} = \int u^{-2} du = \frac{u^{-1}}{-1} = -\frac{1}{u}$.
Substitute back $u = \sin x$:
$\int \frac{\cos x}{\sin^2 x} dx = -\frac{1}{\sin x} = -\text{cosec} x$
Substitute the Integral Result back into the General Solution:
$\frac{y}{\sin^3 x} = 2 (-\text{cosec} x) + C$
$\frac{y}{\sin^3 x} = -2 \text{cosec} x + C$
where $C$ is the constant of integration.
Apply Given Condition to Find Particular Solution:
Given condition: $y = 2$ when $x = \frac{\pi}{2}$. Substitute these values into the general solution:
$\frac{2}{\sin^3 \left(\frac{\pi}{2}\right)} = -2 \text{cosec}\left(\frac{\pi}{2}\right) + C$
Recall $\sin(\pi/2) = 1$ and $\text{cosec}(\pi/2) = 1/\sin(\pi/2) = 1/1 = 1$.
$\frac{2}{(1)^3} = -2 (1) + C$}
$\frac{2}{1} = -2 + C$
$2 = -2 + C$
$C = 2 + 2 = 4$
Particular Solution:
Substitute the value of $C$ back into the general solution $\frac{y}{\sin^3 x} = -2 \text{cosec} x + C$:
$\frac{y}{\sin^3 x} = -2 \text{cosec} x + 4$
To express $y$ explicitly:
$y = \sin^3 x (-2 \text{cosec} x + 4)$
$y = -2 \sin^3 x \text{cosec} x + 4 \sin^3 x$
$y = -2 \sin^3 x \cdot \frac{1}{\sin x} + 4 \sin^3 x$
$y = -2 \sin^2 x + 4 \sin^3 x$
Final Answer:
The particular solution is:
$y = 4 \sin^3 x - 2 \sin^2 x$
Question 16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Answer:
Problem Statement:
We are given that the curve passes through the origin, i.e., the point (0, 0).
The slope of the tangent to the curve at any point (x, y) is given by $\frac{dy}{dx}$.
According to the problem, the slope is equal to the sum of the coordinates of the point (x, y).
Slope = x-coordinate + y-coordinate
$\frac{dy}{dx} = x + y$
Differential Equation:
The differential equation representing the curve is:
$\frac{dy}{dx} = x + y$
Rearrange into the standard linear first-order form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} - y = x$
Comparing this with the standard form, we identify:
$P(x) = -1$
$Q(x) = x$
Integrating Factor (IF):
The integrating factor is given by $e^{\int P(x) dx}$.
IF = $e^{\int -1 dx}$
IF = $e^{-x}$
General Solution:
The general solution of a linear first-order differential equation is given by:
$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$
Substitute the values of IF and $Q(x)$:
$y e^{-x} = \int x e^{-x} dx + C$
Evaluate the Integral:
We need to evaluate the integral $\int x e^{-x} dx$. We use integration by parts, $\int u dv = uv - \int v du$.
Let $u = x$ and $dv = e^{-x} dx$.
Then $du = dx$ and $v = \int e^{-x} dx = -e^{-x}$.
$\int x e^{-x} dx = x(-e^{-x}) - \int (-e^{-x}) dx$
$= -x e^{-x} + \int e^{-x} dx$
$= -x e^{-x} + (-e^{-x})$
$= -x e^{-x} - e^{-x}$
$= -e^{-x}(x + 1)$
Substitute the Integral Result back into the General Solution:
$y e^{-x} = -e^{-x}(x + 1) + C$
where $C$ is the constant of integration.
Apply Given Condition to Find Particular Solution:
The curve passes through the origin (0, 0). Substitute $x=0$ and $y=0$ into the general solution:
$0 \cdot e^{-0} = -e^{-0}(0 + 1) + C$
$0 \cdot 1 = -1(1) + C$
$0 = -1 + C$
$C = 1$
Particular Solution:
Substitute the value of $C$ back into the general solution $y e^{-x} = -e^{-x}(x + 1) + C$:
$y e^{-x} = -e^{-x}(x + 1) + 1$
To express $y$ explicitly, divide by $e^{-x}$:
$y = \frac{-e^{-x}(x + 1) + 1}{e^{-x}}$
$y = \frac{-e^{-x}(x + 1)}{e^{-x}} + \frac{1}{e^{-x}}$
$y = -(x + 1) + e^{x}$
$y = -x - 1 + e^{x}$
Final Answer:
The equation of the curve passing through the origin is:
$y = e^{x} - x - 1$
Question 17. Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Answer:
Let the equation of the curve be $y = f(x)$. A point on the curve is $(x, y)$ and the slope of the tangent to the curve at this point is $\frac{dy}{dx}$.
According to the given condition, the sum of the coordinates ($x+y$) exceeds the magnitude of the slope ($|\frac{dy}{dx}|$) by 5.
This can be written as:
$x+y = |\frac{dy}{dx}| + 5$
Rearranging the equation, we get:
$|\frac{dy}{dx}| = x+y - 5$
For the magnitude of the slope to be equal to $x+y-5$, it is required that $x+y-5 \ge 0$, i.e., $x+y \ge 5$. However, the curve passes through the point (0, 2), where $x+y = 0+2 = 2$, which does not satisfy $x+y \ge 5$. This indicates a potential inconsistency in the problem statement as written with the given initial point, as $|\frac{dy}{dx}|$ cannot be negative.
Given the structure of such problems, it is likely that the intended differential equation was one of the two forms arising from the absolute value, specifically the one whose solution passes through the given point.
The two possibilities arising from $|\frac{dy}{dx}| = x+y-5$ are $\frac{dy}{dx} = x+y-5$ and $\frac{dy}{dx} = -(x+y-5)$. Let us assume the intended differential equation for the curve passing through (0, 2) is $\frac{dy}{dx} = x+y-5$.
The differential equation is $\frac{dy}{dx} = x+y-5$.
We can rewrite this as a first-order linear differential equation:
$\frac{dy}{dx} - y = x-5$
This is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -1$ and $Q(x) = x-5$.
The integrating factor (IF) is given by $e^{\int P(x) dx}$.
$IF = e^{\int -1 dx} = e^{-x}$
Multiply the differential equation by the integrating factor:
$\frac{d}{dx}(y \cdot e^{-x}) = (x-5)e^{-x}$
Integrate both sides with respect to $x$:
$y \cdot e^{-x} = \int (x-5)e^{-x} dx$
We evaluate the integral $\int (x-5)e^{-x} dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$ with $u=x$ and $dv=e^{-x}dx$ for the term $\int xe^{-x} dx$, we have $du=dx$ and $v=-e^{-x}$.
$\int xe^{-x} dx = x(-e^{-x}) - \int (-e^{-x}) dx = -xe^{-x} + \int e^{-x} dx = -xe^{-x} - e^{-x}$.
Now integrate the full expression:
$\int (x-5)e^{-x} dx = \int xe^{-x} dx - 5\int e^{-x} dx$
$= (-xe^{-x} - e^{-x}) - 5(-e^{-x}) + C$
$= -xe^{-x} - e^{-x} + 5e^{-x} + C$
$= -xe^{-x} + 4e^{-x} + C$
So, the general solution in implicit form is $y e^{-x} = -xe^{-x} + 4e^{-x} + C$.
Multiply by $e^x$ to solve for $y$ explicitly:
$y = -x + 4 + C e^x$
$\mathbf{y = C e^x - x + 4}$
This is the general equation of the curve satisfying the assumed differential equation. We are given that the curve passes through the point (0, 2).
Substitute $x=0$ and $y=2$ into the general solution to find the value of the constant $C$:
$2 = C e^0 - 0 + 4$
$2 = C \cdot 1 + 4$
$2 = C + 4$
$C = 2 - 4$
$\mathbf{C = -2}$
Substitute the value of $C$ back into the general solution to get the equation of the specific curve passing through (0, 2).
$\mathbf{y = -2 e^x - x + 4}$
This curve passes through (0, 2) and satisfies the differential equation $\frac{dy}{dx} = x+y-5$. While the initial point creates an issue with the strict interpretation of $x+y = |\frac{dy}{dx}|+5$, this solution is derived from a plausible underlying linear DE.
Question 18. The Integrating Factor of the differential equation $x \frac{dy}{dx} - y = 2x^2$ is
(A) e–x
(B) e–y
(C) $\frac{1}{x}$
(D) x
Answer:
The given differential equation is $x \frac{dy}{dx} - y = 2x^2$.
To find the Integrating Factor (IF), we first need to rewrite the differential equation in the standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$.
Divide the entire equation by $x$ (assuming $x \neq 0$):
$\frac{dy}{dx} - \frac{1}{x}y = 2x$
Now, compare this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.
We identify $P(x) = -\frac{1}{x}$ and $Q(x) = 2x$.
The Integrating Factor (IF) is given by the formula $e^{\int P(x) dx}$.
IF = $e^{\int -\frac{1}{x} dx}$
Evaluate the integral in the exponent:
$\int -\frac{1}{x} dx = - \int \frac{1}{x} dx = -\ln|x|$
Substitute this back into the IF formula:
IF = $e^{-\ln|x|}$
Using the properties of logarithms and exponents ($e^{\ln a} = a$ and $a^{-1} = \frac{1}{a}$), we have:
IF = $e^{\ln(|x|^{-1})} = |x|^{-1} = \frac{1}{|x|}$
For practical purposes in solving the linear differential equation, we usually take $IF = \frac{1}{x}$, assuming $x > 0$ or without the absolute value as it serves the purpose of making the left side integrable.
IF = $\frac{1}{x}$
Comparing this result with the given options, we find that option (C) matches our calculated Integrating Factor.
The correct option is (C) $\frac{1}{x}$.
Question 19. The Integrating Factor of the differential equation $(1 - y^2) \frac{dy}{dx} + yx = ay \;(-1 < y < 1)$ is
(A) $\frac{1}{y^2 − 1}$
(B) $\frac{1}{\sqrt{y^2 − 1}}$
(C) $\frac{1}{1 − y^2}$
(D) $\frac{1}{\sqrt{1 − y^2}}$
Answer:
The given differential equation is $(1 - y^2) \frac{dy}{dx} + yx = ay$.
We are asked to find the Integrating Factor (IF). The form of the given options for the IF are all functions of $y$. This strongly suggests that the differential equation should be treated as a linear equation in $x$, where $y$ is the independent variable and $x$ is the dependent variable.
The standard form of a linear differential equation in $x$ is $\frac{dx}{dy} + P(y)x = Q(y)$. The Integrating Factor for this form is given by $e^{\int P(y) dy}$.
Let's rearrange the given equation $(1 - y^2) \frac{dy}{dx} + yx = ay$ to identify the coefficient of $x$ when the equation is potentially viewed as linear in $x$. Although the derivative is given as $\frac{dy}{dx}$, the structure and the options imply we should relate it to the form $\frac{dx}{dy} + P(y)x = Q(y)$.
Divide the given equation by $(1 - y^2)$ (Since $-1 < y < 1$, $1 - y^2 > 0$, so this division is valid):
$\frac{dy}{dx} + \frac{y}{1 - y^2} x = \frac{ay}{1 - y^2}$
This equation involves $\frac{dy}{dx}$ and $x$. If we consider the related linear equation in $x$ (treating $y$ as independent), its standard form would be $\frac{dx}{dy} + P(y)x = Q(y)$. The term $\frac{y}{1-y^2}x$ in the rearranged equation above suggests that $P(y)$ in the standard form for $x(y)$ is $\frac{y}{1-y^2}$.
So, let's take $P(y) = \frac{y}{1 - y^2}$. Now we calculate the Integrating Factor using this $P(y)$.
IF = $e^{\int P(y) dy} = e^{\int \frac{y}{1 - y^2} dy}$
To evaluate the integral $\int \frac{y}{1 - y^2} dy$, we use a substitution. Let $u = 1 - y^2$. Then, the differential $du = \frac{d}{dy}(1 - y^2) dy = -2y \, dy$. This means $y \, dy = -\frac{1}{2} du$.
$\int \frac{y}{1 - y^2} dy = \int \frac{-\frac{1}{2} du}{u} = -\frac{1}{2} \int \frac{1}{u} du$
$= -\frac{1}{2} \ln|u| + C_1$
Substitute back $u = 1 - y^2$. Since $-1 < y < 1$, $1 - y^2$ is positive, so $|1 - y^2| = 1 - y^2$. We can omit the constant of integration $C_1$ when calculating the Integrating Factor.
$\int \frac{y}{1 - y^2} dy = -\frac{1}{2} \ln(1 - y^2)$
Now, substitute this result into the formula for the Integrating Factor:
IF = $e^{-\frac{1}{2} \ln(1 - y^2)}$
Using the logarithm property $a \ln b = \ln b^a$ and the exponential property $e^{\ln c} = c$:
IF = $e^{\ln((1 - y^2)^{-1/2})}$
IF = $(1 - y^2)^{-1/2}$
IF = $\frac{1}{\sqrt{1 - y^2}}$
This calculated Integrating Factor matches option (D).
The correct option is (D) $\frac{1}{\sqrt{1 − y^2}}$.
Example 24 to 28 - Miscellaneous Examples
Example 24: Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are arbitrary constants is a solution of the differential equation
$\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y = 0$
Answer:
The given function is $y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$.
The given differential equation is $\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y = 0$.
To verify that the given function is a solution, we need to find the first and second derivatives of $y$ with respect to $x$, and then substitute them into the differential equation.
First derivative $\frac{dy}{dx}$:
$y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$
Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$:
$\frac{dy}{dx} = \frac{d}{dx}(c_1 e^{ax} \cos bx) + \frac{d}{dx}(c_2 e^{ax} \sin bx)$
$\frac{dy}{dx} = c_1 (ae^{ax} \cos bx + e^{ax}(-b \sin bx)) + c_2 (ae^{ax} \sin bx + e^{ax}(b \cos bx))$
$\frac{dy}{dx} = c_1 ae^{ax} \cos bx - c_1 be^{ax} \sin bx + c_2 ae^{ax} \sin bx + c_2 be^{ax} \cos bx$
Grouping terms with $\cos bx$ and $\sin bx$:
$\mathbf{\frac{dy}{dx} = (c_1 a + c_2 b) e^{ax} \cos bx + (c_2 a - c_1 b) e^{ax} \sin bx}$
Second derivative $\frac{d^2y}{dx^2}$:
Differentiate $\frac{dy}{dx}$ with respect to $x$. Let $A = (c_1 a + c_2 b)$ and $B = (c_2 a - c_1 b)$.
$\frac{d^2y}{dx^2} = \frac{d}{dx}(A e^{ax} \cos bx) + \frac{d}{dx}(B e^{ax} \sin bx)$
$\frac{d^2y}{dx^2} = A(ae^{ax} \cos bx + e^{ax}(-b \sin bx)) + B(ae^{ax} \sin bx + e^{ax}(b \cos bx))$
$\frac{d^2y}{dx^2} = Aae^{ax} \cos bx - Abe^{ax} \sin bx + Bae^{ax} \sin bx + Bbe^{ax} \cos bx$
Grouping terms with $\cos bx$ and $\sin bx$:
$\frac{d^2y}{dx^2} = (Aa + Bb) e^{ax} \cos bx + (Ba - Ab) e^{ax} \sin bx$
Substitute back $A = (c_1 a + c_2 b)$ and $B = (c_2 a - c_1 b)$:
$Aa + Bb = (c_1 a + c_2 b)a + (c_2 a - c_1 b)b = c_1 a^2 + c_2 ab + c_2 ab - c_1 b^2 = c_1 a^2 + 2c_2 ab - c_1 b^2$
$Ba - Ab = (c_2 a - c_1 b)a - (c_1 a + c_2 b)b = c_2 a^2 - c_1 ab - (c_1 ab + c_2 b^2) = c_2 a^2 - c_1 ab - c_1 ab - c_2 b^2 = c_2 a^2 - 2c_1 ab - c_2 b^2$
$\mathbf{\frac{d^2y}{dx^2} = (c_1 a^2 + 2c_2 ab - c_1 b^2) e^{ax} \cos bx + (c_2 a^2 - 2c_1 ab - c_2 b^2) e^{ax} \sin bx}$
Now substitute $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the left side of the differential equation: $\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y$
LHS = $\Big[ (c_1 a^2 + 2c_2 ab - c_1 b^2) e^{ax} \cos bx + (c_2 a^2 - 2c_1 ab - c_2 b^2) e^{ax} \sin bx \Big]$
$- 2a \Big[ (c_1 a + c_2 b) e^{ax} \cos bx + (c_2 a - c_1 b) e^{ax} \sin bx \Big]$
$+ (a^2 + b^2) \Big[ c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx \Big]$
Collect the terms multiplying $e^{ax} \cos bx$:
$(c_1 a^2 + 2c_2 ab - c_1 b^2) - 2a(c_1 a + c_2 b) + (a^2 + b^2)c_1$
$= c_1 a^2 + 2c_2 ab - c_1 b^2 - 2c_1 a^2 - 2c_2 ab + c_1 a^2 + c_1 b^2$
$= (c_1 a^2 - 2c_1 a^2 + c_1 a^2) + (2c_2 ab - 2c_2 ab) + (-c_1 b^2 + c_1 b^2) = 0$
Collect the terms multiplying $e^{ax} \sin bx$:
$(c_2 a^2 - 2c_1 ab - c_2 b^2) - 2a(c_2 a - c_1 b) + (a^2 + b^2)c_2$
$= c_2 a^2 - 2c_1 ab - c_2 b^2 - 2c_2 a^2 + 2c_1 ab + c_2 a^2 + c_2 b^2$
$= (c_2 a^2 - 2c_2 a^2 + c_2 a^2) + (-2c_1 ab + 2c_1 ab) + (-c_2 b^2 + c_2 b^2) = 0$
So, LHS = $0 \cdot e^{ax} \cos bx + 0 \cdot e^{ax} \sin bx = 0$.
LHS = $0$ = RHS.
Since substituting the function and its derivatives into the differential equation results in $0$, which is the right side of the equation, the function $y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$ is indeed a solution to the given differential equation.
Hence Verified.
Example 25: Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.
Answer:
The family of circles in the second quadrant touching the coordinate axes have their center at a point $(-a, a)$ and radius $a$, where $a$ is a positive constant.
The equation of such a circle is given by:
$(x - (-a))^2 + (y - a)^2 = a^2$
$(x + a)^2 + (y - a)^2 = a^2$
This is the general equation of the family of circles, where $a$ is the arbitrary constant to be eliminated.
Since there is only one arbitrary constant, we need to differentiate the equation once with respect to $x$ to form the differential equation.
Differentiate $(x + a)^2 + (y - a)^2 = a^2$ with respect to $x$:
$\frac{d}{dx}((x + a)^2) + \frac{d}{dx}((y - a)^2) = \frac{d}{dx}(a^2)$
Applying the chain rule for the terms involving $y$ (where $\frac{dy}{dx} = y'$):
$2(x + a) \cdot 1 + 2(y - a) \cdot \frac{dy}{dx} = 0$
$(x + a) + (y - a) \frac{dy}{dx} = 0$
Now, we need to eliminate the constant $a$ from this differentiated equation and the original equation. From the differentiated equation, solve for $a$ (assuming $\frac{dy}{dx} \neq 1$):
$x + a = -(y - a) \frac{dy}{dx}$
$x + a = -y \frac{dy}{dx} + a \frac{dy}{dx}$
$x + y \frac{dy}{dx} = a \frac{dy}{dx} - a$
$x + y \frac{dy}{dx} = a (\frac{dy}{dx} - 1)$
$\mathbf{a = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx} - 1}}$
Substitute this expression for $a$ back into the original equation $(x + a)^2 + (y - a)^2 = a^2$.
Let $y' = \frac{dy}{dx}$ for simplicity in the substitution.
$(x + \frac{x + yy'}{y' - 1})^2 + (y - \frac{x + yy'}{y' - 1})^2 = (\frac{x + yy'}{y' - 1})^2$
Simplify the terms inside the parentheses:
$x + \frac{x + yy'}{y' - 1} = \frac{x(y' - 1) + (x + yy')}{y' - 1} = \frac{xy' - x + x + yy'}{y' - 1} = \frac{xy' + yy'}{y' - 1} = \frac{y'(x+y)}{y' - 1}$
$y - \frac{x + yy'}{y' - 1} = \frac{y(y' - 1) - (x + yy')}{y' - 1} = \frac{yy' - y - x - yy'}{y' - 1} = \frac{-x - y}{y' - 1} = -\frac{x+y}{y' - 1}$
Substitute these back into the equation:
$(\frac{y'(x+y)}{y' - 1})^2 + (-\frac{x+y}{y' - 1})^2 = (\frac{x + yy'}{y' - 1})^2$
$\frac{(y')^2 (x+y)^2}{(y' - 1)^2} + \frac{(x+y)^2}{(y' - 1)^2} = \frac{(x + yy')^2}{(y' - 1)^2}$
Assuming $y' \neq 1$, multiply both sides by $(y' - 1)^2$:
$(y')^2 (x+y)^2 + (x+y)^2 = (x + yy')^2$
$(x+y)^2 ((y')^2 + 1) = (x + yy')^2$
Replace $y'$ with $\frac{dy}{dx}$ to write the differential equation:
$\mathbf{(x+y)^2 \left( \left(\frac{dy}{dx}\right)^2 + 1 \right) = \left(x + y\frac{dy}{dx}\right)^2}$
This is the required differential equation of the family of circles in the second quadrant touching the coordinate axes.
Example 26: Find the particular solution of the differential equation $\log \left( \frac{dy}{dx} \right) = 3x + 4y$ given that y = 0 when x = 0.
Answer:
The given differential equation is $\log \left( \frac{dy}{dx} \right) = 3x + 4y$.
This is a logarithmic equation involving the derivative. We can rewrite it in exponential form.
$\frac{dy}{dx} = e^{3x + 4y}$
Using the property of exponents $e^{a+b} = e^a \cdot e^b$, we can separate the terms involving $x$ and $y$ on the right side.
$\frac{dy}{dx} = e^{3x} \cdot e^{4y}$
This is a separable differential equation. We can separate the variables by moving all terms involving $y$ to one side with $dy$ and all terms involving $x$ to the other side with $dx$.
Divide both sides by $e^{4y}$ (or multiply by $e^{-4y}$) and multiply both sides by $dx$:
$\frac{dy}{e^{4y}} = e^{3x} dx$
Rewrite the left side:
$e^{-4y} dy = e^{3x} dx$
Now, integrate both sides of the equation:
$\int e^{-4y} dy = \int e^{3x} dx$
Evaluate the integrals:
$\int e^{-4y} dy = -\frac{1}{4} e^{-4y} + C_1$
$\int e^{3x} dx = \frac{1}{3} e^{3x} + C_2$
So, the general solution is:
$-\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} + C$
where $C = C_2 - C_1$ is the constant of integration.
We are given the initial condition that $y = 0$ when $x = 0$. We use this condition to find the particular solution, which means finding the specific value of $C$.
Substitute $x=0$ and $y=0$ into the general solution:
$-\frac{1}{4} e^{-4(0)} = \frac{1}{3} e^{3(0)} + C$
$-\frac{1}{4} e^0 = \frac{1}{3} e^0 + C$
$-\frac{1}{4} (1) = \frac{1}{3} (1) + C$
$-\frac{1}{4} = \frac{1}{3} + C$
Solve for $C$:
$C = -\frac{1}{4} - \frac{1}{3}$
Find a common denominator (12):
$C = -\frac{3}{12} - \frac{4}{12}$
$\mathbf{C = -\frac{7}{12}}$
Substitute the value of $C$ back into the general solution to obtain the particular solution:
$-\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} - \frac{7}{12}$
Multiply the entire equation by $-12$ to eliminate fractions and move terms:
$-12 \left(-\frac{1}{4} e^{-4y}\right) = -12 \left(\frac{1}{3} e^{3x}\right) - 12 \left(-\frac{7}{12}\right)$
$\frac{\cancel{-12}^{3}}{\cancel{-4}_{1}} e^{-4y} = \frac{\cancel{-12}^{4}}{\cancel{3}_{1}} e^{3x} + \frac{\cancel{12}^{1}}{\cancel{12}_{1}} (7)$
$\mathbf{3 e^{-4y} = -4 e^{3x} + 7}$
We can also write it as:
$\mathbf{3 e^{-4y} + 4 e^{3x} = 7}$
The particular solution of the differential equation passing through (0, 0) is $3 e^{-4y} + 4 e^{3x} = 7$.
Example 27: Solve the differential equation
$(x \;dy \;–\; y \;dx) \;y \sin \left( \frac{y}{x} \right) = (y \;dx + x \;dy) \;x \cos \left( \frac{y}{x} \right)$ .
Answer:
The given differential equation is $(x \;dy \;–\; y \;dx) \;y \sin \left( \frac{y}{x} \right) = (y \;dx + x \;dy) \;x \cos \left( \frac{y}{x} \right)$.
Expand the terms:
$(xy \;dy - y^2 \;dx) \sin \left( \frac{y}{x} \right) = (xy \;dx + x^2 \;dy) \cos \left( \frac{y}{x} \right)$
Rearrange the equation by grouping the terms with $dx$ and $dy$:
$(xy \sin \left( \frac{y}{x} \right)) dy - (y^2 \sin \left( \frac{y}{x} \right)) dx = (xy \cos \left( \frac{y}{x} \right)) dx + (x^2 \cos \left( \frac{y}{x} \right)) dy$
$(xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)) dy = (xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)) dx$
From this, we can write $\frac{dy}{dx}$ as:
$\frac{dy}{dx} = \frac{xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)}{xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)}$
Divide the numerator and the denominator by $x^2$:
$\frac{dy}{dx} = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + \left( \frac{y}{x} \right)^2 \sin \left( \frac{y}{x} \right)}{\frac{y}{x} \sin \left( \frac{y}{x} \right) - \cos \left( \frac{y}{x} \right)}$
This is a homogeneous differential equation, as $\frac{dy}{dx}$ is a function of $\frac{y}{x}$ only.
Let $y = vx$, where $v$ is a function of $x$. Then, differentiate with respect to $x$ using the product rule:
$\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + \left( \frac{y}{x} \right)^2 \sin \left( \frac{y}{x} \right)}{\frac{y}{x} \sin \left( \frac{y}{x} \right) - \cos \left( \frac{y}{x} \right)}$:
$v + x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}$
Separate the variables $v$ and $x$:
$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v$
$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v}$
$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v}$
$x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v}$
Now, separate the variables completely:
$\frac{v \sin v - \cos v}{2v \cos v} dv = \frac{dx}{x}$
Split the fraction on the left side:
$\left( \frac{v \sin v}{2v \cos v} - \frac{\cos v}{2v \cos v} \right) dv = \frac{dx}{x}$
$\left( \frac{1}{2} \tan v - \frac{1}{2v} \right) dv = \frac{dx}{x}$
$\frac{1}{2} \tan v \, dv - \frac{1}{2v} dv = \frac{dx}{x}$
Integrate both sides:
$\int \limits \left( \frac{1}{2} \tan v - \frac{1}{2v} \right) dv = \int \limits \frac{1}{x} dx$
$\frac{1}{2} \int \limits \tan v \, dv - \frac{1}{2} \int \limits \frac{1}{v} dv = \int \limits \frac{1}{x} dx$
Recall that $\int \tan v \, dv = -\ln|\cos v|$ and $\int \frac{1}{v} dv = \ln|v|$, $\int \frac{1}{x} dx = \ln|x|$.
$\frac{1}{2} (-\ln|\cos v|) - \frac{1}{2} \ln|v| = \ln|x| + C'$
$-\frac{1}{2} (\ln|\cos v| + \ln|v|) = \ln|x| + C'$
Using logarithm properties $\ln A + \ln B = \ln(AB)$ and $a \ln A = \ln(A^a)$:
$-\frac{1}{2} \ln|v \cos v| = \ln|x| + C'$
$\ln|v \cos v|^{-1/2} = \ln|x| + C'$
$\ln \left| \frac{1}{\sqrt{v \cos v}} \right| = \ln|x| + C'$
Exponentiate both sides:
$\left| \frac{1}{\sqrt{v \cos v}} \right| = e^{\ln|x| + C'}$
$\frac{1}{\sqrt{|v \cos v|}} = e^{C'} |x|$
Let $e^{C'} = K_1 > 0$. Square both sides:
$\frac{1}{|v \cos v|} = K_1^2 x^2$
$|v \cos v| = \frac{1}{K_1^2 x^2} = K_2 x^{-2}$
where $K_2 = 1/K_1^2$ is a positive constant. Removing the absolute value, we introduce an arbitrary non-zero constant $K = \pm K_2$.
$\mathbf{v \cos v = K x^{-2}}$
Substitute back $v = \frac{y}{x}$:
$\frac{y}{x} \cos \left( \frac{y}{x} \right) = K x^{-2}$
Multiply by $x$ to simplify:
$\mathbf{y \cos \left( \frac{y}{x} \right) = K x^{-1}}$
This can also be written as:
$\mathbf{y \cos \left( \frac{y}{x} \right) = \frac{K}{x}}$
Multiplying by $x$ gives a common form:
$\mathbf{xy \cos \left( \frac{y}{x} \right) = K}$
where $K$ is an arbitrary constant (including 0, which corresponds to singular solutions like $y=0$).
Example 28: Solve the differential equation
(tan–1 y – x) dy = (1 + y2) dx.
Answer:
The given differential equation is $(\tan^{-1} y - x) dy = (1 + y^2) dx$.
We can rewrite this equation in the form $\frac{dx}{dy}$. Divide both sides by $(1 + y^2) dy$:
$\frac{(\tan^{-1} y - x) dy}{(1 + y^2) dy} = \frac{(1 + y^2) dx}{(1 + y^2) dy}$
$\frac{\tan^{-1} y - x}{1 + y^2} = \frac{dx}{dy}$
Rearrange the equation to get it into the standard form of a linear differential equation in $x$ as a function of $y$, which is $\frac{dx}{dy} + P(y)x = Q(y)$.
$\frac{dx}{dy} = \frac{\tan^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$
$\mathbf{\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan^{-1} y}{1 + y^2}}$
Comparing this to the standard form, we identify $P(y) = \frac{1}{1 + y^2}$ and $Q(y) = \frac{\tan^{-1} y}{1 + y^2}$.
The Integrating Factor (IF) is given by $e^{\int P(y) dy}$.
IF = $e^{\int \frac{1}{1 + y^2} dy}$
Evaluate the integral:
$\int \frac{1}{1 + y^2} dy = \tan^{-1} y$
So, the Integrating Factor is:
$\mathbf{IF = e^{\tan^{-1} y}}$
Multiply the linear differential equation $\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan^{-1} y}{1 + y^2}$ by the Integrating Factor $e^{\tan^{-1} y}$:
$e^{\tan^{-1} y} \left( \frac{dx}{dy} + \frac{1}{1 + y^2} x \right) = e^{\tan^{-1} y} \left( \frac{\tan^{-1} y}{1 + y^2} \right)$
The left side of the equation is the derivative of the product of the dependent variable ($x$) and the Integrating Factor:
$\frac{d}{dy} (x \cdot e^{\tan^{-1} y}) = \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y}$
Now, integrate both sides with respect to $y$:
$\int \frac{d}{dy} (x \cdot e^{\tan^{-1} y}) dy = \int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy$
The left side integral is simply the function inside the derivative:
$x \cdot e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy$
To evaluate the integral on the right side, use the substitution method. Let $u = \tan^{-1} y$. Then, the differential $du = \frac{1}{1 + y^2} dy$.
The integral becomes $\int u e^u du$.
Use integration by parts, $\int v \, dw = vw - \int w \, dv$, with $v = u$ and $dw = e^u du$. Then $dv = du$ and $w = e^u$.
$\int u e^u du = u e^u - \int e^u du$
$= u e^u - e^u + C$
Substitute back $u = \tan^{-1} y$:
$\int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy = (\tan^{-1} y) e^{\tan^{-1} y} - e^{\tan^{-1} y} + C$
Substitute this result back into the integrated differential equation:
$x \cdot e^{\tan^{-1} y} = (\tan^{-1} y) e^{\tan^{-1} y} - e^{\tan^{-1} y} + C$
Finally, solve for $x$. Divide the entire equation by $e^{\tan^{-1} y}$ (since $e^{\tan^{-1} y} > 0$):
$\frac{x \cdot e^{\tan^{-1} y}}{e^{\tan^{-1} y}} = \frac{(\tan^{-1} y) e^{\tan^{-1} y}}{e^{\tan^{-1} y}} - \frac{e^{\tan^{-1} y}}{e^{\tan^{-1} y}} + \frac{C}{e^{\tan^{-1} y}}$
$\mathbf{x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}}$
This is the general solution to the given differential equation.
Miscellaneous Exercise on Chapter 9
Question 1. For each of the differential equations given below, indicate its order and degree (if defined).
(i) $\frac{d^2y}{dx^2} + 5x \left( \frac{dy}{dx} \right)^2 - 6y = \log x$
(ii) $\left( \frac{dy}{dx} \right)^3 - 4\left( \frac{dy}{dx} \right)^2 + 7y = \sin x$
(iii) $\frac{d^4y}{dx^4} - \sin \left( \frac{d^3y}{dx^3} \right) = 0$
Answer:
(i) The given differential equation is $\frac{d^2y}{dx^2} + 5x \left( \frac{dy}{dx} \right)^2 - 6y = \log x$.
The highest order derivative present in the equation is $\frac{d^2y}{dx^2}$.
Therefore, the order of the differential equation is 2.
The given equation is a polynomial in derivatives $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$.
The highest power of the highest order derivative ($\frac{d^2y}{dx^2}$) is 1.
Therefore, the degree of the differential equation is 1.
(ii) The given differential equation is $\left( \frac{dy}{dx} \right)^3 - 4\left( \frac{dy}{dx} \right)^2 + 7y = \sin x$.
The highest order derivative present in the equation is $\frac{dy}{dx}$.
Therefore, the order of the differential equation is 1.
The given equation is a polynomial in the derivative $\frac{dy}{dx}$.
The highest power of the highest order derivative ($\frac{dy}{dx}$) is 3.
Therefore, the degree of the differential equation is 3.
(iii) The given differential equation is $\frac{d^4y}{dx^4} - \sin \left( \frac{d^3y}{dx^3} \right) = 0$.
The highest order derivative present in the equation is $\frac{d^4y}{dx^4}$.
Therefore, the order of the differential equation is 4.
The given equation is not a polynomial in derivatives due to the presence of the term $\sin \left( \frac{d^3y}{dx^3} \right)$, where the derivative is inside a transcendental function (sine).
Therefore, the degree of the differential equation is not defined.
Question 2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i)
$xy = a e^x + b e^{-x} + x^2$
:
$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2y = 0$
(ii)
$y = e^x (a \cos x + b \sin x)$
:
$\frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$
(iii)
$y = x \sin 3x$
:
$\frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$
(iv)
$x^2 = 2y^2 \log y$
:
$(x^2 + y^2) \frac{dy}{dx} - xy = 0$
Answer:
(i) Given function: $xy = a e^x + b e^{-x} + x^2$
Given differential equation: $x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2y = 0$
Differentiate the given function implicitly with respect to $x$:
$y + x \frac{dy}{dx} = a e^x - b e^{-x} + 2x$
... (1)
Differentiate equation (1) with respect to $x$:
$\frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = a e^x + b e^{-x} + 2$
$2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = a e^x + b e^{-x} + 2$
... (2)
From the original function, we have $a e^x + b e^{-x} = xy - x^2$. Substitute this into equation (2):
$2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = (xy - x^2) + 2$
Rearrange the terms to match the differential equation:
$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$
This does not exactly match the given differential equation $x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2y = 0$. There seems to be a discrepancy in the coefficient of the constant term or $y$. Let's recheck the differentiation.
Let's use equation (1) and (2) more carefully.
$y + x \frac{dy}{dx} = a e^x - b e^{-x} + 2x$
... (1)
$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = a e^x + b e^{-x} + 2$
... (2)
From the original function: $a e^x + b e^{-x} = xy - x^2$.
Substitute this into the right side of equation (2):
$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = (xy - x^2) + 2$
Rearranging terms:
$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$
Comparing with the given differential equation $x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2y = 0$, there is a term $-2$ instead of $-2y$. This suggests a possible error in the question's stated differential equation or the provided solution function.
Assuming the question's differential equation is correct, the given function is not a solution. Assuming the function is correct, the differential equation should be $x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$.
Let's assume there is a typo in the question and the differential equation should indeed have a $-2$ instead of $-2y$. In that case, the verification shows:
$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = (xy - x^2 + 2) - xy + x^2 - 2 = 0$
If we strictly follow the provided differential equation, the verification fails.
Let's proceed with the given differential equation and see if the function fits. Substituting the derivatives into the given differential equation:
LHS = $x \left( \frac{xy - x^2 + 2 - 2 \frac{dy}{dx}}{x} \right) + 2 \frac{dy}{dx} - xy + x^2 - 2y$
This becomes circular. Let's use equations (1) and (2) directly.
From (2), $x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = a e^x + b e^{-x} + 2$
Substitute this into the LHS of the given differential equation:
LHS = $(a e^x + b e^{-x} + 2) - xy + x^2 - 2y$
Substitute $a e^x + b e^{-x} = xy - x^2$ from the original function:
LHS = $(xy - x^2 + 2) - xy + x^2 - 2y$
LHS = $xy - x^2 + 2 - xy + x^2 - 2y$
LHS = $2 - 2y$
For the function to be a solution, the LHS must be 0. So, we would need $2 - 2y = 0$, which means $y=1$. However, $y$ is a function of $x$, not a constant 1. Thus, the LHS $2-2y$ is generally not equal to 0.
There appears to be an inconsistency in the problem statement for part (i). Assuming the function is correct, the differential equation is likely incorrect.
Given the instructions are to verify, we must show that the substitution leads to the stated equation being true. Since it does not, the given function is not a solution to the given differential equation.
(ii) Given function: $y = e^x (a \cos x + b \sin x)$
Given differential equation: $\frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$
Calculate the first derivative with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (e^x (a \cos x + b \sin x))$
$\frac{dy}{dx} = e^x (a \cos x + b \sin x) + e^x (-a \sin x + b \cos x)$
$\frac{dy}{dx} = y + e^x (-a \sin x + b \cos x)$
... (1)
Calculate the second derivative with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} (y + e^x (-a \sin x + b \cos x))$
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{d}{dx} (e^x (-a \sin x + b \cos x))$
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x (-a \sin x + b \cos x) + e^x (-a \cos x - b \sin x)$
From equation (1), $e^x (-a \sin x + b \cos x) = \frac{dy}{dx} - y$. Substitute this and the original function $y = e^x (a \cos x + b \sin x)$ into the expression for $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^x (a \cos x + b \sin x)$
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{dy}{dx} - y - y$
$\frac{d^2y}{dx^2} = 2 \frac{dy}{dx} - 2y$
Rearrange the terms:
$\frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$
This matches the given differential equation.
Therefore, the given function is a solution to the corresponding differential equation.
(iii) Given function: $y = x \sin 3x$
Given differential equation: $\frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$
Calculate the first derivative with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (x \sin 3x)$
$\frac{dy}{dx} = 1 \cdot \sin 3x + x \cdot (3 \cos 3x)$
$\frac{dy}{dx} = \sin 3x + 3x \cos 3x$
Calculate the second derivative with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} (\sin 3x + 3x \cos 3x)$
$\frac{d^2y}{dx^2} = (3 \cos 3x) + (3 \cdot \cos 3x + 3x \cdot (-3 \sin 3x))$
$\frac{d^2y}{dx^2} = 3 \cos 3x + 3 \cos 3x - 9x \sin 3x$
$\frac{d^2y}{dx^2} = 6 \cos 3x - 9x \sin 3x$
Now substitute $\frac{d^2y}{dx^2}$ and $y$ into the LHS of the given differential equation:
LHS = $\frac{d^2y}{dx^2} + 9y - 6\cos 3x$
LHS = $(6 \cos 3x - 9x \sin 3x) + 9(x \sin 3x) - 6\cos 3x$
LHS = $6 \cos 3x - 9x \sin 3x + 9x \sin 3x - 6\cos 3x$
LHS = $(6 \cos 3x - 6 \cos 3x) + (-9x \sin 3x + 9x \sin 3x)$
LHS = $0 + 0 = 0$
This matches the RHS of the given differential equation.
Therefore, the given function is a solution to the corresponding differential equation.
(iv) Given function: $x^2 = 2y^2 \log y$
Given differential equation: $(x^2 + y^2) \frac{dy}{dx} - xy = 0$
Differentiate the given function implicitly with respect to $x$:
$\frac{d}{dx} (x^2) = \frac{d}{dx} (2y^2 \log y)$
$2x = 2 \left( \frac{d}{dx}(y^2) \log y + y^2 \frac{d}{dx}(\log y) \right)$
$2x = 2 \left( (2y \frac{dy}{dx}) \log y + y^2 \frac{1}{y} \frac{dy}{dx} \right)$
$2x = 2 \left( 2y \log y \frac{dy}{dx} + y \frac{dy}{dx} \right)$
$2x = 2y \frac{dy}{dx} (2 \log y + 1)$
Divide by 2 on both sides:
$x = y \frac{dy}{dx} (2 \log y + 1)$
From the original function, $\log y = \frac{x^2}{2y^2}$. Substitute this into the equation:
$x = y \frac{dy}{dx} \left( 2 \left( \frac{x^2}{2y^2} \right) + 1 \right)$
$x = y \frac{dy}{dx} \left( \frac{x^2}{y^2} + 1 \right)$
$x = y \frac{dy}{dx} \left( \frac{x^2 + y^2}{y^2} \right)$
$x = \frac{1}{y} \frac{dy}{dx} (x^2 + y^2)$
Rearrange the terms to isolate $\frac{dy}{dx}$ or to match the differential equation form:
$xy = (x^2 + y^2) \frac{dy}{dx}$
Rearrange to match the given differential equation:
$(x^2 + y^2) \frac{dy}{dx} - xy = 0$
This matches the given differential equation.
Therefore, the given function is a solution to the corresponding differential equation.
Question 3. Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2 , where a is an arbitrary constant.
Answer:
Given:
The family of curves is given by the equation:
$(x - a)^2 + 2y^2 = a^2$
...(i)
where 'a' is an arbitrary constant.
To Find:
The differential equation representing the given family of curves.
Solution:
First, expand the given equation (i):
$x^2 - 2ax + a^2 + 2y^2 = a^2$
Simplifying the equation by canceling $a^2$ from both sides:
$x^2 - 2ax + 2y^2 = 0$
$x^2 + 2y^2 = 2ax$
...(ii)
Now, differentiate equation (ii) with respect to $x$:
$\frac{d}{dx}(x^2 + 2y^2) = \frac{d}{dx}(2ax)$
$2x + 2(2y)\frac{dy}{dx} = 2a(1)$
$2x + 4y\frac{dy}{dx} = 2a$
Divide by 2:
$x + 2y\frac{dy}{dx} = a$
...(iii)
Now, substitute the value of 'a' from equation (iii) into equation (ii) to eliminate the arbitrary constant 'a':
$x^2 + 2y^2 = 2x \left( x + 2y\frac{dy}{dx} \right)$
$x^2 + 2y^2 = 2x^2 + 4xy\frac{dy}{dx}$
Rearrange the terms to find $\frac{dy}{dx}$:
$2y^2 - 2x^2 = 4xy\frac{dy}{dx}$
$4xy\frac{dy}{dx} = 2(y^2 - x^2)$
$\frac{dy}{dx} = \frac{2(y^2 - x^2)}{4xy}$
$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$
Alternatively, we can write it as:
$4xy\frac{dy}{dx} = 2y^2 - 2x^2$
$2xy\frac{dy}{dx} = y^2 - x^2$
Alternatively using equation (i):
Differentiate equation (i) with respect to $x$:
$\frac{d}{dx}((x - a)^2 + 2y^2) = \frac{d}{dx}(a^2)$
$2(x - a)(1) + 4y\frac{dy}{dx} = 0$
$2(x - a) = -4y\frac{dy}{dx}$
$x - a = -2y\frac{dy}{dx}$
$a = x + 2y\frac{dy}{dx}$
...(iv)
Substitute this value of $a$ back into the original equation (i):
$(x - (x + 2y\frac{dy}{dx}))^2 + 2y^2 = (x + 2y\frac{dy}{dx})^2$
$(-2y\frac{dy}{dx})^2 + 2y^2 = x^2 + 2(x)(2y\frac{dy}{dx}) + (2y\frac{dy}{dx})^2$
$4y^2(\frac{dy}{dx})^2 + 2y^2 = x^2 + 4xy\frac{dy}{dx} + 4y^2(\frac{dy}{dx})^2$
Cancel $4y^2(\frac{dy}{dx})^2$ from both sides:
$2y^2 = x^2 + 4xy\frac{dy}{dx}$
$4xy\frac{dy}{dx} = 2y^2 - x^2$
$\frac{dy}{dx} = \frac{2y^2 - x^2}{4xy}$
Thus, the required differential equation is:
$\mathbf{\frac{dy}{dx} = \frac{2y^2 - x^2}{4xy}}$ or $\mathbf{4xy\frac{dy}{dx} = 2y^2 - x^2}$
Question 4. Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter.
Answer:
Given:
The proposed general solution is:
$x^2 - y^2 = c (x^2 + y^2)^2$
...(i)
The differential equation is:
$(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$
This can be rewritten as:
$\frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y}$
...(ii)
where $c$ is an arbitrary parameter.
To Prove:
$x^2 - y^2 = c (x^2 + y^2)^2$ is the general solution of the given differential equation.
Proof:
We will differentiate the proposed general solution (i) with respect to $x$ and show that it leads to the given differential equation (ii).
Differentiating equation (i) implicitly with respect to $x$:
$\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}[c (x^2 + y^2)^2]$
$2x - 2y\frac{dy}{dx} = c \cdot 2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2)$
$2x - 2y\frac{dy}{dx} = 2c(x^2 + y^2) (2x + 2y\frac{dy}{dx})$
Divide both sides by 2:
$x - y\frac{dy}{dx} = c(x^2 + y^2) (2x + 2y\frac{dy}{dx})$
$x - y\frac{dy}{dx} = 2c(x^2 + y^2) (x + y\frac{dy}{dx})$
From the original equation (i), we can express $c$ as:
$c = \frac{x^2 - y^2}{(x^2 + y^2)^2}$
[Provided $(x^2+y^2) \ne 0$]
Substitute this expression for $c$ back into the differentiated equation:
$x - y\frac{dy}{dx} = 2 \left( \frac{x^2 - y^2}{(x^2 + y^2)^2} \right) (x^2 + y^2) (x + y\frac{dy}{dx})$
$x - y\frac{dy}{dx} = \frac{2(x^2 - y^2)}{x^2 + y^2} (x + y\frac{dy}{dx})$
Multiply both sides by $(x^2 + y^2)$ to clear the denominator:
$(x - y\frac{dy}{dx})(x^2 + y^2) = 2(x^2 - y^2)(x + y\frac{dy}{dx})$
Expand both sides:
$x(x^2 + y^2) - y(x^2 + y^2)\frac{dy}{dx} = 2x(x^2 - y^2) + 2y(x^2 - y^2)\frac{dy}{dx}$
$x^3 + xy^2 - (x^2y + y^3)\frac{dy}{dx} = 2x^3 - 2xy^2 + (2x^2y - 2y^3)\frac{dy}{dx}$
Rearrange the terms to group $\frac{dy}{dx}$ terms on one side and other terms on the other side:
$x^3 + xy^2 - 2x^3 + 2xy^2 = (2x^2y - 2y^3)\frac{dy}{dx} + (x^2y + y^3)\frac{dy}{dx}$
$-x^3 + 3xy^2 = (2x^2y - 2y^3 + x^2y + y^3)\frac{dy}{dx}$
$-(x^3 - 3xy^2) = (3x^2y - y^3)\frac{dy}{dx}$
$(x^3 - 3xy^2) = -(3x^2y - y^3)\frac{dy}{dx}$
$(x^3 - 3xy^2) = (y^3 - 3x^2y)\frac{dy}{dx}$
Separating the differentials $dx$ and $dy$:
$(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$
This is the original differential equation given in the question.
Conclusion:
Since differentiating the given relation $x^2 - y^2 = c (x^2 + y^2)^2$ with respect to $x$ and eliminating the parameter $c$ leads to the differential equation $(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$, the given relation is indeed the general solution of the differential equation.
Question 5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Answer:
To Find:
The differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Solution:
Let a circle touch both the coordinate axes (x-axis and y-axis) in the first quadrant.
For such a circle, the distance of the center from the x-axis is equal to the distance of the center from the y-axis, and this distance is equal to the radius of the circle.
Let the center of the circle be $(h, k)$ and the radius be $r$.
Since the circle touches the x-axis, $k = r$.
Since the circle touches the y-axis, $h = r$.
Since the circle is in the first quadrant, $h > 0$ and $k > 0$.
Let $h = k = r = a$, where $a > 0$ is an arbitrary constant (the radius).
The center of the circle is $(a, a)$ and the radius is $a$.
The equation of such a circle is given by:
$(x - h)^2 + (y - k)^2 = r^2$
Substituting $h=a$, $k=a$, and $r=a$, we get the equation of the family of circles:
$(x - a)^2 + (y - a)^2 = a^2$
...(i)
This equation represents the family of circles touching both axes in the first quadrant, with 'a' as the parameter.
To find the differential equation, we need to eliminate the parameter 'a'. We differentiate equation (i) with respect to $x$:
$\frac{d}{dx}[(x - a)^2 + (y - a)^2] = \frac{d}{dx}[a^2]$
$2(x - a)\frac{d}{dx}(x - a) + 2(y - a)\frac{d}{dx}(y - a) = 0$
$2(x - a)(1) + 2(y - a)\frac{dy}{dx} = 0$
Divide by 2:
$(x - a) + (y - a)\frac{dy}{dx} = 0$
Let $y' = \frac{dy}{dx}$.
$(x - a) + (y - a)y' = 0$
$x - a + yy' - ay' = 0$
$x + yy' = a + ay'$
$x + yy' = a(1 + y')$
Now, we solve for 'a':
$a = \frac{x + yy'}{1 + y'}$
...(ii)
Substitute this expression for 'a' back into the equation $(x - a) + (y - a)y' = 0$ (or equation (i)). Let's use $(x - a) + (y - a)y' = 0$ as it's simpler, but substituting into (i) is the standard procedure.
Calculate $(x-a)$ and $(y-a)$ using equation (ii):
$x - a = x - \frac{x + yy'}{1 + y'} = \frac{x(1 + y') - (x + yy')}{1 + y'} = \frac{x + xy' - x - yy'}{1 + y'} = \frac{xy' - yy'}{1 + y'} = \frac{y'(x - y)}{1 + y'}$
$y - a = y - \frac{x + yy'}{1 + y'} = \frac{y(1 + y') - (x + yy')}{1 + y'} = \frac{y + yy' - x - yy'}{1 + y'} = \frac{y - x}{1 + y'}$
Now substitute these into the original equation (i): $(x - a)^2 + (y - a)^2 = a^2$
$\left(\frac{y'(x - y)}{1 + y'}\right)^2 + \left(\frac{y - x}{1 + y'}\right)^2 = \left(\frac{x + yy'}{1 + y'}\right)^2$
$\frac{(y')^2(x - y)^2}{(1 + y')^2} + \frac{(y - x)^2}{(1 + y')^2} = \frac{(x + yy')^2}{(1 + y')^2}$
Since $(y - x)^2 = (x - y)^2$, we can combine the terms on the left:
$\frac{(x - y)^2 [(y')^2 + 1]}{(1 + y')^2} = \frac{(x + yy')^2}{(1 + y')^2}$
Assuming $(1 + y')^2 \neq 0$, we can multiply both sides by $(1 + y')^2$:
$(x - y)^2 (1 + (y')^2) = (x + yy')^2$
Replacing $y'$ with $\frac{dy}{dx}$:
$(x - y)^2 \left(1 + \left(\frac{dy}{dx}\right)^2\right) = \left(x + y\frac{dy}{dx}\right)^2$
Conclusion:
The required differential equation for the family of circles in the first quadrant touching the coordinate axes is:
$\mathbf{(x - y)^2 \left[1 + \left(\frac{dy}{dx}\right)^2\right] = \left(x + y\frac{dy}{dx}\right)^2}$
Question 6. Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 − y^2}{1 − x^2}} = 0$.
Answer:
Given:
The differential equation is:
$\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$
To Find:
The general solution of the given differential equation.
Solution:
The given differential equation can be rewritten as:
$\frac{dy}{dx} = - \sqrt{\frac{1 - y^2}{1 - x^2}}$
$\frac{dy}{dx} = - \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}$
This is a variable separable differential equation. We can separate the variables $x$ and $y$ as follows:
$\frac{dy}{\sqrt{1 - y^2}} = - \frac{dx}{\sqrt{1 - x^2}}$
Integrating both sides with respect to their respective variables:
$\int \frac{1}{\sqrt{1 - y^2}} dy = \int - \frac{1}{\sqrt{1 - x^2}} dx$
$\int \frac{1}{\sqrt{1 - y^2}} dy = - \int \frac{1}{\sqrt{1 - x^2}} dx$
We know the standard integral $\int \frac{1}{\sqrt{1 - u^2}} du = \arcsin(u)$. Applying this formula:
$\arcsin(y) = - \arcsin(x) + C$
where $C$ is the constant of integration.
Rearranging the terms, we get the general solution:
$\arcsin(y) + \arcsin(x) = C$
Conclusion:
The general solution of the given differential equation $\frac{dy}{dx} + \sqrt{\frac{1 − y^2}{1 − x^2}} = 0$ is:
$\mathbf{\arcsin(x) + \arcsin(y) = C}$
Question 7. Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.
Answer:
Given:
The differential equation is:
$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$
...(i)
The proposed general solution is:
$(x + y + 1) = A (1 - x - y - 2xy)$
...(ii)
where A is a parameter.
To Show:
Equation (ii) is the general solution of the differential equation (i).
Solution:
We start by solving the given differential equation (i).
Rewrite the equation as:
$\frac{dy}{dx} = - \frac{y^2 + y + 1}{x^2 + x + 1}$
This is a variable separable equation. Separating the variables:
$\frac{dy}{y^2 + y + 1} = - \frac{dx}{x^2 + x + 1}$
Integrate both sides:
$\int \frac{1}{y^2 + y + 1} dy = - \int \frac{1}{x^2 + x + 1} dx$
To evaluate the integral $\int \frac{1}{u^2 + u + 1} du$, we complete the square in the denominator:
$u^2 + u + 1 = u^2 + u + (\frac{1}{2})^2 - (\frac{1}{2})^2 + 1 = (u + \frac{1}{2})^2 + 1 - \frac{1}{4} = (u + \frac{1}{2})^2 + \frac{3}{4} = (u + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2$
The integral becomes:
$\int \frac{1}{(u + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} du$
Using the standard integral formula $\int \frac{1}{z^2 + a^2} dz = \frac{1}{a} \arctan(\frac{z}{a})$, with $z = u + \frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$:
$\int \frac{1}{u^2 + u + 1} du = \frac{1}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{u + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \arctan\left(\frac{2u + 1}{\sqrt{3}}\right)$
Applying this result to our integration:
$\frac{2}{\sqrt{3}} \arctan\left(\frac{2y + 1}{\sqrt{3}}\right) = - \frac{2}{\sqrt{3}} \arctan\left(\frac{2x + 1}{\sqrt{3}}\right) + C_1$
Divide by $\frac{2}{\sqrt{3}}$:
$\arctan\left(\frac{2y + 1}{\sqrt{3}}\right) = - \arctan\left(\frac{2x + 1}{\sqrt{3}}\right) + C_2 \quad$ (where $C_2 = \frac{\sqrt{3}}{2} C_1$ is an arbitrary constant)
Rearrange the terms:
$\arctan\left(\frac{2y + 1}{\sqrt{3}}\right) + \arctan\left(\frac{2x + 1}{\sqrt{3}}\right) = C_2$
Use the arctangent addition formula: $\arctan(P) + \arctan(Q) = \arctan\left(\frac{P+Q}{1-PQ}\right)$
Let $P = \frac{2y+1}{\sqrt{3}}$ and $Q = \frac{2x+1}{\sqrt{3}}$.
$P + Q = \frac{2y+1 + 2x+1}{\sqrt{3}} = \frac{2x + 2y + 2}{\sqrt{3}}$
$P \cdot Q = \frac{(2y+1)(2x+1)}{(\sqrt{3})^2} = \frac{4xy + 2x + 2y + 1}{3}$
$1 - PQ = 1 - \frac{4xy + 2x + 2y + 1}{3} = \frac{3 - (4xy + 2x + 2y + 1)}{3} = \frac{2 - 4xy - 2x - 2y}{3}$
$\frac{P+Q}{1-PQ} = \frac{\frac{2(x+y+1)}{\sqrt{3}}}{\frac{2(1 - 2xy - x - y)}{3}} = \frac{2(x+y+1)}{\sqrt{3}} \times \frac{3}{2(1 - x - y - 2xy)} = \frac{\sqrt{3}(x+y+1)}{1 - x - y - 2xy}$
Substituting back into the equation:
$\arctan\left(\frac{\sqrt{3}(x+y+1)}{1 - x - y - 2xy}\right) = C_2$
Take the tangent of both sides:
$\frac{\sqrt{3}(x+y+1)}{1 - x - y - 2xy} = \tan(C_2)$
Let $K = \tan(C_2)$. Since $C_2$ is an arbitrary constant, $K$ is also an arbitrary constant.
$\sqrt{3}(x+y+1) = K(1 - x - y - 2xy)$
Divide by $\sqrt{3}$:
$x+y+1 = \frac{K}{\sqrt{3}}(1 - x - y - 2xy)$
Let $A = \frac{K}{\sqrt{3}}$. Since $K$ is an arbitrary constant, $A$ is also an arbitrary constant (parameter).
$x+y+1 = A(1 - x - y - 2xy)$
This derived solution matches the form given in equation (ii).
Conclusion:
By solving the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$, we obtained the general solution $(x + y + 1) = A (1 – x – y – 2xy)$, where A is an arbitrary parameter. This confirms that the given expression is indeed the general solution of the differential equation.
Question 8. Find the equation of the curve passing through the point $\left( 0, \frac{π}{4} \right)$ whose differential equation is sin x cos y dx + cos x sin y dy = 0.
Answer:
Given:
The differential equation of the curve is:
$ \sin x \cos y dx + \cos x \sin y dy = 0 $
The curve passes through the point $\left( 0, \frac{\pi}{4} \right)$.
To Find:
The equation of the particular curve.
Solution:
The given differential equation is:
$\sin x \cos y dx + \cos x \sin y dy = 0$
Rearranging the terms:
$\cos x \sin y dy = - \sin x \cos y dx$
This is a variable separable differential equation. We can separate the variables by dividing both sides by $\cos x \cos y$ (assuming $\cos x \neq 0$ and $\cos y \neq 0$):
$\frac{\sin y}{\cos y} dy = - \frac{\sin x}{\cos x} dx$
$\tan y dy = - \tan x dx$
Now, integrate both sides:
$\int \tan y dy = - \int \tan x dx$
We know that $\int \tan u du = \ln|\sec u| + C$ or $\int \tan u du = -\ln|\cos u| + C$. Using the second form:
$-\ln|\cos y| = - (-\ln|\cos x|) + C_1$
$-\ln|\cos y| = \ln|\cos x| + C_1$
Rearranging the terms:
$-\ln|\cos y| - \ln|\cos x| = C_1$
$-(\ln|\cos y| + \ln|\cos x|) = C_1$
$\ln|\cos x \cos y| = -C_1$
Let $C = -C_1$.
$\ln|\cos x \cos y| = C$
Exponentiating both sides:
$|\cos x \cos y| = e^C$
Let $A = \pm e^C$. A is an arbitrary constant.
$\cos x \cos y = A$
[General Solution]
Now, we use the given point $\left( 0, \frac{\pi}{4} \right)$ to find the value of the constant A.
Substitute $x = 0$ and $y = \frac{\pi}{4}$ into the general solution:
$\cos(0) \cos\left(\frac{\pi}{4}\right) = A$
$(1) \left(\frac{1}{\sqrt{2}}\right) = A$
$A = \frac{1}{\sqrt{2}}$
Substitute the value of A back into the general solution to get the particular solution:
$\cos x \cos y = \frac{1}{\sqrt{2}}$
Conclusion:
The equation of the curve passing through the point $\left( 0, \frac{\pi}{4} \right)$ is:
$\mathbf{\cos x \cos y = \frac{1}{\sqrt{2}}}$
Question 9. Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.
Answer:
Given:
The differential equation is:
$(1 + e^{2x}) dy + (1 + y^2) e^x dx = 0$
The initial condition is $y = 1$ when $x = 0$.
To Find:
The particular solution of the given differential equation.
Solution:
The given differential equation is:
$(1 + e^{2x}) dy + (1 + y^2) e^x dx = 0$
Rearranging the terms to separate the variables:
$(1 + e^{2x}) dy = -(1 + y^2) e^x dx$
Divide both sides by $(1 + e^{2x})$ and $(1 + y^2)$:
$\frac{dy}{1 + y^2} = - \frac{e^x}{1 + e^{2x}} dx$
$\frac{dy}{1 + y^2} = - \frac{e^x}{1 + (e^x)^2} dx$
Now, integrate both sides:
$\int \frac{1}{1 + y^2} dy = - \int \frac{e^x}{1 + (e^x)^2} dx$
The integral on the left side is a standard integral:
$\int \frac{1}{1 + y^2} dy = \arctan(y)$
For the integral on the right side, let $u = e^x$. Then $du = e^x dx$.
$- \int \frac{e^x}{1 + (e^x)^2} dx = - \int \frac{1}{1 + u^2} du = -\arctan(u) + C$
Substituting back $u = e^x$:
$- \int \frac{e^x}{1 + (e^x)^2} dx = -\arctan(e^x) + C$
Combining the results from both sides:
$\arctan(y) = -\arctan(e^x) + C$
This is the general solution. We can write it as:
$\arctan(y) + \arctan(e^x) = C$
[General Solution]
Now, apply the given initial condition: $y = 1$ when $x = 0$.
Substitute $x = 0$ and $y = 1$ into the general solution:
$\arctan(1) + \arctan(e^0) = C$
$\arctan(1) + \arctan(1) = C$
$\frac{\pi}{4} + \frac{\pi}{4} = C$
$C = \frac{2\pi}{4} = \frac{\pi}{2}$
Substitute the value of $C$ back into the general solution to find the particular solution:
$\arctan(y) + \arctan(e^x) = \frac{\pi}{2}$
Conclusion:
The particular solution of the differential equation $(1 + e^{2x}) dy + (1 + y^{2}) e^{x} dx = 0$, given that $y = 1$ when $x = 0$, is:
$\mathbf{\arctan(y) + \arctan(e^x) = \frac{\pi}{2}}$
(Note: Using the identity $\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)$ is not directly applicable here since the sum is $\frac{\pi}{2}$. However, we know $\arctan(a) + \arctan(1/a) = \frac{\pi}{2}$ for $a>0$. So, $\arctan(y) + \arctan(e^x) = \frac{\pi}{2}$ implies $\arctan(y) = \frac{\pi}{2} - \arctan(e^x) = \arctan(1/e^x) = \arctan(e^{-x})$. Therefore, $y = e^{-x}$ or $ye^x=1$ is another form of the solution.)
Question 10. Solve the differential equation $y \;e^{\frac{x}{y}} \;dx = \left( x \;e^{\frac{x}{y}} + y^2 \right) \;dy\; (y ≠ 0)$.
Answer:
Given:
The differential equation is:
$y e^{\frac{x}{y}} dx = (x e^{\frac{x}{y}} + y^2) dy$, with the condition $y \neq 0$.
To Find:
The general solution of the given differential equation.
Solution:
The given differential equation is:
$y e^{\frac{x}{y}} dx = (x e^{\frac{x}{y}} + y^2) dy$
Since $y \neq 0$, we can rearrange the equation to find $\frac{dx}{dy}$:
$\frac{dx}{dy} = \frac{x e^{\frac{x}{y}} + y^2}{y e^{\frac{x}{y}}}$
$\frac{dx}{dy} = \frac{x e^{\frac{x}{y}}}{y e^{\frac{x}{y}}} + \frac{y^2}{y e^{\frac{x}{y}}}$
$\frac{dx}{dy} = \frac{x}{y} + \frac{y}{e^{\frac{x}{y}}}$
...(i)
This appears to be a homogeneous differential equation in terms of $x$ as a function of $y$. Let's use the substitution $x = vy$.
Differentiating $x = vy$ with respect to $y$:
$\frac{dx}{dy} = v \frac{d}{dy}(y) + y \frac{dv}{dy}$
$\frac{dx}{dy} = v + y \frac{dv}{dy}$
Also, from $x = vy$, we have $v = \frac{x}{y}$.
Substitute $\frac{dx}{dy} = v + y \frac{dv}{dy}$ and $\frac{x}{y} = v$ into equation (i):
$v + y \frac{dv}{dy} = v + \frac{y}{e^v}$
Subtract $v$ from both sides:
$y \frac{dv}{dy} = \frac{y}{e^v}$
Since $y \neq 0$, we can divide both sides by $y$:
$\frac{dv}{dy} = \frac{1}{e^v}$
$\frac{dv}{dy} = e^{-v}$
This is a variable separable equation. Separate the variables $v$ and $y$:
$\frac{dv}{e^{-v}} = dy$
$e^v dv = dy$
Integrate both sides:
$\int e^v dv = \int dy$
$e^v = y + C$
where C is the constant of integration.
Finally, substitute back $v = \frac{x}{y}$ into the equation:
$e^{\frac{x}{y}} = y + C$
Conclusion:
The general solution of the differential equation $y \;e^{\frac{x}{y}} \;dx = \left( x \;e^{\frac{x}{y}} + y^2 \right) \;dy\; (y ≠ 0)$ is:
$\mathbf{e^{\frac{x}{y}} = y + C}$
Question 11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)
Answer:
Given:
The differential equation is:
$(x - y) (dx + dy) = dx - dy$
The initial condition is $y = -1$ when $x = 0$.
Hint: Use the substitution $t = x - y$.
To Find:
The particular solution of the given differential equation.
Solution:
First, rewrite the given differential equation:
$(x - y) dx + (x - y) dy = dx - dy$
Rearrange the terms to group $dx$ and $dy$:
$(x - y) dx - dx = -dy - (x - y) dy$
$(x - y - 1) dx = -(1 + x - y) dy$
$(x - y - 1) dx + (x - y + 1) dy = 0$
Now, use the suggested substitution $t = x - y$.
Differentiate $t = x - y$ with respect to $x$:
$\frac{dt}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(y)$
$\frac{dt}{dx} = 1 - \frac{dy}{dx}$
So, $\frac{dy}{dx} = 1 - \frac{dt}{dx}$.
From the rearranged equation $(x - y - 1) dx + (x - y + 1) dy = 0$, we can write:
$(x - y + 1) dy = -(x - y - 1) dx$
$(x - y + 1) dy = (1 - (x - y)) dx$
$\frac{dy}{dx} = \frac{1 - (x - y)}{1 + (x - y)}$
Substitute $t = x - y$ and $\frac{dy}{dx} = 1 - \frac{dt}{dx}$:
$1 - \frac{dt}{dx} = \frac{1 - t}{1 + t}$
$\frac{dt}{dx} = 1 - \frac{1 - t}{1 + t}$
$\frac{dt}{dx} = \frac{(1 + t) - (1 - t)}{1 + t}$
$\frac{dt}{dx} = \frac{1 + t - 1 + t}{1 + t}$
$\frac{dt}{dx} = \frac{2t}{1 + t}$
This is a variable separable equation. Separate variables $t$ and $x$:
$\frac{1 + t}{2t} dt = dx$
$\left(\frac{1}{2t} + \frac{t}{2t}\right) dt = dx$
$\left(\frac{1}{2t} + \frac{1}{2}\right) dt = dx$
Integrate both sides:
$\int \left(\frac{1}{2t} + \frac{1}{2}\right) dt = \int dx$
$\frac{1}{2} \int \frac{1}{t} dt + \frac{1}{2} \int 1 dt = \int 1 dx$
$\frac{1}{2} \ln|t| + \frac{1}{2} t = x + C_1$
Multiply by 2:
$\ln|t| + t = 2x + 2C_1$
Let $C = 2C_1$.
$\ln|t| + t = 2x + C$
Substitute back $t = x - y$:
$\ln|x - y| + (x - y) = 2x + C$
$\ln|x - y| = 2x - x + y + C$
$\ln|x - y| = x + y + C$
[General Solution]
Now, apply the given initial condition: $y = -1$ when $x = 0$.
Substitute $x = 0$ and $y = -1$ into the general solution:
$\ln|0 - (-1)| = 0 + (-1) + C$
$\ln|1| = -1 + C$
$0 = -1 + C$
$C = 1$
Substitute the value of $C$ back into the general solution to find the particular solution:
$\ln|x - y| = x + y + 1$
Conclusion:
The particular solution of the differential equation $(x – y) (dx + dy) = dx – dy$, given that $y = –1$ when $x = 0$, is:
$\mathbf{\ln|x - y| = x + y + 1}$
Question 12. Solve the differential equation $\left[ \frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}} \right] \frac{dy}{dx} = 1 \;(x ≠ 0)$.
Answer:
Given:
The differential equation is:
$\left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1$, with the condition $x \neq 0$.
To Find:
The general solution of the given differential equation.
Solution:
The given differential equation can be written as:
$\frac{dx}{dy} = \frac{1}{\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}}$
$\frac{dx}{dy} = \frac{\sqrt{x}}{e^{-2\sqrt{x}} - y}$
$(e^{-2\sqrt{x}} - y) dx = \sqrt{x} dy$
Let's use the substitution $u = \sqrt{x}$. Since $x \neq 0$, $u \neq 0$.
Then $x = u^2$.
Differentiating $x$ with respect to $u$, we get $\frac{dx}{du} = 2u$. So, $dx = 2u du$.
Substitute $x = u^2$, $\sqrt{x} = u$, and $dx = 2u du$ into the rearranged equation $(e^{-2\sqrt{x}} - y) dx = \sqrt{x} dy$:
$(e^{-2u} - y) (2u du) = u dy$
Since $u = \sqrt{x} \neq 0$, we can divide both sides by $u$:
$(e^{-2u} - y) 2 du = dy$
$2e^{-2u} du - 2y du = dy$
$2e^{-2u} du = dy + 2y du$
Rearrange to get a linear differential equation in $y$ with $u$ as the independent variable:
$\frac{dy}{du} + 2y = 2e^{-2u}$
This is a linear differential equation of the form $\frac{dy}{du} + P(u)y = Q(u)$, where $P(u) = 2$ and $Q(u) = 2e^{-2u}$.
The integrating factor (I.F.) is given by:
I.F. $= e^{\int P(u) du} = e^{\int 2 du} = e^{2u}$
Multiply the linear differential equation by the integrating factor $e^{2u}$:
$e^{2u} \left( \frac{dy}{du} + 2y \right) = e^{2u} (2e^{-2u})$
$e^{2u} \frac{dy}{du} + 2e^{2u} y = 2e^{2u - 2u}$
$e^{2u} \frac{dy}{du} + 2e^{2u} y = 2e^{0}$
$e^{2u} \frac{dy}{du} + 2e^{2u} y = 2$
The left side is the derivative of the product of $y$ and the integrating factor:
$\frac{d}{du} (y \cdot e^{2u}) = 2$
Integrate both sides with respect to $u$:
$\int \frac{d}{du} (y e^{2u}) du = \int 2 du$
$y e^{2u} = 2u + C$
where C is the constant of integration.
Finally, substitute back $u = \sqrt{x}$ into the equation:
$y e^{2\sqrt{x}} = 2\sqrt{x} + C$
Conclusion:
The general solution of the differential equation $\left[ \frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1 \;(x ≠ 0)$ is:
$\mathbf{y e^{2\sqrt{x}} = 2\sqrt{x} + C}$
Question 13. Find a particular solution of the differential equation $\frac{dy}{dx} + y \;\cot x = 4x \;cosec \;x \;(x ≠ 0)$, given that y = 0 when x = $\frac{π}{2}$ .
Answer:
Given:
The differential equation is:
$\frac{dy}{dx} + y \cot x = 4x \;\text{cosec}\; x$
The initial condition is $y = 0$ when $x = \frac{\pi}{2}$.
To Find:
The particular solution of the given differential equation.
Solution:
The given differential equation is of the linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Here, $P(x) = \cot x$ and $Q(x) = 4x \;\text{cosec}\; x$.
First, find the integrating factor (I.F.):
I.F. $= e^{\int P(x) dx} = e^{\int \cot x dx}$
$\int \cot x dx = \int \frac{\cos x}{\sin x} dx = \ln|\sin x|$
So, I.F. $= e^{\ln|\sin x|} = |\sin x|$.
Since the initial condition is given at $x = \frac{\pi}{2}$, where $\sin(\frac{\pi}{2}) = 1 > 0$, we can consider the interval where $\sin x > 0$. Thus, we take I.F. $= \sin x$.
The general solution of a linear differential equation is given by:
$y \times (\text{I.F.}) = \int (Q(x) \times \text{I.F.}) dx + C$
$y \sin x = \int (4x \;\text{cosec}\; x \times \sin x) dx + C$
$y \sin x = \int \left(4x \frac{1}{\sin x} \times \sin x\right) dx + C$
$y \sin x = \int 4x dx + C$
$y \sin x = 4 \int x dx + C$
$y \sin x = 4 \left(\frac{x^2}{2}\right) + C$
$y \sin x = 2x^2 + C$
[General Solution]
Now, apply the initial condition: $y = 0$ when $x = \frac{\pi}{2}$.
Substitute these values into the general solution:
$(0) \sin\left(\frac{\pi}{2}\right) = 2\left(\frac{\pi}{2}\right)^2 + C$
$0 \times 1 = 2\left(\frac{\pi^2}{4}\right) + C$
$0 = \frac{\pi^2}{2} + C$
$C = -\frac{\pi^2}{2}$
Substitute the value of C back into the general solution to obtain the particular solution:
$y \sin x = 2x^2 - \frac{\pi^2}{2}$
Conclusion:
The particular solution of the differential equation $\frac{dy}{dx} + y \;\cot x = 4x \;\text{cosec} \;x$, given that $y = 0$ when $x = \frac{π}{2}$, is:
$\mathbf{y \sin x = 2x^2 - \frac{\pi^2}{2}}$
Question 14. Find a particular solution of the differential equation $(x + 1) \frac{dy}{dx} = 2 \;e^{-y} - 1$, given that y = 0 when x = 0.
Answer:
Given:
The differential equation is:
$(x + 1) \frac{dy}{dx} = 2 e^{-y} - 1$
The initial condition is $y = 0$ when $x = 0$.
To Find:
The particular solution of the given differential equation.
Solution:
The given differential equation is:
$(x + 1) \frac{dy}{dx} = 2 e^{-y} - 1$
Assuming $2e^{-y} - 1 \neq 0$, we can separate the variables:
$\frac{dy}{2e^{-y} - 1} = \frac{dx}{x + 1}$
To integrate the left side, multiply the numerator and denominator by $e^y$:
$\frac{e^y dy}{e^y(2e^{-y} - 1)} = \frac{dx}{x + 1}$
$\frac{e^y dy}{2e^y e^{-y} - e^y} = \frac{dx}{x + 1}$
$\frac{e^y dy}{2 - e^y} = \frac{dx}{x + 1}$
Integrate both sides:
$\int \frac{e^y}{2 - e^y} dy = \int \frac{1}{x + 1} dx$
For the left integral, let $u = 2 - e^y$. Then $du = -e^y dy$, so $e^y dy = -du$.
$\int \frac{-du}{u} = \int \frac{1}{x + 1} dx$
$-\ln|u| = \ln|x + 1| + C_1$
Substitute back $u = 2 - e^y$:
$-\ln|2 - e^y| = \ln|x + 1| + C_1$
$\ln|x + 1| + \ln|2 - e^y| = -C_1$
Let $C = -C_1$.
$\ln|(x + 1)(2 - e^y)| = C$
Exponentiate both sides:
$|(x + 1)(2 - e^y)| = e^C$
Let $A = \pm e^C$. A is an arbitrary constant.
$(x + 1)(2 - e^y) = A$
[General Solution]
Now, apply the initial condition: $y = 0$ when $x = 0$.
Substitute these values into the general solution:
$(0 + 1)(2 - e^0) = A$
$(1)(2 - 1) = A$
$(1)(1) = A$
$A = 1$
Substitute the value of A back into the general solution to get the particular solution:
$(x + 1)(2 - e^y) = 1$
We can optionally rearrange this equation to solve for y:
$2 - e^y = \frac{1}{x + 1}$
$e^y = 2 - \frac{1}{x + 1}$
$e^y = \frac{2(x + 1) - 1}{x + 1}$
$e^y = \frac{2x + 2 - 1}{x + 1}$
$e^y = \frac{2x + 1}{x + 1}$
$y = \ln\left(\frac{2x + 1}{x + 1}\right)$
Conclusion:
The particular solution of the differential equation $(x + 1) \frac{dy}{dx} = 2 \;e^{-y} - 1$, given that $y = 0$ when $x = 0$, is:
$\mathbf{(x + 1)(2 - e^y) = 1}$ or equivalently $\mathbf{y = \ln\left(\frac{2x + 1}{x + 1}\right)}$
Question 15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?
Answer:
Given:
Let $P(t)$ be the population of the village at time $t$ (in years).
The rate of increase of population, $\frac{dP}{dt}$, is proportional to the population $P$ present at time $t$.
This can be modeled by the differential equation:
$\frac{dP}{dt} \propto P$
$\frac{dP}{dt} = kP$, where $k$ is the constant of proportionality.
Let $t=0$ correspond to the year 1999.
Population in 1999 ($t=0$) was $P(0) = 20,000$.
Population in 2004 ($t=2004-1999=5$) was $P(5) = 25,000$.
To Find:
The population of the village in 2009 ($t=2009-1999=10$), i.e., $P(10)$.
Solution:
The differential equation is $\frac{dP}{dt} = kP$.
Separating variables:
$\frac{dP}{P} = k dt$
Integrating both sides:
$\int \frac{1}{P} dP = \int k dt$
$\ln|P| = kt + C_1$
Since population P must be positive, $|P|=P$.
$\ln P = kt + C_1$
Exponentiating both sides:
$P = e^{kt + C_1} = e^{C_1} e^{kt}$
Let $A = e^{C_1}$ (where A is a positive constant). The general solution is:
$P(t) = A e^{kt}$
...(i)
Now use the given conditions to find the values of A and k.
Condition 1: $P(0) = 20,000$. Substitute $t=0$ and $P=20000$ into equation (i):
$20000 = A e^{k(0)} = A e^0 = A \times 1$
$A = 20000$
So, the population model becomes:
$P(t) = 20000 e^{kt}$
...(ii)
Condition 2: $P(5) = 25,000$. Substitute $t=5$ and $P=25000$ into equation (ii):
$25000 = 20000 e^{k(5)}$
$e^{5k} = \frac{25000}{20000} = \frac{25}{20} = \frac{5}{4}$
We need to find the population at $t=10$, i.e., $P(10)$.
Using equation (ii) with $t=10$:
$P(10) = 20000 e^{k(10)} = 20000 e^{10k}$
We can write $e^{10k}$ as $(e^{5k})^2$. We already found $e^{5k} = \frac{5}{4}$.
$P(10) = 20000 \left(e^{5k}\right)^2$
$P(10) = 20000 \left(\frac{5}{4}\right)^2$
$P(10) = 20000 \times \frac{5^2}{4^2}$
$P(10) = 20000 \times \frac{25}{16}$
Calculate the value:
$P(10) = \frac{20000}{16} \times 25$
$P(10) = 1250 \times 25$
$P(10) = 31250$
Conclusion:
The population of the village in the year 2009 will be 31,250.
Question 16. The general solution of the differential equation $\frac{y \;dx − x \;dy}{y} = 0$ is
(A) xy = C
(B) x = Cy2
(C) y = Cx
(D) y = Cx2
Answer:
Given:
The differential equation is:
$\frac{y dx - x dy}{y} = 0$
Since the term is divided by $y$, we assume $y \neq 0$.
To Find:
The general solution of the given differential equation.
Solution:
Multiply the given equation by $y$ (since $y \neq 0$):
$y dx - x dy = 0$
Rearrange the terms:
$y dx = x dy$
This is a variable separable equation. Assuming $x \neq 0$, separate the variables by dividing both sides by $xy$:
$\frac{y dx}{xy} = \frac{x dy}{xy}$
$\frac{1}{x} dx = \frac{1}{y} dy$
Integrate both sides:
$\int \frac{1}{x} dx = \int \frac{1}{y} dy$
$\ln|x| = \ln|y| + C_1$
where $C_1$ is the constant of integration.
Rearrange the equation:
$\ln|y| = \ln|x| - C_1$
Exponentiate both sides:
$|y| = e^{\ln|x| - C_1} = e^{\ln|x|} \cdot e^{-C_1}$
$|y| = |x| \cdot e^{-C_1}$
Let $C = \pm e^{-C_1}$. $C$ is an arbitrary non-zero constant.
$y = C x$
(Note: If $x=0$, then from $y dx = x dy$, we get $y \cdot 0 = 0 \cdot dy$, implying $0=0$. If $y=0$, then $0 \cdot dx = x \cdot 0$, implying $0=0$. The trivial solutions $x=0$ or $y=0$ are included in the general solution $y=Cx$ when $C=0$ or for specific points).
Comparing this solution with the given options:
(A) $xy = C \implies y = C/x$
(B) $x = Cy^2$
(C) $y = Cx$
(D) $y = Cx^2$
The derived solution $y = Cx$ matches option (C).
Conclusion:
The general solution of the differential equation $\frac{y \;dx − x \;dy}{y} = 0$ is $\mathbf{y = Cx}$.
Therefore, option (C) is correct.
Question 17. The general solution of a differential equation of the type $\frac{dy}{dx} + P_1x = Q_1$ is
(A) $y \;e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$
(B) $y . e^{\int P_1 \;dx} = \int \left( Q_1 e^{\int P_1\;dx} \right) dx + C$
(C) $x \;e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$
(D) $x \;e^{\int P_1 \;dx} = \int \left( Q_1 e^{\int P_1\;dx} \right) dx + C$
Answer:
Given:
The standard form of a first-order linear differential equation in $y$ is:
$\frac{dy}{dx} + P(x) y = Q(x)$
The question likely intended to ask for the solution of an equation of this type, written as:
$\frac{dy}{dx} + P_1 y = Q_1$
where $P_1$ and $Q_1$ are functions of $x$ (or constants).
To Find:
The formula for the general solution of this type of differential equation.
Solution:
To solve the linear differential equation $\frac{dy}{dx} + P_1 y = Q_1$, we first find the integrating factor (I.F.).
The integrating factor is given by:
I.F. $= e^{\int P_1 dx}$
Multiply the differential equation by the integrating factor:
$e^{\int P_1 dx} \left( \frac{dy}{dx} + P_1 y \right) = Q_1 e^{\int P_1 dx}$
$e^{\int P_1 dx} \frac{dy}{dx} + P_1 y e^{\int P_1 dx} = Q_1 e^{\int P_1 dx}$
The left side of the equation is the derivative of the product of $y$ and the integrating factor with respect to $x$:
$\frac{d}{dx} \left( y \cdot e^{\int P_1 dx} \right) = Q_1 e^{\int P_1 dx}$
Integrate both sides with respect to $x$:
$\int \frac{d}{dx} \left( y e^{\int P_1 dx} \right) dx = \int \left( Q_1 e^{\int P_1 dx} \right) dx + C$
$y e^{\int P_1 dx} = \int \left( Q_1 e^{\int P_1 dx} \right) dx + C$
This formula represents the general solution of the differential equation.
Comparing this result with the given options:
(A) $y \;e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$ (Incorrect I.F. and integration variable for the standard form in y)
(B) $y . e^{\int P_1 \;dx} = \int \left( Q_1 e^{\int P_1\;dx} \right) dx + C$ (Matches the derived general solution)
(C) $x \;e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$ (This is the solution form for $\frac{dx}{dy} + P_1(y) x = Q_1(y)$)
(D) $x \;e^{\int P_1 \;dx} = \int \left( Q_1 e^{\int P_1\;dx} \right) dx + C$ (Incorrect dependent variable $x$ with integration wrt $x$)
Thus, option (B) correctly represents the general solution for the standard linear first-order differential equation $\frac{dy}{dx} + P_1 y = Q_1$.
Conclusion:
Assuming the differential equation is of the standard linear form $\frac{dy}{dx} + P_1 y = Q_1$ (where $P_1, Q_1$ are functions of $x$), the general solution is given by $\mathbf{y . e^{\int P_1 \;dx} = \int \left( Q_1 e^{\int P_1\;dx} \right) dx + C}$.
Therefore, option (B) is correct.
Question 18. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
(A) x ey + x2 = C
(B) x ey + y2 = C
(C) y ex + x2 = C
(D) y ey + x2 = C
Answer:
Given:
The differential equation is:
$e^x dy + (y e^x + 2x) dx = 0$
To Find:
The general solution of the given differential equation.
Solution:
The given differential equation is:
$e^x dy + (y e^x + 2x) dx = 0$
Method 1: Rearranging into Linear Form
Rearrange the equation:
$e^x dy = -(y e^x + 2x) dx$
Divide by $dx$ and $e^x$ (assuming $e^x \neq 0$, which is always true):
$\frac{dy}{dx} = -\frac{y e^x + 2x}{e^x}$
$\frac{dy}{dx} = -\frac{y e^x}{e^x} - \frac{2x}{e^x}$
$\frac{dy}{dx} = -y - 2x e^{-x}$
Rewrite in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} + y = -2x e^{-x}$
Here, $P(x) = 1$ and $Q(x) = -2x e^{-x}$.
The integrating factor (I.F.) is:
I.F. $= e^{\int P(x) dx} = e^{\int 1 dx} = e^x$
The general solution is given by $y \times (\text{I.F.}) = \int (Q(x) \times \text{I.F.}) dx + C$:
$y e^x = \int (-2x e^{-x} \times e^x) dx + C$
$y e^x = \int -2x dx + C$
$y e^x = -2 \int x dx + C$
$y e^x = -2 \left(\frac{x^2}{2}\right) + C$
$y e^x = -x^2 + C$
$y e^x + x^2 = C$
Method 2: Exact Differential Equation
The equation is $(y e^x + 2x) dx + e^x dy = 0$.
This is in the form $M dx + N dy = 0$, where $M = y e^x + 2x$ and $N = e^x$.
Check if the equation is exact by comparing partial derivatives:
$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (y e^x + 2x) = e^x$
$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (e^x) = e^x$
Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the differential equation is exact.
The general solution of an exact equation is given by $\int M dx \text{ (treating y as constant)} + \int (\text{terms in N not containing x}) dy = C$.
$\int (y e^x + 2x) dx + \int (0) dy = C$
($N = e^x$ contains x, so there are no terms in N not containing x).
$y \int e^x dx + \int 2x dx + 0 = C$
$y e^x + 2 \left(\frac{x^2}{2}\right) = C$
$y e^x + x^2 = C$
Conclusion:
Both methods yield the same general solution: $\mathbf{y e^x + x^2 = C}$.
Comparing this with the given options:
(A) $x e^y + x^2 = C$
(B) $x e^y + y^2 = C$
(C) $y e^x + x^2 = C$
(D) $y e^y + x^2 = C$
The derived solution matches option (C).
Therefore, option (C) is correct.