| Classwise Concept with Examples | ||||||
|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
|---|---|---|
| Introduction to Differential Equations | Solution of Differential Equation | Formation of Differential Equations |
| Solving a Differential Equation | Integrating Factor | |
Chapter 9 Differential Equations (Concepts)
Welcome to the fascinating and powerful world of Differential Equations. This chapter introduces a fundamental concept in mathematics that extends the ideas of calculus to model dynamic systems where quantities change in relation to each other and their rates of change. While algebra deals with static relationships and basic calculus provides tools to find rates of change (derivatives) or accumulations (integrals), differential equations are the language used to describe processes involving change itself. They form the mathematical backbone for describing phenomena across physics, engineering, biology, economics, chemistry, and countless other disciplines, enabling us to predict the behavior of systems over time.
A differential equation is formally defined as an equation that involves an independent variable (often $x$ or $t$), a dependent variable (often $y$), and one or more derivatives of the dependent variable with respect to the independent variable. Examples range from the simple $\frac{dy}{dx} = 2x$ to more complex forms like $\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = \sin(x)$. Understanding the basic characteristics of a differential equation is crucial:
- The Order of a differential equation is determined by the highest order derivative present in the equation. For instance, $\frac{dy}{dx} + y = 0$ is of first order, while $\frac{d^2y}{dx^2} + ky = 0$ is of second order.
- The Degree of a differential equation (if defined) is the power (exponent) of the highest order derivative, provided the equation is expressed as a polynomial in its derivatives (meaning it's cleared of any radicals or fractional powers involving derivatives). For example, $\left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} = x$ has order 2 and degree 3.
The primary goal when faced with a differential equation is usually to find its solution. A solution is a function (e.g., $y = f(x)$) that, when substituted along with its derivatives into the differential equation, satisfies the equation identically. We distinguish between two types of solutions:
- A General Solution: This represents a family of functions that satisfy the differential equation and contains a number of arbitrary constants (integration constants) equal to the order of the equation.
- A Particular Solution: This is a specific solution obtained from the general solution by assigning definite values to the arbitrary constants. These values are typically determined using given initial conditions or boundary conditions (e.g., knowing the value of $y$ and/or its derivatives at a specific point $x$).
The major focus of this introductory chapter, however, lies in developing techniques for solving first-order, first-degree differential equations. We concentrate on three principal methods:
- Variables Separable: This method applies if the equation can be algebraically rearranged into a form where all terms involving $y$ (and $dy$) are on one side, and all terms involving $x$ (and $dx$) are on the other: $f(y) dy = g(x) dx$. The solution is then obtained by integrating both sides independently: $\int f(y) \, dy = \int g(x) \, dx + C$.
- 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 technique is the substitution $y = vx$ (or $x = vy$). This substitution cleverly transforms the original equation into a new differential equation involving $v$ and $x$ (or $v$ and $y$) which is typically solvable using the variables separable method.
- Linear Differential Equations: This important class covers equations of the specific form $\mathbf{\frac{dy}{dx} + P(x)y = Q(x)}$ (or the alternative form $\frac{dx}{dy} + P(y)x = Q(y)$). The solution method involves calculating an Integrating Factor (IF), defined as $\mathbf{\text{IF} = e^{\int P(x) dx}}$ (or $e^{\int P(y) dy}$). Multiplying the entire linear equation by the IF cleverly transforms the left-hand side into the derivative of the product of the dependent variable and the IF (i.e., $\frac{d}{dx}(y \cdot \text{IF})$). Integrating both sides then yields the general solution: $\mathbf{y(\text{IF}) = \int [Q(x) \times \text{IF}] \, dx + C}$.
Finally, the practical significance of differential equations is often highlighted through discussions of their applications in modeling real-world phenomena such as population dynamics (growth and decay), radioactive decay rates, Newton's law of cooling, simple circuits, and other processes involving rates of change.
Introduction to Differential Equations
In many real-world problems in physics, engineering, biology, economics, and other fields, the relationships between quantities are described not just by the quantities themselves, but by their rates of change. For instance, the rate of growth of a population depends on its current size, the velocity of an object depends on its position, and the rate of decay of a radioactive substance depends on the amount present. These relationships are often expressed in the form of equations that involve derivatives. Such equations are called differential equations.
Definition: A differential equation is an equation that involves an independent variable, a dependent variable, and the derivatives of the dependent variable with respect to the independent variable.
For example, if $y$ is a dependent variable and $x$ is an independent variable, a differential equation involving $y$ and $x$ might look like:
- $\frac{dy}{dx} = 2x$ (The rate of change of $y$ with respect to $x$ is $2x$)
- $\frac{d^2y}{dx^2} + y = 0$ (A relationship between the second derivative of $y$ and $y$ itself)
- $x^2 \frac{dy}{dx} + y^2 = xy$ (An equation involving $x, y,$ and $\frac{dy}{dx}$)
Differential equations can also involve more than one independent variable (Partial Differential Equations, not typically covered in Class 12) or more than one dependent variable (Systems of Differential Equations). In Class 12, we focus on Ordinary Differential Equations (ODEs), which involve only one independent variable and its ordinary derivatives.
Order of a Differential Equation
Differential equations are classified based on their order and degree. The order of a differential equation is the order of the highest derivative that appears in the equation.
If $\frac{dy}{dx}$ is the highest derivative, the order is 1. If $\frac{d^2y}{dx^2}$ is the highest derivative, the order is 2. If $\frac{d^ny}{dx^n}$ is the highest derivative, the order is $n$.
Example 1. Determine the order of the differential equation $\frac{dy}{dx} + y = x$.
Answer:
Given:
The differential equation $\frac{dy}{dx} + y = x$.
To Determine:
The order of the differential equation.
Solution:
Identify the derivatives present in the equation: The only derivative appearing is $\frac{dy}{dx}$.
Determine the order of this highest derivative: $\frac{dy}{dx}$ is the first derivative, so its order is 1.
Since the highest derivative is of order 1, the order of the differential equation is 1.
The order of the differential equation is $\mathbf{1}$.
Example 2. Determine the order of the differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$.
Answer:
Given:
The differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$.
To Determine:
The order of the differential equation.
Solution:
Identify the derivatives present in the equation: We have $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.
Determine the order of each derivative:
- $\frac{dy}{dx}$ is a first-order derivative.
- $\frac{d^2y}{dx^2}$ is a second-order derivative.
The highest order derivative appearing in the equation is $\frac{d^2y}{dx^2}$, which has order 2.
The power to which the derivatives are raised does not affect the order of the equation. The term $\left(\frac{dy}{dx}\right)^3$ involves a first derivative raised to the power 3, but the highest order derivative is still the second derivative.
Therefore, the order of the differential equation is 2.
The order of the differential equation is $\mathbf{2}$.
Degree of a Differential Equation
The degree of a differential equation is defined if the equation can be expressed as a polynomial in the derivatives. When it can be expressed as a polynomial in derivatives, the degree is the highest power of the highest order derivative present in the equation.
Before finding the degree, the differential equation must be cleared of any radicals (like square roots, cube roots, etc.) or fractions in which the derivatives are involved, so that it is a polynomial equation in terms of the derivatives.
If a differential equation cannot be written as a polynomial in terms of its derivatives (e.g., if a derivative appears inside a function like $\sin(\frac{dy}{dx})$ or $e^{\frac{d^2y}{dx^2}}$), then the degree of the differential equation is undefined.
To find the degree:
- Ensure the equation is a polynomial in derivatives (clear fractions/radicals involving derivatives).
- Identify the highest order derivative.
- Find the power of this highest order derivative in the equation. This power is the degree.
Example 3. Find the degree of the differential equation $\frac{dy}{dx} + y = x$.
Answer:
Given:
The differential equation $\frac{dy}{dx} + y = x$.
To Find:
The degree of the differential equation.
Solution:
The equation $\frac{dy}{dx} + y - x = 0$ is a polynomial equation in the derivative $\frac{dy}{dx}$.
The highest order derivative is $\frac{dy}{dx}$, which has order 1.
The power of this highest order derivative $\frac{dy}{dx}$ in the equation is 1.
Therefore, the degree of the differential equation is 1.
The degree of the differential equation is $\mathbf{1}$.
Example 4. Find the degree of the differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$.
Answer:
Given:
The differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$.
To Find:
The degree of the differential equation.
Solution:
The equation is a polynomial in the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.
The highest order derivative is $\frac{d^2y}{dx^2}$, which has order 2.
The power of the highest order derivative $\frac{d^2y}{dx^2}$ in the equation is 1.
The power of the lower order derivative $\left(\frac{dy}{dx}\right)^3$ does not affect the degree, which is determined solely by the power of the highest order derivative.
Therefore, the degree of the differential equation is 1.
The degree of the differential equation is $\mathbf{1}$.
Example 5. Find the degree of the differential equation $\frac{d^3y}{dx^3} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$.
Answer:
Given:
The differential equation $\frac{d^3y}{dx^3} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$.
To Find:
The degree of the differential equation.
Solution:
The equation contains a square root involving a derivative. To find the degree, we must eliminate the radical sign by squaring both sides of the equation.
$\left(\frac{d^3y}{dx^3}\right)^2 = \left(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\right)^2$
(Squaring both sides)
$\left(\frac{d^3y}{dx^3}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2$
... (A)
Equation (A) is a polynomial equation in terms of the derivatives. The derivatives present are $\frac{dy}{dx}$ and $\frac{d^3y}{dx^3}$. The highest order derivative is $\frac{d^3y}{dx^3}$, which has order 3.
The power of this highest order derivative $\frac{d^3y}{dx^3}$ in the equation is 2.
Therefore, the degree of the differential equation is 2.
The order of the differential equation is 3, and the degree is 2.
The degree of the differential equation is $\mathbf{2}$.
Example 6. Find the degree of the differential equation $\sin\left(\frac{dy}{dx}\right) = x + y$.
Answer:
Given:
The differential equation $\sin\left(\frac{dy}{dx}\right) = x + y$.
To Find:
The degree of the differential equation.
Solution:
In this equation, the derivative $\frac{dy}{dx}$ appears as the argument of the trigonometric function $\sin$. This means the equation cannot be expressed as a polynomial in terms of the derivatives $\frac{dy}{dx}, \frac{d^2y}{dx^2}, \dots$.
For example, if we try to express $\sin(\frac{dy}{dx})$ using its Taylor series expansion, we would get an infinite series involving powers of $\frac{dy}{dx}$, not a finite polynomial.
Since the equation is not a polynomial in its derivatives, its degree is undefined.
The highest order derivative is $\frac{dy}{dx}$, so the order is 1. However, the degree is not defined.
The degree of this differential equation is undefined.
Solution of Differential Equation
The primary goal when working with differential equations is typically to find its solution. Unlike algebraic equations where the solution is a number or a set of numbers, the solution to a differential equation is a function (or a set of functions) that satisfies the equation.
Definition: A solution (or integral) of a differential equation is a function, or a relation between the variables, that when substituted into the differential equation along with its derivatives, reduces the equation to an identity. Essentially, it's a function that "solves" the differential equation.
Finding the solution of a differential equation means determining a relationship between the independent and dependent variables that does not involve any derivatives and is consistent with the given differential equation. The process of finding a solution is called solving the differential equation or integrating the differential equation.
General Solution
When we solve a differential equation, the solution we obtain usually contains one or more arbitrary constants.
Definition: A general solution (or complete integral or complete primitive) of a differential equation is a solution that contains as many arbitrary independent constants as the order of the differential equation.
For example, a first-order differential equation will generally have a general solution involving one arbitrary constant. A second-order differential equation will generally have a general solution involving two arbitrary constants, and so on. These arbitrary constants arise during the process of integration. The general solution represents a family of curves, where each specific curve in the family corresponds to a particular assignment of values to the arbitrary constants.
Consider the simple first-order differential equation $\frac{dy}{dx} = 2x$. We can solve this by direct integration:
$dy = 2x dx$
$\int dy = \int 2x dx$
y = $x^2 + C$
... (A)
The general solution is $y = x^2 + C$, where $C$ is an arbitrary constant. Since the differential equation $\frac{dy}{dx} = 2x$ is of order 1, its general solution contains one arbitrary constant $C$. This solution $y = x^2 + C$ represents a family of parabolas, each shifted vertically depending on the value of $C$.
Particular Solution
Often, we are interested in a specific solution from the family of solutions represented by the general solution. A particular solution is obtained by giving specific values to the arbitrary constant(s) in the general solution.
Definition: A particular solution of a differential equation is a solution obtained from the general solution by assigning specific values to one or more of the arbitrary constants.
To find a particular solution, we need additional information, usually in the form of initial conditions or boundary conditions. An initial condition specifies the value of the dependent variable (and/or its derivatives) at a particular value of the independent variable. The number of conditions required to uniquely determine a particular solution is typically equal to the order of the differential equation.
- For a first-order ODE, one condition (e.g., $y(x_0) = y_0$) is needed to find the value of the single arbitrary constant.
- For a second-order ODE, two conditions (e.g., $y(x_0) = y_0$ and $y'(x_0) = y_1$, or $y(x_0) = y_0$ and $y(x_1) = y_1$) are needed to find the values of the two arbitrary constants.
These conditions allow us to determine the exact values of the arbitrary constants in the general solution, thereby yielding a specific curve from the family.
Example 1. Verify that $y = e^x$ is a solution to the differential equation $\frac{dy}{dx} - y = 0$.
Answer:
Given:
The differential equation $\frac{dy}{dx} - y = 0$.
A proposed solution $y = e^x$.
To Verify:
Check if the proposed solution $y = e^x$ satisfies the given differential equation.
Solution:
To check if $y = e^x$ is a solution, we need to substitute $y$ and its derivatives into the differential equation. The differential equation involves the first derivative $\frac{dy}{dx}$.
Find the first derivative of $y = e^x$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$
... (i)
Now, substitute $y = e^x$ and $\frac{dy}{dx} = e^x$ into the given differential equation $\frac{dy}{dx} - y = 0$:
[Substituting $\frac{dy}{dx}$ and $y$]
= $0$
The left side of the differential equation becomes 0, which equals the right side.
Since the substitution of $y = e^x$ and its derivative into the differential equation results in a true statement ($0=0$), $y = e^x$ is indeed a solution to the differential equation $\frac{dy}{dx} - y = 0$.
Thus, $y = e^x$ is a solution to the given differential equation.
Example 2. Verify that $y = A \cos x + B \sin x$ is a general solution to the differential equation $\frac{d^2y}{dx^2} + y = 0$, where A and B are arbitrary constants.
Answer:
Given:
The differential equation $\frac{d^2y}{dx^2} + y = 0$.
A proposed solution $y = A \cos x + B \sin x$, where A and B are arbitrary constants.
The given differential equation is of order 2. The proposed solution contains two arbitrary constants, A and B. This is consistent with it being a potential general solution.
To Verify:
Check if the proposed solution satisfies the given differential equation for any values of the arbitrary constants A and B.
Solution:
The differential equation involves the second derivative $\frac{d^2y}{dx^2}$ and the function $y$. We need to find these from the proposed solution $y = A \cos x + B \sin x$.
Find the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{d}{dx}(A \cos x + B \sin x)$
$\frac{dy}{dx} = A \frac{d}{dx}(\cos x) + B \frac{d}{dx}(\sin x) = A(-\sin x) + B(\cos x) = -A \sin x + B \cos x$
... (i)
Find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-A \sin x + B \cos x)$
$\frac{d^2y}{dx^2} = -A \frac{d}{dx}(\sin x) + B \frac{d}{dx}(\cos x) = -A(\cos x) + B(-\sin x) = -A \cos x - B \sin x$
... (ii)
Now, substitute $y = A \cos x + B \sin x$ and $\frac{d^2y}{dx^2} = -A \cos x - B \sin x$ into the given differential equation $\frac{d^2y}{dx^2} + y = 0$:
$(-A \cos x - B \sin x) + (A \cos x + B \sin x)$
[Substituting $\frac{d^2y}{dx^2}$ and $y$]
Group like terms:
$= (-A \cos x + A \cos x) + (-B \sin x + B \sin x)$
$= 0 + 0 = 0$
The left side of the differential equation becomes 0, which equals the right side. This holds true for any real values of the constants A and B.
Since the equation is satisfied for arbitrary constants A and B, $y = A \cos x + B \sin x$ is indeed a general solution to the differential equation $\frac{d^2y}{dx^2} + y = 0$.
Thus, $y = A \cos x + B \sin x$ is the general solution to the given differential equation.
Example 3. For the differential equation $\frac{dy}{dx} = 2x$, the general solution is $y = x^2 + C$. Find the particular solution satisfying the condition $y(0) = 1$.
Answer:
Given:
The differential equation $\frac{dy}{dx} = 2x$, with general solution $y = x^2 + C$.
The initial condition $y(0) = 1$. This condition means that when the independent variable $x$ is 0, the dependent variable $y$ is 1.
To Find:
The particular solution that satisfies the initial condition $y(0) = 1$.
Solution:
The general solution is given by $y = x^2 + C$. This equation represents a family of parabolas.
We use the initial condition $y(0) = 1$ to find the specific value of the constant $C$ for the particular solution that passes through the point $(0, 1)$.
Substitute $x=0$ and $y=1$ into the general solution:
1 = $(0)^2 + C$
1 = $0 + C$
C = 1
[Value of the arbitrary constant]
Now, substitute this value of $C=1$ back into the general solution $y = x^2 + C$:
y = $x^2 + 1$
This equation represents a single parabola from the family, specifically the one that passes through the point $(0, 1)$. This is the particular solution satisfying the given initial condition.
The particular solution is $\mathbf{y = x^2 + 1}$.
Formation of Differential Equations
In the previous section, we learned about the solution of a differential equation. Here, we consider the reverse process: how to form a differential equation when given a family of curves. A family of curves is typically represented by an equation containing one or more arbitrary constants. The process of forming the differential equation involves eliminating these arbitrary constants from the given equation using differentiation.
The key principle is that the order of the resulting differential equation will be equal to the number of essential arbitrary constants in the given equation of the family of curves. Essential arbitrary constants are those that cannot be combined or reduced to a smaller number of constants through algebraic manipulation.
Procedure for Forming a Differential Equation
The procedure depends on the number of arbitrary constants present in the given equation representing the family of curves.
Case 1: The given equation contains one arbitrary constant. Let the given equation be $F(x, y, c) = 0$, where $c$ is the single arbitrary constant.
1. Differentiate the given equation with respect to the independent variable, which is usually $x$. This will introduce the first derivative $\frac{dy}{dx}$ into the equation. The resulting equation will involve $x$, $y$, $\frac{dy}{dx}$, and the constant $c$.
2. Eliminate the arbitrary constant $c$ from the original equation $F(x, y, c) = 0$ and the new equation obtained in step 1 (which involves $\frac{dy}{dx}$). The result will be a first-order differential equation involving $x$, $y$, and $\frac{dy}{dx}$.
Case 2: The given equation contains $n$ arbitrary constants. Let the given equation be $F(x, y, c_1, c_2, \dots, c_n) = 0$, where $c_1, c_2, \dots, c_n$ are $n$ arbitrary constants.
1. Differentiate the given equation successively $n$ times with respect to the independent variable $x$. This will produce equations involving $x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \dots, \frac{d^ny}{dx^n}$, and the constants $c_1, c_2, \dots, c_n$. We will have a total of $n+1$ equations (the original equation plus $n$ derivative equations).
2. Eliminate the $n$ arbitrary constants $c_1, c_2, \dots, c_n$ from the $n+1$ equations obtained in step 1. This elimination process will result in an equation that contains $x$, $y$, and derivatives of $y$ up to order $n$, but no arbitrary constants. This equation will be the required $n$-th order differential equation.
The method of elimination depends on the form of the equations. It might involve substitution, solving systems of equations, or using determinants.
Example 1. Form the differential equation from the equation $y = Ax$, where A is an arbitrary constant.
Answer:
Given:
The equation of a family of curves: $y = Ax$.
This equation contains one arbitrary constant, A.
To Form:
The differential equation by eliminating the arbitrary constant A.
Solution:
Since there is one arbitrary constant (A), the resulting differential equation will be of order 1.
1. Differentiate the given equation with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(Ax)$
$\frac{dy}{dx} = A \cdot \frac{d}{dx}(x) = A \cdot 1 = A$
... (i)
2. Eliminate the arbitrary constant A from the original equation $y=Ax$ and the differentiated equation (i) ($\frac{dy}{dx} = A$).
From equation (i), we have $A = \frac{dy}{dx}$. Substitute this expression for A into the original equation $y = Ax$:
y = $\left(\frac{dy}{dx}\right) \cdot x$
[Substituting A]
Rearrange the equation to get the differential equation in a standard form:
y = x $\frac{dy}{dx}$
x $\frac{dy}{dx} - y = 0$
... (ii)
This is a first-order differential equation involving $x$, $y$, and $\frac{dy}{dx}$, with no arbitrary constants. It is the differential equation for the family of curves $y = Ax$ (which represents all straight lines passing through the origin). Note that $y=0$ (when $A=0$) is also part of the family and satisfies the differential equation $x(0) - 0 = 0$.
The differential equation is $\mathbf{x \frac{dy}{dx} - y = 0}$.
Example 2. Form the differential equation from the equation $y = a \cos x + b \sin x$, where a and b are arbitrary constants.
Answer:
Given:
The equation of a family of curves: $y = a \cos x + b \sin x$.
This equation contains two arbitrary constants, a and b.
To Form:
The differential equation by eliminating the arbitrary constants a and b.
Solution:
Since there are two arbitrary constants (a and b), the resulting differential equation will be of order 2.
1. Differentiate the given equation successively two times with respect to $x$.
Original equation: $y = a \cos x + b \sin x$ ... (A)
First derivative:
$\frac{dy}{dx} = \frac{d}{dx}(a \cos x + b \sin x)$
$\frac{dy}{dx} = -a \sin x + b \cos x$
... (i)
Second derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-a \sin x + b \cos x)$
$\frac{d^2y}{dx^2} = -a \frac{d}{dx}(\sin x) + b \frac{d}{dx}(\cos x) = -a \cos x + b (-\sin x) = -a \cos x - b \sin x$
... (ii)
2. Eliminate a and b from the three equations (A), (i), and (ii).
Observe equation (ii). We can factor out -1:
$\frac{d^2y}{dx^2} = -(a \cos x + b \sin x)$
The expression in the parenthesis $(a \cos x + b \sin x)$ is exactly the original function $y$ from equation (A).
Substitute $y$ from equation (A) into this equation:
$\frac{d^2y}{dx^2} = -y$
[Substituting from (A)]
Rearranging this gives the differential equation:
$\frac{d^2y}{dx^2} + y = 0$
... (iii)
This is a second-order differential equation involving $y$ and its second derivative, with no arbitrary constants. This is the differential equation for the family of curves $y = a \cos x + b \sin x$ (which describes simple harmonic motion, for example). Note that the arbitrary constants $a$ and $b$ were eliminated using the original equation and its second derivative. The first derivative equation (i) was used in the process of finding the second derivative, but not directly in the final elimination step in this particular case.
The differential equation is $\mathbf{\frac{d^2y}{dx^2} + y = 0}$.
Solving a Differential Equation
Solving a differential equation means finding the function or the relationship between the variables that satisfies the equation. As discussed, this solution can be a general solution (containing arbitrary constants) or a particular solution (with specific values for constants). Finding the solution often involves integration.
Different types of first-order differential equations require specific methods for finding their solutions. At the Class 12 level, we primarily study three main types of first-order ordinary differential equations and their corresponding solution techniques.
Methods for Solving First-Order Differential Equations
The common methods for solving first-order differential equations are:
1. Variable Separable Method: This method applies when the equation can be algebraically manipulated such that all terms involving the dependent variable ($y$) and its differential ($dy$) are on one side of the equation, and all terms involving the independent variable ($x$) and its differential ($dx$) are on the other side.
2. Homogeneous Differential Equations: These are equations where the function of $x$ and $y$ can be expressed as a function of the ratio $\frac{y}{x}$. They can be transformed into variable separable form using a specific substitution.
3. Linear Differential Equations of the First Order: These equations have a specific linear structure in $y$ and $\frac{dy}{dx}$. They are solved using a technique involving an "integrating factor". This method is discussed in the next section (I5).
(Note: Other types like Exact Differential Equations might be covered in some syllabi but are less common at this introductory level).
Solving using Variable Separable Method
A first-order differential equation of the form $\frac{dy}{dx} = f(x, y)$ is called variable separable if the function $f(x, y)$ can be expressed as a product or quotient of a function of $x$ only and a function of $y$ only. That is, if $f(x, y) = h(x) \cdot g(y)$ or $f(x, y) = \frac{h(x)}{g(y)}$ or $f(x, y) = \frac{g(y)}{h(x)}$.
If the equation can be written in the form $g(y) \frac{dy}{dx} = f(x)$ or, using differentials, $g(y) dy = f(x) dx$.
Once the variables are separated in the form $g(y) dy = f(x) dx$, the general solution is obtained by integrating both sides of the equation with respect to their respective variables:
$\int g(y) dy = \int f(x) dx + C$
... (1)
Here, $C$ is the arbitrary constant of integration, which we usually add to only one side (conventionally, the side involving the independent variable $x$).
Example 1. Solve the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.
Answer:
Given:
The differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.
To Solve:
Find the general solution of the given differential equation.
Solution:
The given equation is $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$. We can see that the right side is a product of a function of $x$ only ($\frac{1}{1+x^2}$) and a function of $y$ only ($1+y^2$). Thus, it is a variable separable equation.
We can separate the variables by cross-multiplication, bringing all terms involving $y$ to the left side with $dy$, and all terms involving $x$ to the right side with $dx$:
$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$
[Variables separated]
Now, integrate both sides of the equation:
$\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$
These are standard integrals. The integral of $\frac{1}{1+u^2}$ with respect to $u$ is $\tan^{-1} u + C'$.
$\tan^{-1} y = \tan^{-1} x + C$
[Integrating both sides and adding constant C]
where $C$ is the arbitrary constant of integration.
This implicitly defines $y$ as a function of $x$. We can explicitly solve for $y$ by taking the tangent of both sides, but the implicit form is also a valid general solution.
The general solution is $\mathbf{\tan^{-1} y = \tan^{-1} x + C}$.
(Sometimes, the constant of integration might be written in a specific form, like $\tan^{-1} K$, to allow further simplification using inverse trigonometric identities, e.g., $\tan^{-1} y - \tan^{-1} x = \tan^{-1} K \implies \tan^{-1}\left(\frac{y-x}{1+xy}\right) = \tan^{-1} K \implies \frac{y-x}{1+xy} = K \implies y-x = K(1+xy)$. However, $\tan^{-1} y = \tan^{-1} x + C$ is a complete general solution).
Solving Homogeneous Differential Equations
A first-order differential equation $\frac{dy}{dx} = f(x, y)$ is called homogeneous if the function $f(x, y)$ is a homogeneous function of degree zero. This means that $f(\lambda x, \lambda y) = \lambda^0 f(x, y) = f(x, y)$ for any non-zero constant $\lambda$.
Alternatively, a first-order differential equation is homogeneous if it can be expressed in the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.
Examples of homogeneous functions of degree zero are $\frac{y}{x}$, $\sin(\frac{y}{x})$, $\frac{x+y}{x-y}$ (since $\frac{\lambda x + \lambda y}{\lambda x - \lambda y} = \frac{\lambda(x+y)}{\lambda(x-y)} = \frac{x+y}{x-y}$), etc.
The standard method for solving a homogeneous differential equation is to use the substitution $y = vx$, where $v$ is a new dependent variable and $x$ is the independent variable. This substitution implies that $v = \frac{y}{x}$.
To use this substitution in the differential equation, we need to express $\frac{dy}{dx}$ in terms of $v$, $x$, and $\frac{dv}{dx}$. Differentiate $y = vx$ with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(vx) = v \frac{d}{dx}(x) + x \frac{d}{dx}(v)$
$\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} = v + x \frac{dv}{dx}$
... (2)
Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the homogeneous differential equation $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.
v + x $\frac{dv}{dx} = F\left(\frac{vx}{x}\right)$
v + x $\frac{dv}{dx} = F(v)$
... (3)
Rearrange equation (3) to separate the variables $v$ and $x$:
x $\frac{dv}{dx} = F(v) - v$
If $F(v) - v \neq 0$, we can separate the variables:
$\frac{dv}{F(v) - v} = \frac{dx}{x}$
... (4)
This is now a variable separable equation in terms of $v$ and $x$. Integrate both sides to find the solution relating $v$ and $x$:
$\int \frac{dv}{F(v) - v} = \int \frac{dx}{x} + C$
After evaluating the integrals, substitute back $v = \frac{y}{x}$ into the resulting equation to get the general solution in terms of the original variables $y$ and $x$.
Example 2. Solve the differential equation $(x^2 + xy) dy = x^2 dx$.
Answer:
Given:
The differential equation $(x^2 + xy) dy = x^2 dx$.
To Solve:
Find the general solution of the given differential equation.
Solution:
First, rewrite the equation in the form $\frac{dy}{dx} = f(x, y)$.
$\frac{dy}{dx} = \frac{x^2}{x^2 + xy}$
... (A)
Now, check if this is a homogeneous differential equation. The function $f(x, y) = \frac{x^2}{x^2 + xy}$. Replace $x$ with $\lambda x$ and $y$ with $\lambda y$:
f$(\lambda x, \lambda y) = \frac{(\lambda x)^2}{(\lambda x)^2 + (\lambda x)(\lambda y)} = \frac{\lambda^2 x^2}{\lambda^2 x^2 + \lambda^2 xy} = \frac{\lambda^2 x^2}{\lambda^2 (x^2 + xy)} = \frac{x^2}{x^2 + xy}$
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 homogeneous of degree zero. Alternatively, we can express $\frac{dy}{dx}$ as a function of $\frac{y}{x}$ by dividing the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$:
$\frac{dy}{dx} = \frac{x^2/x^2}{(x^2 + xy)/x^2} = \frac{1}{1 + xy/x^2} = \frac{1}{1 + y/x}$
... (B)
This is of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$, where $F(v) = \frac{1}{1+v}$. So, it is a homogeneous differential equation.
Use the substitution $y = vx$. Differentiating with respect to $x$ gives $\frac{dy}{dx} = v + x \frac{dv}{dx}$ (from equation 2).
Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into equation (B):
v + x $\frac{dv}{dx} = \frac{1}{1+v}$
Separate the variables $v$ and $x$. Move $v$ to the right side:
x $\frac{dv}{dx} = \frac{1}{1+v} - v$
x $\frac{dv}{dx} = \frac{1 - v(1+v)}{1+v} = \frac{1 - v - v^2}{1+v}$
Now, separate the variables, getting terms with $v$ and $dv$ on one side and terms with $x$ and $dx$ on the other:
$\frac{1+v}{1 - v - v^2} dv = \frac{dx}{x}$
... (C)
Integrate both sides of the equation:
$\int \frac{1+v}{1 - v - v^2} dv = \int \frac{dx}{x}$
Evaluate the integral on the right side:
$\int \frac{dx}{x} = \ln|x| + C_1$
For the integral on the left side, consider the substitution method. Let the denominator $u = 1 - v - v^2$. The derivative is $\frac{du}{dv} = -1 - 2v$. So, $du = (-1 - 2v) dv$. We have $1+v$ in the numerator.
We can write the numerator $1+v$ in terms of the derivative of the denominator $-1-2v$. $1+v = -\frac{1}{2}(-2-2v) = -\frac{1}{2}(-1-2v - 1) = -\frac{1}{2}((-1-2v) - 1)$. This doesn't quite work directly. Instead, observe that the numerator $1+v$ is related to the derivative of the denominator. $d(1-v-v^2) = (-1-2v)dv$. We have $(1+v)dv$. We can rewrite $1+v$ as $-\frac{1}{2}(-2-2v) = -\frac{1}{2}(-1-2v - 1)$. Let $u = 1 - v - v^2$, so $du = (-1 - 2v) dv$. The numerator $1+v$ can be written as $A(-1 - 2v) + B$. $1+v = -A - 2Av + B$. Comparing coefficients of $v$: $1 = -2A \implies A = -1/2$. Comparing constant terms: $1 = -A + B \implies 1 = -(-1/2) + B = 1/2 + B \implies B = 1/2$. So, $1+v = -\frac{1}{2}(-1-2v) + \frac{1}{2}$.
$\int \frac{1+v}{1 - v - v^2} dv = \int \frac{-\frac{1}{2}(-1-2v) + \frac{1}{2}}{1 - v - v^2} dv$
$= -\frac{1}{2} \int \frac{-1-2v}{1 - v - v^2} dv + \frac{1}{2} \int \frac{dv}{1 - v - v^2}$
The first integral is of the form $\int \frac{g'(v)}{g(v)} dv = \ln|g(v)|$. Let $u = 1-v-v^2$, $du = (-1-2v)dv$. This integral is $-\frac{1}{2} \int \frac{du}{u} = -\frac{1}{2} \ln|u| = -\frac{1}{2} \ln|1-v-v^2|$.
The second integral $\int \frac{dv}{1 - v - v^2}$ requires completing the square in the denominator. $1 - v - v^2 = -(v^2 + v - 1) = -\left(\left(v + \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 1\right) = -\left(\left(v + \frac{1}{2}\right)^2 - \frac{1}{4} - 1\right) = -\left(\left(v + \frac{1}{2}\right)^2 - \frac{5}{4}\right)$ $1 - v - v^2 = \frac{5}{4} - \left(v + \frac{1}{2}\right)^2 = \left(\frac{\sqrt{5}}{2}\right)^2 - \left(v + \frac{1}{2}\right)^2$.
The second integral is $\frac{1}{2} \int \frac{dv}{(\sqrt{5}/2)^2 - (v + 1/2)^2}$. This is of the standard form $\int \frac{du}{a^2 - u^2} = \frac{1}{2a} \ln\left|\frac{a+u}{a-u}\right| + C$. Here $a = \frac{\sqrt{5}}{2}$ and $u = v + \frac{1}{2}$.
$\frac{1}{2} \cdot \left(\frac{1}{2(\sqrt{5}/2)} \ln\left|\frac{\sqrt{5}/2 + (v + 1/2)}{\sqrt{5}/2 - (v + 1/2)}\right|\right) = \frac{1}{2\sqrt{5}} \ln\left|\frac{(\sqrt{5}+1)/2 + v}{(\sqrt{5}-1)/2 - v}\right|$
So, integrating equation (C):
$-\frac{1}{2} \ln|1 - v - v^2| + \frac{1}{2\sqrt{5}} \ln\left|\frac{(\sqrt{5}+1)/2 + v}{(\sqrt{5}-1)/2 - v}\right| = \ln|x| + C_1$
Finally, substitute back $v = y/x$ to get the solution in terms of $x$ and $y$. This would result in a complicated expression involving $\ln|1 - y/x - y^2/x^2|$ and the other logarithmic term.
Example 3. Solve the differential equation $(x+y) dy + (x-y) dx = 0$.
Answer:
Given: $(x+y) dy + (x-y) dx = 0$.
To Solve: Find the general solution.
Solution:
Rewrite in $\frac{dy}{dx}$ form: $(x+y) dy = -(x-y) dx = (y-x) dx$.
$\frac{dy}{dx} = \frac{y-x}{x+y}$
Divide numerator and denominator by $x$ to check for homogeneity:
$\frac{dy}{dx} = \frac{(y/x) - 1}{1 + (y/x)}$
... (D)
This is of the form $\frac{dy}{dx} = F(y/x)$, where $F(v) = \frac{v-1}{v+1}$. It is homogeneous.
Substitute $y=vx$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
v + x $\frac{dv}{dx} = \frac{v-1}{v+1}$
Separate variables:
x $\frac{dv}{dx} = \frac{v-1}{v+1} - v = \frac{v-1 - v(v+1)}{v+1} = \frac{v-1 - v^2 - v}{v+1} = \frac{-1 - v^2}{v+1} = -\frac{v^2+1}{v+1}$
$\frac{v+1}{v^2+1} dv = -\frac{dx}{x}$
Integrate both sides:
$\int \frac{v+1}{v^2+1} dv = \int -\frac{dx}{x}$
$\int \left(\frac{v}{v^2+1} + \frac{1}{v^2+1}\right) dv = -\int \frac{dx}{x}$
Split the integral on the left:
$\int \frac{v}{v^2+1} dv + \int \frac{1}{v^2+1} dv = -\int \frac{dx}{x}$
For the first integral on the left, use substitution $u = v^2+1$, $du = 2v dv \implies v dv = \frac{1}{2} du$.
$\int \frac{1}{u} \frac{1}{2} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(v^2+1)$ (since $v^2+1$ is always positive)
The second integral on the left is a standard form $\tan^{-1} v$. The integral on the right is $-\ln|x|$.
$\frac{1}{2} \ln(v^2+1) + \tan^{-1} v = -\ln|x| + C$
Substitute back $v = y/x$:
$\frac{1}{2} \ln\left(\left(\frac{y}{x}\right)^2+1\right) + \tan^{-1} \left(\frac{y}{x}\right) = -\ln|x| + C$
$\frac{1}{2} \ln\left(\frac{y^2}{x^2}+1\right) + \tan^{-1} \left(\frac{y}{x}\right) = -\ln|x| + C$
$\frac{1}{2} \ln\left(\frac{y^2+x^2}{x^2}\right) + \tan^{-1} \left(\frac{y}{x}\right) = -\ln|x| + C$
$\frac{1}{2} (\ln(y^2+x^2) - \ln(x^2)) + \tan^{-1} \left(\frac{y}{x}\right) = -\ln|x| + C$
$\frac{1}{2} \ln(x^2+y^2) - \frac{1}{2} (2\ln|x|) + \tan^{-1} \left(\frac{y}{x}\right) = -\ln|x| + C$
$\frac{1}{2} \ln(x^2+y^2) - \ln|x| + \tan^{-1} \left(\frac{y}{x}\right) = -\ln|x| + C$
Cancel $-\ln|x|$ from both sides:
$\frac{1}{2} \ln(x^2+y^2) + \tan^{-1} \left(\frac{y}{x}\right) = C$
Multiply by 2 and let $2C = K$ (another arbitrary constant):
$\ln(x^2+y^2) + 2\tan^{-1} \left(\frac{y}{x}\right) = K$
The general solution is $\mathbf{\ln(x^2+y^2) + 2\tan^{-1} \left(\frac{y}{x}\right) = K}$.
Integrating Factor
Among the various types of first-order differential equations, linear differential equations form an important class because there is a systematic method to find their general solution. This method involves multiplying the entire equation by a special function called the Integrating Factor (IF).
A first-order differential equation is said to be linear if the dependent variable and its derivative appear only in the first power and are not multiplied together. The standard form of a first-order linear differential equation is:
$\frac{dy}{dx} + P(x)y = Q(x)$
... (1)
where $P(x)$ and $Q(x)$ are continuous functions of the independent variable $x$ (or constants) on a given interval. The term $P(x)$ is the coefficient of $y$, and $Q(x)$ is the term that does not involve $y$ or $\frac{dy}{dx}$.
Examples of linear differential equations:
- $\frac{dy}{dx} + y \cos x = \sin x$ (Here $P(x) = \cos x$, $Q(x) = \sin x$)
- $\frac{dy}{dx} - 2y = e^x$ (Here $P(x) = -2$, $Q(x) = e^x$)
- $x \frac{dy}{dx} + y = x^2$. To get this into standard form, divide by $x$: $\frac{dy}{dx} + \frac{1}{x} y = x$. (Here $P(x) = 1/x$, $Q(x) = x$)
- $\frac{dy}{dx} + y^2 = x$ (due to $y^2$)
- $\frac{dy}{dx} + xy = 1$ (due to $xy$)
- $\frac{dy}{dx} = \cos y$ (due to $\cos y$)
The Concept and Derivation of the Integrating Factor
The idea behind the integrating factor is to find a function, let's call it $\mu(x)$, such that when we multiply the standard form of the linear differential equation (1) by $\mu(x)$, the left side of the resulting equation becomes the exact derivative of the product of the integrating factor and the dependent variable, i.e., $\frac{d}{dx}(\mu(x)y)$.
Let's multiply equation (1) by $\mu(x)$:
$\mu(x) \frac{dy}{dx} + \mu(x) P(x)y = \mu(x) Q(x)$
... (2)
Now, let's look at the derivative of the product $\mu(x)y$ using the product rule:
$\frac{d}{dx}(\mu(x)y) = \mu(x) \frac{dy}{dx} + \frac{d\mu}{dx} y$
... (3)
We want the left side of equation (2) to be equal to the right side of equation (3). Comparing the terms involving $\frac{dy}{dx}$, we see that they are already identical ($\mu(x) \frac{dy}{dx}$). For the terms involving $y$ to be equal, we must have:
$\mu(x) P(x)y = \frac{d\mu}{dx} y$
Assuming $y$ is not identically zero on the interval (if $y=0$ is a solution, it's usually trivial), we can divide by $y$:
$\mu(x) P(x) = \frac{d\mu}{dx}$
... (4)
This is a first-order differential equation involving $\mu(x)$ and $x$. We can solve this equation for $\mu(x)$ because it is a variable separable equation.
From equation (4), rearrange to separate the variables:
$\frac{d\mu}{\mu} = P(x) dx$
Integrate both sides with respect to their respective variables:
$\int \frac{d\mu}{\mu} = \int P(x) dx$
The integral on the left side is a standard logarithmic integral:
$\ln|\mu(x)| = \int P(x) dx$
... (5)
To find $\mu(x)$, take the exponential of both sides:
$|\mu(x)| = e^{\int P(x) dx}$
Since we only need *one* function $\mu(x)$ that satisfies the condition, we can choose the simplest one. We can drop the absolute value sign (as the exponential is always positive) and set the constant of integration that would arise from $\int P(x) dx$ to zero. This gives the formula for the Integrating Factor:
Integrating Factor (IF) = $e^{\int P(x) dx}$
... (6)
Solution of the Linear Differential Equation using IF
Once we have found the Integrating Factor $\mu(x) = e^{\int P(x) dx}$, we multiply the standard form of the linear differential equation (1) by this IF.
e$^{\int P(x) dx} \frac{dy}{dx} + P(x) e^{\int P(x) dx} y = Q(x) e^{\int P(x) dx}$
... (7)
As designed, the left side of this equation is the derivative of the product of the dependent variable $y$ and the Integrating Factor $e^{\int P(x) dx}$:
$\frac{d}{dx} (y \cdot e^{\int P(x) dx}) = Q(x) e^{\int P(x) dx}$
... (8)
Now, to find the general solution for $y$, we integrate both sides of equation (8) with respect to $x$:
$\int \frac{d}{dx} (y \cdot e^{\int P(x) dx}) dx = \int \left(Q(x) e^{\int P(x) dx}\right) dx$
The integral of a derivative is the function itself (plus the arbitrary constant of integration, which represents the general solution):
y $\cdot e^{\int P(x) dx} = \int Q(x) e^{\int P(x) dx} dx + C$
... (9)
This equation gives the general solution of the linear first-order differential equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. It is often written as $y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$.
Summary of Steps to Solve $\frac{dy}{dx} + P(x)y = Q(x)$
1. Standard Form: Ensure the given differential equation is in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$. Identify the functions $P(x)$ and $Q(x)$. Make sure the coefficient of $\frac{dy}{dx}$ is 1.
2. Calculate Integrating Factor: Compute the integrating factor (IF) using the formula: IF $= e^{\int P(x) dx}$. Remember to omit the constant of integration in the exponent when calculating the IF.
3. Apply Solution Formula: Write down the general solution using the derived formula: $y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$.
4. Evaluate the Integral: Evaluate the integral on the right side, $\int Q(x) \cdot (\text{IF}) dx$. Add the arbitrary constant of integration $C$ at this stage.
5. Isolate y (Optional but Recommended): If possible, solve the resulting equation for $y$ to express the general solution explicitly as $y = \dots$.
6. Find Particular Solution (if needed): If an initial or boundary condition is given (e.g., $y(x_0) = y_0$), substitute the values of $x_0$ and $y_0$ into the general solution (from step 4 or 5) to find the specific value of $C$. Substitute this value of $C$ back into the general solution to get the particular solution.
Example 1. Find the general solution of the differential equation $\frac{dy}{dx} + 2y = \sin x$.
Answer:
Given:
The differential equation $\frac{dy}{dx} + 2y = \sin x$.
To Find:
The general solution of the given differential equation.
Solution:
1. Standard Form: The equation is already in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Identify $P(x)$ and $Q(x)$: $P(x) = 2$ and $Q(x) = \sin x$.
2. Calculate the Integrating Factor (IF):
IF = $e^{\int P(x) dx}$
[Formula for IF]
IF = $e^{\int 2 dx}$
Evaluate the integral in the exponent: $\int 2 dx = 2x$ (omitting the constant).
IF = $e^{2x}$
... (A)
3. Apply the general solution formula: $y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$.
y $\cdot e^{2x} = \int (\sin x) \cdot (e^{2x}) dx + C$
... (B)
4. Evaluate the integral on the right side: $\int e^{2x} \sin x dx$. This integral requires integration by parts. Let $I_{int} = \int e^{2x} \sin x dx$. We use integration by parts formula $\int u \, dv = uv - \int v \, du$. Using ILATE rule, we choose $u = \sin x$ and $dv = e^{2x} dx$.
Let $u = \sin x \implies du = \frac{d}{dx}(\sin x) dx = \cos x dx$.
Let $dv = e^{2x} dx \implies v = \int e^{2x} dx = \frac{1}{2} e^{2x}$ (no constant needed for $v$).
$I_{int} = (\sin x)\left(\frac{1}{2} e^{2x}\right) - \int \left(\frac{1}{2} e^{2x}\right)(\cos x dx)$
$= \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \int e^{2x} \cos x dx$
... (C)
The new integral $\int e^{2x} \cos x dx$ also requires integration by parts. Let $u = \cos x$ and $dv = e^{2x} dx$.
Let $u = \cos x \implies du = \frac{d}{dx}(\cos x) dx = -\sin x dx$.
Let $dv = e^{2x} dx \implies v = \frac{1}{2} e^{2x}$.
$\int e^{2x} \cos x dx = (\cos x)\left(\frac{1}{2} e^{2x}\right) - \int \left(\frac{1}{2} e^{2x}\right)(-\sin x dx)$
$= \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \int e^{2x} \sin x dx$
Notice that $\int e^{2x} \sin x dx$ is our original integral $I_{int}$. Substitute this back into the equation:
$\int e^{2x} \cos x dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I_{int}$
... (D)
Now substitute the result from equation (D) back into equation (C):
$I_{int} = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left(\frac{1}{2} e^{2x} \cos x + \frac{1}{2} I_{int}\right)$
$I_{int} = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} I_{int}$
Collect terms with $I_{int}$ on the left side:
$I_{int} + \frac{1}{4} I_{int} = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x$
$\frac{5}{4} I_{int} = \frac{1}{4} e^{2x} (2 \sin x - \cos x)$
Solve for $I_{int}$ by multiplying both sides by $\frac{4}{5}$:
$I_{int} = \frac{4}{5} \cdot \frac{1}{4} e^{2x} (2 \sin x - \cos x) = \frac{1}{5} e^{2x} (2 \sin x - \cos x)$
... (E)
Now, substitute the result of the integral $I_{int}$ (equation E) back into the general solution equation (B):
y $e^{2x} = \frac{1}{5} e^{2x} (2 \sin x - \cos x) + C$
5. Isolate $y$. Divide the entire equation by $e^{2x}$ (note that $e^{2x}$ is never zero):
y = $\frac{1}{e^{2x}} \left(\frac{1}{5} e^{2x} (2 \sin x - \cos x) + C\right)$
y = $\frac{1}{5} (2 \sin x - \cos x) + C e^{-2x}$
... (F)
This is the general solution of the given differential equation.
The general solution is $\mathbf{y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x}}$.
Equation Reducible to Linear Form (Bernoulli's Equation)
Some differential equations that are not linear in $y$ can be transformed into a linear equation by using an appropriate substitution. A common example is the Bernoulli's equation.
A first-order differential equation of the form:
$\frac{dy}{dx} + P(x)y = Q(x)y^n$
... (10)
where $P(x)$ and $Q(x)$ are functions of $x$, and $n$ is a real number, is called a Bernoulli's equation.
- If $n=0$, the equation becomes $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear equation.
- If $n=1$, the equation becomes $\frac{dy}{dx} + P(x)y = Q(x)y$, which can be written as $\frac{dy}{dx} = (Q(x) - P(x))y$. This is a variable separable equation.
If $n \neq 0$ and $n \neq 1$, we can reduce equation (10) to a linear equation using the following procedure:
1. Divide the entire equation by $y^n$:
$\frac{1}{y^n} \frac{dy}{dx} + P(x) \frac{y}{y^n} = Q(x)$
y$^{-n} \frac{dy}{dx} + P(x) y^{1-n} = Q(x)$
... (11)
2. Make the substitution $v = y^{1-n}$. This is the key substitution for Bernoulli's equation.
3. Find $\frac{dv}{dx}$ by differentiating $v = y^{1-n}$ with respect to $x$ using the chain rule (since $y$ is a function of $x$):
$\frac{dv}{dx} = \frac{d}{dx}(y^{1-n}) = (1-n) y^{(1-n)-1} \frac{dy}{dx}$
$\frac{dv}{dx} = (1-n) y^{-n} \frac{dy}{dx}$
... (12)
4. From equation (12), we can express the term $y^{-n} \frac{dy}{dx}$ in terms of $\frac{dv}{dx}$ and $(1-n)$:
y$^{-n} \frac{dy}{dx} = \frac{1}{1-n} \frac{dv}{dx}$
... (13)
5. Substitute $y^{-n} \frac{dy}{dx}$ from (13) and $y^{1-n} = v$ into equation (11):
$\left(\frac{1}{1-n} \frac{dv}{dx}\right) + P(x) (v) = Q(x)$
Multiply the entire equation by $(1-n)$ (since $n \neq 1$, $1-n \neq 0$):
$\frac{dv}{dx} + (1-n) P(x) v = (1-n) Q(x)$
... (14)
Equation (14) is a first-order linear differential equation in the variable $v$. It is in the standard form $\frac{dv}{dx} + P_{new}(x) v = Q_{new}(x)$, where $P_{new}(x) = (1-n)P(x)$ and $Q_{new}(x) = (1-n)Q(x)$.
This linear equation in $v$ can be solved using the integrating factor method described earlier in this section. Once the general solution for $v$ in terms of $x$ is found, substitute back $v = y^{1-n}$ to obtain the general solution in terms of the original variables $y$ and $x$.
Solving Bernoulli's equation involves converting it to a linear equation, then applying the integrating factor method.