Chapter 9 Differential Equations (Additional Questions)

The order of a differential equation is the order of the highest order Derivative appearing in the equation.

(B) Derivative

The given differential equation is $(\frac{dy}{dx})^3 + (\frac{d^2y}{dx^2})^2 + y = 0$.

The order of the highest derivative is 2 (from $\frac{d^2y}{dx^2}$).

The power of the highest order derivative ($\frac{d^2y}{dx^2}$) is 2.

Therefore, the degree of the differential equation is 2.

(B) 2

The given differential equation is $\frac{d^3y}{dx^3} + (\frac{dy}{dx})^4 + y^2 = 0$.

The highest order derivative present in the equation is $\frac{d^3y}{dx^3}$. Therefore, the order of the differential equation is 3.

The highest power of the highest order derivative ($\frac{d^3y}{dx^3}$) is 1. Therefore, the degree of the differential equation is 1.

(C) Order 3, Degree 1

A general solution of a differential equation contains arbitrary constants equal to the order of the differential equation.

(B) Order

Given the family of curves $y = mx + c$. Here, $m$ and $c$ are arbitrary constants.

To find the differential equation, we need to eliminate these arbitrary constants by differentiating the equation with respect to $x$ twice.

Differentiating the given equation with respect to $x$:

Differentiating equation (i) with respect to $x$ again:

Since $m$ is a constant, its derivative is 0.

This is the differential equation of the family of straight lines $y = mx + c$.

(B) $\frac{d^2y}{dx^2} = 0$

We are asked to find the general solution of the differential equation $\frac{dy}{dx} = e^x$.

To find the general solution, we integrate both sides with respect to $x$:

The left side simplifies to $y$. The integral of $e^x$ is $e^x$. We also add an arbitrary constant of integration, $C$.

Therefore, the general solution of the differential equation $\frac{dy}{dx} = e^x$ is $y = e^x + C$.

(B) $y = e^x + C$

The differential equation is given as $\frac{dy}{dx} = \frac{x}{y}$.

By separating the variables, we get $y \, dy = x \, dx$.

Integrating both sides:

Performing the integration:

where $C_1$ is the constant of integration.

Multiplying both sides by 2, we get:

Let $C = 2C_1$. Then the general solution is:

We can rearrange this equation to $y^2 - x^2 = C$.

Looking at the options:

(A) $\frac{y^2}{2} = \frac{x^2}{2}$ is incorrect as it omits the constant of integration.

(B) $\frac{y^2}{2} = \frac{x^2}{2} + C$ is a valid form of the solution before multiplying by 2.

(C) $y^2 = x^2 + C$ is the simplified form of the solution.

(D) $y^2 - x^2 = C'$ is also a valid form of the solution where $C' = C$.

Option (B) represents the direct result of integration before simplifying the constant. Option (C) and (D) are equivalent forms after manipulation of the constant. Based on the immediate result of integration, option (B) is the most direct representation of the integrated form.

(B) $\frac{y^2}{2} = \frac{x^2}{2} + C$

A differential equation $\frac{dy}{dx} = f(x, y)$ is homogeneous if $f(\lambda x, \lambda y) = f(x, y)$ (i.e., $n=0$ in the given definition, which is the standard definition of homogeneous in this context, where the function is a ratio of homogeneous functions of the same degree). Let's test each option:

(A) $\frac{dy}{dx} = \frac{x+y}{x-y}$

Let $f(x, y) = \frac{x+y}{x-y}$.

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$:

So, $f(\lambda x, \lambda y) = f(x, y)$. This means $n=0$, and the differential equation is homogeneous.

(B) $\frac{dy}{dx} = x+y$

Let $f(x, y) = x+y$.

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$:

So, $f(\lambda x, \lambda y) = \lambda f(x, y)$. This means $n=1$, and the differential equation is homogeneous.

(C) $\frac{dy}{dx} = \frac{x+y+1}{x-y+2}$

Let $f(x, y) = \frac{x+y+1}{x-y+2}$.

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$:

This cannot be simplified to $\lambda^n \left(\frac{x+y+1}{x-y+2}\right)$ for any integer $n$. Thus, it is not homogeneous.

(D) $\frac{dy}{dx} = x^2+y^2$

Let $f(x, y) = x^2+y^2$.

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$:

So, $f(\lambda x, \lambda y) = \lambda^2 f(x, y)$. This means $n=2$, and the differential equation is homogeneous.

The question asks which of the following is a homogeneous differential equation. Options (A), (B), and (D) are all homogeneous differential equations with different values of $n$. However, usually in such multiple-choice questions, there is a specific type of homogeneity expected, often a ratio of homogeneous functions of the same degree. Option (A) fits this common interpretation, and the definition provided ($f(\lambda x, \lambda y) = \lambda^n f(x, y)$) implies that all these options could be considered homogeneous depending on the interpretation of $n$. Given the options, option (A) is the most common example of a homogeneous differential equation presented in this format.

If the question strictly follows the definition provided, and we are looking for *any* integer $n$, then (A), (B), and (D) are all correct. In a typical context, the definition of homogeneous often implies $n=0$ when $f(x,y)$ is a ratio of homogeneous functions. If we consider the possibility of multiple correct answers, we should re-examine common problem patterns. Option (A) is a quintessential example of a homogeneous differential equation solvable by the substitution $y=vx$.

Let's assume the question expects the most standard form of a homogeneous differential equation, which is when $f(x,y)$ is a ratio of homogeneous functions of the same degree.

(A) $\frac{dy}{dx} = \frac{x+y}{x-y}$ is homogeneous of degree 0.

(B) $\frac{dy}{dx} = x+y$ is homogeneous of degree 1.

(D) $\frac{dy}{dx} = x^2+y^2$ is homogeneous of degree 2.

All three satisfy the definition. However, it is possible that the question implies a specific type of homogeneity or there is a preferred answer among the choices. Often, when "homogeneous" is used without qualification, it refers to the ratio type where $f(x,y)$ is a function of $y/x$ or $x/y$. Option (A) fits this perfectly.

In educational contexts, when presented with such options, (A) is typically the intended answer as it is the most direct representation of a homogeneous function of degree zero in ratio form.

(A) $\frac{dy}{dx} = \frac{x+y}{x-y}$

For a homogeneous differential equation of the form $\frac{dy}{dx} = f(\frac{y}{x})$, the standard substitution used to convert it into a separable differential equation is $y = vx$.

When $y = vx$, we differentiate with respect to $x$ using the product rule:

Also, $\frac{y}{x} = v$.

Substituting these into the original differential equation $\frac{dy}{dx} = f(\frac{y}{x})$:

This results in a separable differential equation in terms of $v$ and $x$:

Therefore, the substitution commonly used is $y = vx$.

(A) $y = vx$

A linear differential equation of the first order is given by the form:

where $P$ and $Q$ are functions of $x$ or constants.

To solve this type of differential equation, we multiply the entire equation by an integrating factor (IF). The integrating factor is chosen such that the left-hand side becomes the derivative of the product of the dependent variable ($y$) and the integrating factor.

The integrating factor for a linear first-order differential equation of the form $\frac{dy}{dx} + Py = Q$ is given by:

Multiplying equation (i) by the integrating factor $e^{\int P dx}$:

The left side of this equation is the derivative of $(y \cdot e^{\int P dx})$ with respect to $x$:

Integrating both sides with respect to $x$ gives the general solution:

Therefore, the integrating factor (IF) for the given linear differential equation is $e^{\int P dx}$.

(B) $e^{\int P dx}$

A first-order linear differential equation has the general form $\frac{dy}{dx} + P(x)y = Q(x)$ or $\frac{dx}{dy} + P(y)x = Q(y)$, where $P$ and $Q$ are functions of the independent variable or constants. The dependent variable and its first derivative should appear linearly, with no powers or products of them.

Let's analyze each option:

(A) $\frac{dy}{dx} + y \tan x = \sec x$

This equation is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \tan x$ and $Q(x) = \sec x$. Both $P(x)$ and $Q(x)$ are functions of $x$. The dependent variable $y$ and its derivative $\frac{dy}{dx}$ appear linearly. Thus, this is a first-order linear differential equation.

(B) $y \frac{dy}{dx} + x = 0$

This equation can be written as $y \frac{dy}{dx} = -x$. This equation is not linear because of the term $y \frac{dy}{dx}$, which is a product of the dependent variable and its derivative. It is also not of the standard linear form.

(C) $\frac{dy}{dx} = \frac{x+y}{x}$

This equation can be rewritten as $\frac{dy}{dx} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}$.

Rearranging it to the standard form: $\frac{dy}{dx} - \frac{1}{x}y = 1$.

This equation is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\frac{1}{x}$ and $Q(x) = 1$. Both $P(x)$ and $Q(x)$ are functions of $x$. The dependent variable $y$ and its derivative $\frac{dy}{dx}$ appear linearly. Thus, this is a first-order linear differential equation.

(D) $\frac{dx}{dy} + x \cot y = \text{cosec } y$

This equation is in the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \cot y$ and $Q(y) = \text{cosec } y$. Both $P(y)$ and $Q(y)$ are functions of $y$. The dependent variable $x$ and its derivative $\frac{dx}{dy}$ appear linearly. Thus, this is a first-order linear differential equation.

Therefore, the first-order linear differential equations are (A), (C), and (D).

(A), (C), (D)

Let's analyze the Assertion (A) first.

Given the equation of the family of curves:

$y = A e^x + B e^{-x}$

We need to find the differential equation by eliminating the arbitrary constants $A$ and $B$.

Differentiating the equation with respect to $x$:

Differentiating again with respect to $x$:

From the original equation, we know that $y = A e^x + B e^{-x}$.

Substituting this into equation (ii):

$\frac{d^2y}{dx^2} = y$

Rearranging the terms, we get the differential equation:

$\frac{d^2y}{dx^2} - y = 0$

Thus, Assertion (A) is true.

Now let's analyze the Reason (R).

Reason (R) states that differentiating $y = A e^x + B e^{-x}$ twice eliminates the arbitrary constants $A$ and $B$.

As shown above, by differentiating the given equation twice, we obtained $\frac{d^2y}{dx^2} = A e^x + B e^{-x}$. We then used the original equation $y = A e^x + B e^{-x}$ to substitute for $A e^x + B e^{-x}$, thereby eliminating $A$ and $B$. This process indeed eliminates the arbitrary constants and leads to the differential equation.

Therefore, Reason (R) is also true and it correctly explains why Assertion (A) is true.

The correct option is (A).

The given differential equation is:

$x \frac{dy}{dx} + 2y = x^2$

To find the integrating factor, we first need to convert this equation into the standard linear form, which is $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide the entire equation by $x$ (assuming $x \neq 0$):

$\frac{dy}{dx} + \frac{2}{x}y = \frac{x^2}{x}$

This simplifies to:

$\frac{dy}{dx} + \frac{2}{x}y = x$

Now, comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we can identify $P(x)$ and $Q(x)$.

Here, $P(x) = \frac{2}{x}$ and $Q(x) = x$.

The integrating factor (I.F.) is given by the formula $e^{\int P(x) dx}$.

Calculate the integral of $P(x)$:

$\int P(x) dx = \int \frac{2}{x} dx$

The integral of $\frac{1}{x}$ is $\log|x|$. So,

$\int \frac{2}{x} dx = 2 \log|x|$

Now, we find the integrating factor:

I.F. $= e^{\int P(x) dx} = e^{2 \log|x|}$

Using the logarithmic property $a \log b = \log b^a$, we have:

$2 \log|x| = \log|x|^2 = \log(x^2)$

So, the integrating factor is:

I.F. $= e^{\log(x^2)}$

Since $e^{\log a} = a$, we get:

I.F. $= x^2$

Therefore, the integrating factor of the given differential equation is $x^2$.

Comparing this with the given options:

(A) $x$

(B) $x^2$

(C) $e^{2/x}$

(D) $e^{2 \log x}$

The correct option is (B).

The given differential equation is in the standard form of a first-order linear differential equation:

$\frac{dy}{dx} + Py = Q$

where $P$ and $Q$ are functions of $x$ or constants.

To solve such an equation, we first find the integrating factor (IF), which is given by:

$\text{IF} = e^{\int P dx}$

The general solution of the linear differential equation is then obtained by multiplying the entire equation by the integrating factor and integrating:

Multiply the standard form by the integrating factor:

$(\text{IF}) \frac{dy}{dx} + P (\text{IF}) y = Q (\text{IF})$

The left side of the equation is the derivative of the product of $y$ and the integrating factor:

$\frac{d}{dx} (y \cdot \text{IF}) = Q \cdot \text{IF}$

Now, integrate both sides with respect to $x$:

$\int \frac{d}{dx} (y \cdot \text{IF}) dx = \int Q \cdot \text{IF} dx$

This gives the general solution:

$y \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dx + C$

where $C$ is the constant of integration.

Comparing this with the given options:

(A) $y \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dx + C$

(B) $y \cdot (\text{IF}) = \int P \cdot (\text{IF}) dx + C$

(C) $x \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dy + C$

(D) $x \cdot (\text{IF}) = \int P \cdot (\text{IF}) dy + C$

The correct option is (A).

A differential equation can have a general solution, which contains arbitrary constants, and a particular solution, which is a specific case of the general solution.

The general solution represents a family of curves that satisfy the differential equation. To find a particular solution, we need additional information, typically in the form of initial conditions or boundary conditions.

Initial conditions usually specify the value of the dependent variable (and possibly its derivatives) at a particular value of the independent variable. Boundary conditions specify values at different points of the independent variable.

By substituting these conditions into the general solution, we can determine the specific values of the arbitrary constants. Once these constants are determined, the resulting equation is the particular solution.

Let's examine the options:

(A) Substituting particular values for the variables (like $x$ or $y$) into the differential equation itself without reference to a general solution or conditions does not yield a particular solution of the differential equation.

(B) Assigning specific values to the arbitrary constants in the general solution, usually based on given initial or boundary conditions. This is the correct method for obtaining a particular solution.

(C) Using the integrating factor method is a technique used to solve certain types of differential equations (specifically, first-order linear differential equations) and leads to the general solution.

(D) Separating the variables is another method for solving specific types of differential equations, which also leads to the general solution.

Therefore, the particular solution of a differential equation is obtained by assigning specific values to the arbitrary constants in the general solution, based on given initial or boundary conditions.

The correct option is (B).

The given differential equation is:

$y = \sqrt{\frac{dy}{dx}} + x$

To determine the order and degree of a differential equation, it must be in a polynomial form with respect to the derivatives. The presence of a square root of a derivative means we need to eliminate it.

First, isolate the term with the derivative:

$y - x = \sqrt{\frac{dy}{dx}}$

Now, square both sides of the equation to eliminate the square root:

$(y - x)^2 = \left(\sqrt{\frac{dy}{dx}}\right)^2$

$(y - x)^2 = \frac{dy}{dx}$

Now the differential equation is in a polynomial form with respect to its derivatives. The order of a differential equation is the order of the highest derivative present in the equation. The degree of a differential equation is the highest power of the highest order derivative, after the equation has been cleared of radicals and fractions with respect to the derivatives.

In the equation $(y - x)^2 = \frac{dy}{dx}$:

The highest order derivative present is $\frac{dy}{dx}$, which is a first-order derivative. Therefore, the order of the differential equation is 1.

The highest power of the highest order derivative ($\frac{dy}{dx}$) is 1 (since $\frac{dy}{dx}$ is raised to the power of 1).

Therefore, the degree of the differential equation is 1.

So, the order is 1 and the degree is 1.

Comparing this with the given options:

(A) Order 1, Degree 1

(B) Order 1, Degree 2

(C) Order 2, Degree 1

(D) Order 1, Degree undefined

The correct option is (A).

The given differential equation is:

$\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$

This is a separable differential equation. We can separate the variables by moving all terms involving $y$ to one side and all terms involving $x$ to the other side.

Multiply both sides by $dx$ and by $\frac{1}{1+y^2}$:

$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$

Now, integrate both sides of the equation:

$\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$

The integral of $\frac{1}{1+y^2}$ with respect to $y$ is $\tan^{-1}y$.

The integral of $\frac{1}{1+x^2}$ with respect to $x$ is $\tan^{-1}x$.

So, upon integration, we get:

$\tan^{-1}y = \tan^{-1}x + C$

where $C$ is the constant of integration.

Comparing this general solution with the given options:

(A) $\tan^{-1}y = \tan^{-1}x + C$

(B) $\tan^{-1}y = \log|1+x^2| + C$

(C) $\log|1+y^2| = \tan^{-1}x + C$

(D) $\frac{y^3}{3} = \frac{x^3}{3} + C$

The obtained general solution matches option (A).

The equation of a circle passing through the origin and having its center on the x-axis is given by:

$(x-a)^2 + y^2 = a^2$

where $(a, 0)$ is the center of the circle and $a$ is the radius (or negative of radius, depending on the center's position).

Expanding this equation, we get:

$x^2 - 2ax + a^2 + y^2 = a^2$

Subtracting $a^2$ from both sides:

$x^2 - 2ax + y^2 = 0$

This equation has one arbitrary constant, $a$. To form the differential equation, we need to eliminate $a$. We can do this by differentiating the equation with respect to $x$ and then substituting the value of $a$ back into the differentiated equation.

Differentiating $x^2 - 2ax + y^2 = 0$ with respect to $x$ (treating $y$ as a function of $x$):

$\frac{d}{dx}(x^2) - \frac{d}{dx}(2ax) + \frac{d}{dx}(y^2) = \frac{d}{dx}(0)$

$2x - 2a(1) + 2y \frac{dy}{dx} = 0$

$2x - 2a + 2y \frac{dy}{dx} = 0$

Now, we need to express $a$ in terms of $x, y,$ and $\frac{dy}{dx}$ from this equation. Let's solve for $a$:

$2a = 2x + 2y \frac{dy}{dx}$

$a = x + y \frac{dy}{dx}$

Substitute this expression for $a$ back into the original simplified equation $x^2 - 2ax + y^2 = 0$:

$x^2 - 2\left(x + y \frac{dy}{dx}\right)x + y^2 = 0$

Expand and simplify:

$x^2 - 2x^2 - 2xy \frac{dy}{dx} + y^2 = 0$

$-x^2 - 2xy \frac{dy}{dx} + y^2 = 0$

Rearrange the terms:

$y^2 - x^2 = 2xy \frac{dy}{dx}$

Now, let's compare this with the given options:

(A) $x - y \frac{dy}{dx} = 0$ (Incorrect)

(B) $y^2 - x^2 = 2xy \frac{dy}{dx}$ (Correct)

(C) $x^2 - y^2 = 2xy \frac{dy}{dx}$ (Incorrect sign for $x^2$ term)

(D) $2x - 2a + 2y \frac{dy}{dx} = 0 \implies a = x + y \frac{dy}{dx}$. Substitute this back into $x^2 - 2ax + y^2 = 0$ to get the differential equation. (This option describes the process correctly and the substitution leads to the correct differential equation, but it's presented as a statement of the process rather than the final equation itself as the answer choice. However, the derived equation matches option B).

The question asks for the resulting differential equation. Following the steps derived, the equation is $y^2 - x^2 = 2xy \frac{dy}{dx}$. This matches option (B).

Let's analyze each differential equation from Column I and determine its type from Column II.

(i) $\frac{dy}{dx} = e^{x+y}$

We can rewrite this equation as:

$\frac{dy}{dx} = e^x e^y$

Now, we can separate the variables:

$\frac{dy}{e^y} = e^x dx$

This is a differential equation where the variables can be separated. Therefore, its type is (b) Variables separable.

(ii) $\frac{dy}{dx} = \frac{x^2+y^2}{xy}$

To check if this is a homogeneous differential equation, we can replace $x$ with $kx$ and $y$ with $ky$ and see if the equation remains the same.

RHS = $\frac{(kx)^2+(ky)^2}{(kx)(ky)} = \frac{k^2x^2+k^2y^2}{k^2xy} = \frac{k^2(x^2+y^2)}{k^2xy} = \frac{x^2+y^2}{xy}$

Since the RHS is unchanged when $x$ and $y$ are replaced by $kx$ and $ky$, the differential equation is homogeneous. Therefore, its type is (c) Homogeneous.

(iii) $\frac{dy}{dx} + \frac{y}{x} = x$

This equation can be written in the standard form of a first-order linear differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$

Here, $P(x) = \frac{1}{x}$ and $Q(x) = x$. Thus, it is a first-order linear differential equation. Therefore, its type is (d) First order linear.

(iv) $\frac{d^2y}{dx^2} + y = 0$

The highest derivative in this equation is $\frac{d^2y}{dx^2}$, which is the second derivative. So, the order is 2. The equation is linear in $y$ and its derivatives, and the coefficients are constants. Therefore, it is a second-order linear differential equation. Its type is (a) Second order linear.

Matching the equations with their types:

(i) $\leftrightarrow$ (b)

(ii) $\leftrightarrow$ (c)

(iii) $\leftrightarrow$ (d)

(iv) $\leftrightarrow$ (a)

The given differential equation is:

$\frac{dy}{dx} + y = \sin x$

This equation is in the standard form of a first-order linear differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$.

By comparing the given equation with the standard form, we can identify $P(x)$ and $Q(x)$.

Here, $P(x) = 1$ (the coefficient of $y$) and $Q(x) = \sin x$.

The integrating factor (IF) of a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ is given by the formula:

$\text{IF} = e^{\int P(x) dx}$

In this case, $P(x) = 1$. So, we need to calculate the integral of $P(x)$:

$\int P(x) dx = \int 1 dx$

The integral of 1 with respect to $x$ is $x$. We do not need to include the constant of integration here when finding the integrating factor.

Now, we can find the integrating factor:

$\text{IF} = e^{\int 1 dx} = e^x$

Thus, the integrating factor of the differential equation $\frac{dy}{dx} + y = \sin x$ is $e^x$.

Let's compare this with the given options:

(A) $e^x$

(B) $e^{-x}$

(C) $\sin x$

(D) $\cos x$

The correct option is (A).

The general solution of a differential equation typically contains one or more arbitrary constants. These constants represent the family of solutions to the differential equation.

A particular solution is a specific member of this family of solutions. To obtain a particular solution, we need additional information, which is usually provided in the form of initial conditions or boundary conditions.

These conditions allow us to determine the specific values of the arbitrary constants present in the general solution.

Let's analyze the options:

(A) No arbitrary constants are present in the solution. This statement describes a particular solution, but it doesn't explain how it is obtained. A particular solution *results* from a process that removes arbitrary constants.

(B) Initial or boundary conditions are used to find the values of the arbitrary constants. This is the precise method by which a particular solution is derived from the general solution.

(C) The equation is solved by separating variables. This is a method for finding the general solution of a separable differential equation, not a particular solution.

(D) The integrating factor method is applied. This is a method for finding the general solution of a first-order linear differential equation, not a particular solution.

Therefore, a particular solution to a differential equation is obtained when initial or boundary conditions are used to find the values of the arbitrary constants.

The correct option is (B).

Let's analyze the given differential equation: $\frac{d^2y}{dx^2} + \sin(\frac{dy}{dx}) = 0$.

For a differential equation to have a defined degree, it must be expressible as a polynomial in its derivatives. This means that the derivatives should not appear within transcendental functions (like sine, cosine, exponential, logarithm, etc.) or in denominators or under radicals.

In the given equation, the term $\sin(\frac{dy}{dx})$ involves the derivative $\frac{dy}{dx}$ as the argument of the sine function. This is not a polynomial form with respect to the derivatives.

Therefore, the degree of this differential equation cannot be determined in the standard way, and it is considered undefined.

Now let's evaluate the Assertion (A) and Reason (R):

Assertion (A): The degree of the differential equation $\frac{d^2y}{dx^2} + \sin(\frac{dy}{dx}) = 0$ is 1.

As explained above, the presence of $\sin(\frac{dy}{dx})$ means the degree is not defined as 1. Thus, Assertion (A) is false.

Reason (R): The differential equation is not a polynomial in derivatives, so its degree is undefined.

This statement accurately describes why the degree cannot be defined in the usual manner. The term $\sin(\frac{dy}{dx})$ prevents the equation from being a polynomial in its derivatives. Thus, Reason (R) is true.

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).

The final answer is $\boxed{D}$

The given differential equation is:

$\frac{dy}{dx} = (x+y)^2$

We can see that the right-hand side of the equation is a function of $(x+y)$. This suggests that a substitution involving $(x+y)$ would simplify the equation.

Let's consider the substitution $v = x+y$.

Now we need to express $\frac{dy}{dx}$ in terms of $v$ and $\frac{dv}{dx}$.

Differentiating $v = x+y$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(y)$

$\frac{dv}{dx} = 1 + \frac{dy}{dx}$

From this, we can express $\frac{dy}{dx}$ as:

$\frac{dy}{dx} = \frac{dv}{dx} - 1$

Now substitute $v = x+y$ and $\frac{dy}{dx} = \frac{dv}{dx} - 1$ into the original differential equation:

$\frac{dv}{dx} - 1 = v^2$

This simplifies to:

$\frac{dv}{dx} = v^2 + 1$

This is now a separable differential equation in terms of $v$ and $x$, which can be easily solved.

The substitution that simplifies the differential equation to a solvable form is $v = x+y$.

Let's look at the options:

(A) $y = vx$. This is typically used for homogeneous differential equations of the form $\frac{dy}{dx} = f(\frac{y}{x})$. This is not directly applicable here.

(B) $y = x+v$. This is equivalent to $v = y-x$, which would lead to $\frac{dy}{dx} = 1 - \frac{dv}{dx}$. Substituting this would give $1 - \frac{dv}{dx} = (x + (x+v))^2 = (2x+v)^2$, which doesn't seem to simplify well.

(C) $v = x+y$. As shown above, this substitution leads to a separable differential equation $\frac{dv}{dx} = v^2 + 1$. This is the correct substitution.

(D) $v = x-y$. This would imply $\frac{dy}{dx} = 1 - \frac{dv}{dx}$. Substituting into the original equation: $1 - \frac{dv}{dx} = (x + (x-v))^2 = (2x-v)^2$, which also doesn't simplify as effectively.

Therefore, the correct substitution to use is $v = x+y$.

The given family of straight lines is $y = mx$.

In this equation, $y$ is the dependent variable, $x$ is the independent variable, and $m$ is an arbitrary constant. This family represents all straight lines passing through the origin.

To find the differential equation of this family, we need to eliminate the arbitrary constant $m$.

First, differentiate the equation $y = mx$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(mx)$

$\frac{dy}{dx} = m \frac{d}{dx}(x)$

$\frac{dy}{dx} = m \cdot 1$

$\frac{dy}{dx} = m$

Now we have an expression for $m$: $m = \frac{dy}{dx}$.

Substitute this expression for $m$ back into the original equation $y = mx$:

$y = \left(\frac{dy}{dx}\right) x$

Rearranging this equation, we get:

$y = x \frac{dy}{dx}$

This is the differential equation representing the family of straight lines passing through the origin.

Let's check the given options:

(A) $\frac{dy}{dx} = m$. This is the first derivative, but it still contains the arbitrary constant $m$, so it's not the differential equation of the family.

(B) $y = x \frac{dy}{dx}$. This is the differential equation we derived after eliminating $m$. It represents the family of lines $y=mx$.

(C) $\frac{d^2y}{dx^2} = 0$. This would be the differential equation for a family of straight lines $y = ax + b$ (where $a$ and $b$ are arbitrary constants), which represents all straight lines, not just those through the origin.

(D) $y = mx + \frac{dy}{dx}$. This does not represent the family of lines $y=mx$.

Therefore, the correct differential equation for the family of straight lines $y = mx$ is $y = x \frac{dy}{dx}$.

The given differential equation is:

$\frac{dx}{dy} + Px = Q$

Here, the dependent variable is $x$, and the independent variable is $y$. $P$ and $Q$ are functions of $y$ or constants.

This is a first-order linear differential equation in terms of $x$ as a function of $y$. The standard form for such an equation is:

$\frac{dx}{dy} + P(y)x = Q(y)$

The integrating factor (IF) for a differential equation of this form is given by:

$\text{IF} = e^{\int P(y) dy}$

Comparing this formula with the given options:

(A) $e^{\int P dx}$: This is the integrating factor for an equation where $y$ is the dependent variable and $x$ is the independent variable ($\frac{dy}{dx} + Py = Q$).

(B) $e^{\int P dy}$: This matches the formula for the integrating factor when $x$ is the dependent variable and $y$ is the independent variable.

(C) $\int P dy$: This is just the integral of $P$ with respect to $y$, not the integrating factor itself.

(D) $\int Q dy$: This is the integral of $Q$ with respect to $y$, which is part of the solution process but not the integrating factor.

Therefore, the integrating factor of the differential equation $\frac{dx}{dy} + Px = Q$ is $e^{\int P dy}$.

The given differential equation is:

$\frac{dy}{dx} = e^{x-y}$

We can rewrite the right-hand side using the property of exponents $a^{m-n} = \frac{a^m}{a^n}$:

$\frac{dy}{dx} = \frac{e^x}{e^y}$

This is a separable differential equation. To solve it, we separate the variables $y$ and $x$. Multiply both sides by $e^y dx$:

$e^y dy = e^x dx$

Now, integrate both sides:

$\int e^y dy = \int e^x dx$

The integral of $e^y$ with respect to $y$ is $e^y$.

The integral of $e^x$ with respect to $x$ is $e^x$.

So, performing the integration gives:

$e^y = e^x + C$

where $C$ is the constant of integration.

Let's compare this solution with the given options:

(A) $e^y = e^x + C$. This matches our derived solution.

(B) $e^y = -e^x + C$. This would imply $\int e^y dy = -\int e^x dx$, which is incorrect.

(C) $e^{-y} = e^x + C$. This would imply $-\int e^{-y} dy = \int e^x dx$, which is incorrect.

(D) $e^{-y} = -e^x + C$. This would imply $-\int e^{-y} dy = -\int e^x dx$, which is incorrect.

Therefore, the correct solution to the differential equation is $e^y = e^x + C$.

The given differential equation is:

$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{d^2y}{dx^2}$

To find the order and degree of a differential equation, we first need to express it in a polynomial form with respect to its derivatives, meaning that the derivatives should not appear under radicals, in denominators, or within transcendental functions.

In the given equation, the derivative $\frac{dy}{dx}$ is under a square root. To eliminate the square root, we square both sides of the equation:

$\left(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2$

$1 + \left(\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2$

Now, the equation is in a polynomial form with respect to its derivatives.

The **order** of a differential equation is the order of the highest derivative present in the equation. In the equation $1 + \left(\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2$, the highest derivative is $\frac{d^2y}{dx^2}$, which is a second-order derivative. Therefore, the order of the differential equation is 2.

The **degree** of a differential equation is the highest power of the highest order derivative, after the equation has been cleared of radicals and fractions with respect to the derivatives. In our cleared equation, $1 + \left(\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2$, the highest order derivative is $\frac{d^2y}{dx^2}$, and its highest power is 2.

Therefore, the degree of the differential equation is 2.

So, the order is 2 and the degree is 2.

Let's check the options:

(A) Order 2, Degree 1

(B) Order 1, Degree 2

(C) Order 2, Degree 2

(D) Order 1, Degree 1

The correct option is (C).

The problem provides the particular solution for the bacterial population growth:

$P(t) = P_0 e^{kt}$

where $P(t)$ is the population at time $t$, $P_0$ is the initial population at $t=0$, and $k$ is the constant of proportionality (growth rate).

We are given that the population doubles in 5 hours. This means that at $t=5$ hours, the population $P(5)$ will be twice the initial population $P_0$.

So, we can write:

$P(5) = 2 P_0$

Substitute $t=5$ into the particular solution:

$P(5) = P_0 e^{k \cdot 5}$

Now, equate the two expressions for $P(5)$:

$2 P_0 = P_0 e^{5k}$

Divide both sides by $P_0$ (assuming $P_0 \neq 0$, which is true for a population):

$2 = e^{5k}$

To solve for $k$, we take the natural logarithm (log base $e$, denoted as $\log$ or $\ln$) of both sides:

$\log(2) = \log(e^{5k})$

Using the logarithm property $\log(a^b) = b \log(a)$ and $\log(e) = 1$:

$\log(2) = 5k \log(e)$

$\log(2) = 5k \cdot 1$

$\log(2) = 5k$

Now, solve for $k$ by dividing by 5:

$k = \frac{\log 2}{5}$

Let's compare this value of $k$ with the given options:

(A) $\log 2$

(B) $\frac{\log 2}{5}$

(C) $5 \log 2$

(D) $\frac{1}{5}\log(\frac{1}{2})$

Our calculated value for $k$ matches option (B).

The given family of curves is $y = A \cos(x+B)$, where $A$ and $B$ are arbitrary constants.

To find the differential equation, we need to eliminate these two arbitrary constants by differentiating the given equation twice with respect to $x$.

First differentiation:

$\frac{dy}{dx} = \frac{d}{dx} [A \cos(x+B)]$

Using the chain rule, $\frac{d}{dx}(\cos u) = -\sin u \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = A [-\sin(x+B) \cdot \frac{d}{dx}(x+B)]$

$\frac{dy}{dx} = -A \sin(x+B) \cdot (1+0)$

$\frac{dy}{dx} = -A \sin(x+B)$            ... (i)

Second differentiation:

$\frac{d^2y}{dx^2} = \frac{d}{dx} [-A \sin(x+B)]$

Using the chain rule, $\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx}$:

$\frac{d^2y}{dx^2} = -A [\cos(x+B) \cdot \frac{d}{dx}(x+B)]$

$\frac{d^2y}{dx^2} = -A [\cos(x+B) \cdot (1+0)]$

$\frac{d^2y}{dx^2} = -A \cos(x+B)$            ... (ii)

Now, we compare the original equation $y = A \cos(x+B)$ with the second derivative in equation (ii).

From equation (ii), we see that $\frac{d^2y}{dx^2} = -[A \cos(x+B)]$.

Since $y = A \cos(x+B)$, we can substitute $y$ into this expression:

$\frac{d^2y}{dx^2} = -y$

Rearranging this equation to the standard form, we get:

$\frac{d^2y}{dx^2} + y = 0$

This is the differential equation of the given family of curves.

Let's check the options:

(A) $\frac{d^2y}{dx^2} + y = 0$

(B) $\frac{d^2y}{dx^2} - y = 0$

(C) $\frac{dy}{dx} + y = 0$

(D) $\frac{dy}{dx} - y = 0$

Our derived differential equation matches option (A).

A differential equation is classified as linear if it can be written in the form:

$a_n(x) \frac{d^ny}{dx^n} + a_{n-1}(x) \frac{d^{n-1}y}{dx^{n-1}} + \dots + a_1(x) \frac{dy}{dx} + a_0(x) y = Q(x)$

In this form, the following conditions must be met:

1. The dependent variable $y$ and all its derivatives ($\frac{dy}{dx}, \frac{d^2y}{dx^2}, \dots, \frac{d^ny}{dx^n}$) appear only to the first power.
2. The dependent variable $y$ and its derivatives are not multiplied together.
3. The dependent variable $y$ and its derivatives do not appear within transcendental functions (like sin, cos, exp, log).
4. The coefficients $a_n(x), a_{n-1}(x), \dots, a_0(x)$ and the term $Q(x)$ are functions of the independent variable (in this case, $x$) or constants.

The statement in the question describes the first three conditions. The blank space refers to the variable upon which the coefficients depend.

Based on the definition of a linear differential equation, the coefficients must be functions of the independent variable.

Let's examine the options:

(A) Dependent: If coefficients were functions of the dependent variable ($y$), the equation might not be linear.

(B) Independent: This matches the definition of a linear differential equation, where coefficients depend on the independent variable (e.g., $x$).

(C) Arbitrary constant: Coefficients are not generally functions of arbitrary constants.

(D) Derivative: Coefficients cannot be functions of the derivatives themselves for the equation to be linear.

Therefore, the correct word to complete the statement is "independent".

The given differential equation is:

$\frac{dy}{dx} + \frac{y}{x} = x^2$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

By comparing, we have $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.

First, we find the integrating factor (IF):

$\text{IF} = e^{\int P(x) dx}$

$\text{IF} = e^{\int \frac{1}{x} dx}$

$\text{IF} = e^{\log|x|}$

$\text{IF} = |x|$

We can take the IF to be $x$ (assuming $x>0$ for simplicity, or considering $x$ in the context of the domain where the equation is defined).

The general solution is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$

Substitute the values of IF, $Q(x)$:

$y \cdot x = \int x^2 \cdot x dx + C$

$xy = \int x^3 dx + C$

Now, perform the integration:

$\int x^3 dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4}$

So, the solution is:

$xy = \frac{x^4}{4} + C$

Let's compare this with the given options:

(A) $xy = \frac{x^3}{3} + C$ (Incorrect integration of $x^3$)

(B) $xy = \frac{x^4}{4} + C$ (Matches our derived solution)

(C) $y = \frac{x^3}{4} + \frac{C}{x}$ (Incorrect format, and incorrect integration of $x^3$ if rewritten as $xy$)

(D) $y = \frac{x^4}{4} + \frac{C}{x}$ (Incorrect format, if we divide option B by x, we get $y = \frac{x^3}{4} + \frac{C}{x}$)

The correct solution is $xy = \frac{x^4}{4} + C$.

The given family of parabolas is $y^2 = 4ax$.

Here, $a$ is the arbitrary constant.

To find the differential equation, we need to eliminate the constant $a$. We do this by differentiating the equation with respect to $x$.

Differentiate both sides of $y^2 = 4ax$ with respect to $x$:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)$

Using the chain rule for $\frac{d}{dx}(y^2)$: $2y \frac{dy}{dx}$.

Differentiating the right side with respect to $x$:

$\frac{d}{dx}(4ax) = 4a \frac{d}{dx}(x) = 4a(1) = 4a$

So, the differentiated equation is:

$2y \frac{dy}{dx} = 4a$            ... (i)

From this equation, we can express $4a$ in terms of $y$ and $\frac{dy}{dx}$.

Alternatively, we can express $a$ from equation (i) as $a = \frac{2y}{4} \frac{dy}{dx} = \frac{y}{2} \frac{dy}{dx}$.

Now, substitute this expression for $a$ back into the original equation $y^2 = 4ax$:

$y^2 = 4 \left(\frac{y}{2} \frac{dy}{dx}\right) x$

$y^2 = 2yx \frac{dy}{dx}$

We can simplify this by dividing both sides by $y$ (assuming $y \neq 0$):

$y = 2x \frac{dy}{dx}$

Now, let's check the options:

(A) $y \frac{dy}{dx} = 2a$. This is not a differential equation as it contains 'a'.

(B) $y = 2x \frac{dy}{dx}$. This matches our derived differential equation after simplification.

(C) $y^2 = 2x y \frac{dy}{dx}$. This is a step before simplification, and it's also a valid form of the differential equation.

(D) $2y \frac{dy}{dx} = 4a$. This is the result of the first differentiation, but it still contains 'a'.

Both (B) and (C) can be considered correct as they are equivalent forms of the differential equation. However, usually, the simplest form is preferred. In multiple-choice questions, we select the one that is explicitly listed or the most simplified form.

Let's re-examine the options. Option (B) is $y = 2x \frac{dy}{dx}$, which is a simplified version of option (C) $y^2 = 2xy \frac{dy}{dx}$ (by dividing by y).

The question asks for "the" differential equation, implying a unique form. Option (C) is a direct result of substituting $a$ from the first differentiation into the original equation without simplification. Option (B) is a further simplification. In many contexts, the simplified form is the expected answer.

Let's check if there's any reason to prefer one over the other. The process of eliminating the constant usually leads to a simplified form.

Consider the differentiation $2y \frac{dy}{dx} = 4a$. If we solve for $a$, we get $a = \frac{y}{2}\frac{dy}{dx}$. Substituting into $y^2 = 4ax$ gives $y^2 = 4x \left(\frac{y}{2}\frac{dy}{dx}\right) = 2xy\frac{dy}{dx}$. This is option (C).

If we divide $y^2 = 2xy\frac{dy}{dx}$ by $y$ (assuming $y \neq 0$), we get $y = 2x\frac{dy}{dx}$, which is option (B).

Both are mathematically equivalent for $y \neq 0$. However, if $y=0$, then the original equation $y^2=4ax$ implies $4ax=0$. If $a \neq 0$, then $x=0$. The point $(0,0)$ is on the parabola. If $a=0$, then $y^2=0$, which means $y=0$, representing the x-axis, not typically considered a family of parabolas in this context.

In the context of differential equations derived from families of curves, the simplified form is generally preferred. Option (B) is the simplified form.

The given differential equation is:

$\frac{dy}{dx} = 1+x+y+xy$

As suggested, we factorize the right-hand side:

$1+x+y+xy = (1+x) + y(1+x) = (1+x)(1+y)$

So the equation becomes:

$\frac{dy}{dx} = (1+x)(1+y)$

This is a separable differential equation. We separate the variables by bringing all terms involving $y$ to the left side and all terms involving $x$ to the right side.

Divide both sides by $(1+y)$ and multiply by $dx$:

$\frac{dy}{1+y} = (1+x) dx$

Now, integrate both sides:

$\int \frac{dy}{1+y} = \int (1+x) dx$

Integrate the left side:

$\int \frac{dy}{1+y} = \log|1+y|$

Integrate the right side:

$\int (1+x) dx = \int 1 dx + \int x dx = x + \frac{x^2}{2}$

Equating the results of the integration, we get:

$\log|1+y| = x + \frac{x^2}{2} + C$

where $C$ is the constant of integration.

Comparing this with the given options:

(A) $\log|1+y| = x + \frac{x^2}{2} + C$ (Matches our derived solution)

(B) $\log|1+x| = y + \frac{y^2}{2} + C$ (Incorrect separation of variables and integration)

(C) $\tan^{-1}y = \tan^{-1}x + C$ (This would be the solution if the equation was $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$)

(D) $e^{x+y} = C$ (This would be the solution if the equation was $\frac{dy}{dx} = -e^{x+y}$ after integration)

The correct general solution is $\log|1+y| = x + \frac{x^2}{2} + C$.

Let's analyze each option in the context of solving first-order, first-degree differential equations:

A first-order differential equation involves only the first derivative ($\frac{dy}{dx}$ or $\frac{dx}{dy}$). A first-degree differential equation means that the highest power of this first derivative is 1.

(A) Variables separable method: This is a fundamental method used to solve first-order differential equations where the variables can be separated, allowing for direct integration. For example, an equation of the form $\frac{dy}{dx} = f(x)g(y)$ can be solved using this method.

(B) Homogeneous differential equations: A differential equation of the form $\frac{dy}{dx} = f(\frac{y}{x})$ is called homogeneous. These can be solved using the substitution $y=vx$, which transforms them into separable equations. So, this is a method for solving certain first-order, first-degree equations.

(C) Linear differential equations: A first-order linear differential equation has the form $\frac{dy}{dx} + P(x)y = Q(x)$. These equations can be solved using an integrating factor. This is a major class of first-order, first-degree equations.

(D) Method of variation of parameters: The method of variation of parameters is primarily used to find a particular solution of a linear non-homogeneous differential equation when the complementary function (general solution of the homogeneous part) is known. While it can be applied to first-order linear equations, its typical and most significant application is for higher-order linear differential equations (second-order and above).

The question asks which method is NOT a method for solving *first-order first-degree* differential equations. While variation of parameters *can* be applied to a first-order linear equation, its primary domain is higher-order equations. The other three options (A, B, C) are direct methods for classifying and solving specific types of first-order, first-degree differential equations.

Given the parenthetical note in option (D), it strongly indicates that this method is not typically categorized as a primary method for solving first-order equations in the same way the others are.

Therefore, the method that is NOT primarily a method for solving first-order first-degree differential equations among the given choices is the Method of Variation of Parameters, especially considering its main use with higher-order equations.

Consider a first-order linear differential equation in the standard form:

$\frac{dy}{dx} + P(x)y = Q(x)$

The integrating factor (IF) for this equation is given by:

$\text{IF} = e^{\int P(x) dx}$

When we multiply the entire differential equation by the integrating factor, the left-hand side (LHS) becomes the derivative of the product of the dependent variable ($y$) and the integrating factor (IF).

Let's demonstrate this:

Multiply the equation by IF:

$\text{IF} \cdot \frac{dy}{dx} + \text{IF} \cdot P(x)y = \text{IF} \cdot Q(x)$

We know that the derivative of a product $(u \cdot v)$ is given by the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.

Let $u = y$ and $v = \text{IF}$. Then, $\frac{du}{dx} = \frac{dy}{dx}$ and $\frac{dv}{dx} = \frac{d}{dx}(\text{IF})$.

Recall that $\text{IF} = e^{\int P(x) dx}$. Differentiating IF with respect to $x$ gives:

$\frac{d}{dx}(\text{IF}) = \frac{d}{dx}(e^{\int P(x) dx}) = e^{\int P(x) dx} \cdot \frac{d}{dx}\left(\int P(x) dx\right)$

Since $\frac{d}{dx}\left(\int P(x) dx\right) = P(x)$, we have:

$\frac{d}{dx}(\text{IF}) = \text{IF} \cdot P(x)$

Now, let's look at the derivative of the product $y \cdot \text{IF}$:

$\frac{d}{dx}(y \cdot \text{IF}) = y \frac{d}{dx}(\text{IF}) + \text{IF} \frac{dy}{dx}$

Substitute $\frac{d}{dx}(\text{IF}) = \text{IF} \cdot P(x)$ into this equation:

$\frac{d}{dx}(y \cdot \text{IF}) = y (\text{IF} \cdot P(x)) + \text{IF} \frac{dy}{dx}$

$\frac{d}{dx}(y \cdot \text{IF}) = \text{IF} \cdot P(x)y + \text{IF} \frac{dy}{dx}$

This matches the left-hand side of the equation after multiplying by the IF:

$\text{IF} \cdot \frac{dy}{dx} + \text{IF} \cdot P(x)y$

Therefore, when the differential equation is multiplied by the integrating factor, the LHS becomes the derivative of the product $y \cdot (\text{IF})$.

Let's check the options:

(A) $y \cdot Q$. This is related to the RHS after multiplying by IF, not the LHS derivative.

(B) $y \cdot (\text{IF})$. This matches our derivation.

(C) $x \cdot y$. This is a product of variables, not necessarily involving the integrating factor or the correct derivative structure.

(D) $y \cdot P$. This is a product involving $y$ and $P(x)$, but not the integrating factor, and doesn't represent the derivative of a product.

The product is $y \cdot (\text{IF})$.

A differential equation of order $n$ has a general solution that contains exactly $n$ arbitrary constants.

A particular solution is obtained from the general solution by specifying values for these arbitrary constants, usually based on initial conditions or boundary conditions.

When we determine the values of these $n$ arbitrary constants using $n$ independent conditions, the constants are no longer arbitrary; they become fixed numerical values.

Therefore, a particular solution does not contain any arbitrary constants.

Let's examine the options:

(A) Equal to the order. This describes the number of arbitrary constants in the *general* solution, not the particular solution.

(B) Equal to the degree. The degree does not determine the number of arbitrary constants.

(C) Zero. This is correct, as all arbitrary constants are determined to find a particular solution.

(D) One. This is generally not true; the number of constants depends on the order of the equation.

The number of arbitrary constants in the particular solution of a differential equation is zero.

The given family of circles is represented by the equation:

$(x-a)^2 + (y-b)^2 = r^2$

In this equation, $a$ and $b$ are arbitrary constants. The radius $r$ is also a constant for a specific circle, but for the entire family of circles, $r$ could also be considered variable, or it might be fixed if the family has a specific characteristic (like all passing through a point or having a fixed radius).

However, the question specifically mentions $a$ and $b$ as arbitrary constants that need to be eliminated to form the differential equation.

The general principle is that to eliminate $n$ arbitrary constants from an equation, we need to differentiate it $n$ times and then use these $n+1$ equations to eliminate the constants.

In this case, we have two arbitrary constants, $a$ and $b$. Therefore, we would need to differentiate the equation twice with respect to $x$ to obtain three equations (the original equation and two differentiated equations). These three equations would then be used to eliminate the two arbitrary constants $a$ and $b$.

The order of the resulting differential equation is equal to the number of arbitrary constants in the original family of curves.

Since there are two arbitrary constants ($a$ and $b$), the order of the differential equation will be 2.

Let's confirm by differentiating:

Original: $(x-a)^2 + (y-b)^2 = r^2$

1st derivative: $2(x-a) + 2(y-b)\frac{dy}{dx} = 0 \implies (x-a) + (y-b)\frac{dy}{dx} = 0$

2nd derivative: $1 + \left(\frac{dy}{dx}\right)^2 + (y-b)\frac{d^2y}{dx^2} = 0$

In these differentiated equations, we can eliminate $a$ and $b$. For instance, from the first derivative, $(x-a) = -(y-b)\frac{dy}{dx}$. Substituting this into the second derivative equation can eliminate $a$ and $b$. The highest derivative present will be $\frac{d^2y}{dx^2}$, indicating an order of 2.

Therefore, the order of the differential equation of the family of circles $(x-a)^2 + (y-b)^2 = r^2$ is 2.

The given differential equation is:

$\frac{dy}{dx} = \frac{y}{x}$

This is a separable differential equation. We can separate the variables $y$ and $x$.

Divide both sides by $y$ and multiply by $dx$ (assuming $y \neq 0$ and $x \neq 0$):

$\frac{dy}{y} = \frac{dx}{x}$

Now, integrate both sides:

$\int \frac{dy}{y} = \int \frac{dx}{x}$

The integral of $\frac{1}{y}$ with respect to $y$ is $\log|y|$.

The integral of $\frac{1}{x}$ with respect to $x$ is $\log|x|$.

So, performing the integration gives:

$\log|y| = \log|x| + C_1$

where $C_1$ is the constant of integration.

Using the logarithm property $\log a + \log b = \log(ab)$, we can write:

$\log|y| = \log|x| + \log(C')$ where $C' = e^{C_1}$ (since $e^{C_1}$ is a positive constant).

$\log|y| = \log(|x| C')$

Exponentiating both sides (taking $e$ to the power of both sides):

$|y| = |x| C'$

We can remove the absolute value signs by introducing a new constant $C$, which can be positive or negative (or zero if $y=0$ is a solution, which it is). Let $C = \pm C'$.

$y = Cx$

This is the general solution.

Let's check the options:

(A) $y = Cx$. This matches our derived solution.

(B) $y = Cx^2$. If $y=Cx^2$, then $\frac{dy}{dx} = 2Cx$. The equation becomes $2Cx = \frac{Cx^2}{x} = Cx$, which is $2Cx = Cx$. This is only true if $Cx=0$. So, this is not the general solution.

(C) $y = C \log x$. If $y=C \log x$, then $\frac{dy}{dx} = \frac{C}{x}$. The equation becomes $\frac{C}{x} = \frac{C \log x}{x}$, which implies $\log x = 1$, which is not true for all $x$. So, this is not the general solution.

(D) $y = C e^x$. If $y=C e^x$, then $\frac{dy}{dx} = C e^x$. The equation becomes $C e^x = \frac{C e^x}{x}$, which implies $1 = \frac{1}{x}$ or $x=1$. This is not true for all $x$. So, this is not the general solution.

The general solution is $y = Cx$.

A differential equation is of the variables separable type if it can be written in the form:

$\frac{dy}{dx} = F(x, y)$

where $F(x, y)$ can be expressed as a product of a function of $x$ only and a function of $y$ only. That is, $F(x, y) = f(x) g(y)$.

Let's examine the given options:

(A) $\frac{dy}{dx} = f(x) + g(y)$

This is the form of an equation where the variables are not separable in the standard way. It might be solvable by other methods, but it's not the definition of a separable equation.

(B) $\frac{dy}{dx} = f(x) g(y)$

This form directly matches the definition of a variables separable differential equation. We can rewrite it as $\frac{dy}{g(y)} = f(x) dx$, allowing for direct integration of both sides.

(C) $\frac{dy}{dx} = f(\frac{y}{x})$

This is the form of a homogeneous differential equation. While homogeneous equations can often be solved using a substitution ($v = y/x$) that leads to a separable equation, this form itself is not the definition of a variables separable equation.

(D) $\frac{dy}{dx} + Py = Q$

This is the standard form of a first-order linear differential equation. These are solved using an integrating factor, not typically by separation of variables directly (unless $P$ and $Q$ are such that it simplifies to a separable form).

Therefore, the condition for a differential equation to be of the variables separable type is that it can be expressed in the form $\frac{dy}{dx} = f(x) g(y)$.

Let's analyze the given differential equation:

$(\frac{d^3y}{dx^3})^2 + (\frac{dy}{dx})^5 + y = 0$

To determine the order and degree of a differential equation, we first ensure that the equation is in a polynomial form with respect to its derivatives, meaning no derivatives appear under radicals, in denominators, or within transcendental functions.

In the given equation, the derivatives are $\frac{d^3y}{dx^3}$ and $\frac{dy}{dx}$. These derivatives are not under radicals, in denominators, or within transcendental functions. Thus, the equation is already in a form suitable for determining its order and degree.

The **order** of a differential equation is the order of the highest derivative present. In this equation, the highest derivative is $\frac{d^3y}{dx^3}$, which is a third-order derivative. Therefore, the order of the differential equation is 3.

The **degree** of a differential equation is the highest power (exponent) of the highest order derivative, after the equation has been cleared of radicals and fractions with respect to the derivatives. In this equation, the highest order derivative is $\frac{d^3y}{dx^3}$, and its power is 2.

Now let's evaluate the Assertion (A) and Reason (R):

Assertion (A): The degree of the differential equation $(\frac{d^3y}{dx^3})^2 + (\frac{dy}{dx})^5 + y = 0$ is 2.

Based on our analysis, the degree of the differential equation is indeed 2. So, Assertion (A) is true.

Reason (R): The highest order derivative is $\frac{d^3y}{dx^3}$ and its power is 2.

The highest order derivative is correctly identified as $\frac{d^3y}{dx^3}$. The power of this highest order derivative is indeed 2. So, Reason (R) is true.

Furthermore, Reason (R) correctly explains why the degree is 2. The degree is determined by the power of the highest order derivative. Since the highest order derivative ($\frac{d^3y}{dx^3}$) has a power of 2, and this is the highest power among all terms involving derivatives in their highest order appearance, the degree is 2.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) provides the correct justification for Assertion (A).

The correct option is (A).

The given differential equation is:

$\frac{dy}{dx} - 3y \cot x = \sin 2x$

This equation is in the standard form of a first-order linear differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$.

By comparing, we identify $P(x)$ and $Q(x)$.

$P(x) = -3 \cot x$

$Q(x) = \sin 2x$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Let's calculate the integral of $P(x)$:

$\int P(x) dx = \int (-3 \cot x) dx$

$\int P(x) dx = -3 \int \cot x dx$

We know that $\int \cot x dx = \log|\sin x|$.

So, $\int P(x) dx = -3 \log|\sin x|$

Using the logarithmic property $a \log b = \log b^a$:

$\int P(x) dx = \log|(\sin x)^{-3}|$

$\int P(x) dx = \log|\sin^{-3} x|$

$\int P(x) dx = \log|\text{cosec}^3 x|$

Now, we find the integrating factor:

$\text{IF} = e^{\int P(x) dx} = e^{\log|\text{cosec}^3 x|}$

Since $e^{\log a} = a$, we get:

$\text{IF} = |\text{cosec}^3 x|$

We can take the integrating factor as $\text{cosec}^3 x$ (assuming $\text{cosec}^3 x > 0$ for the domain, or by convention, we drop the absolute value). Note that $\text{cosec} x = \frac{1}{\sin x}$.

Let's examine the options:

(A) $\sin^3 x$. This is $(\text{cosec}^3 x)^{-1}$, which is incorrect.

(B) $\text{cosec}^3 x$. This matches our calculated integrating factor.

(C) $e^{-3 \cot x}$. This would be the integrating factor if $P(x) = -3 \cot x$ and the integral was taken as $-3 \log|\sin x|$. However, the exponentiation step would result in $e^{\log|\sin x|^{-3}} = |\sin x|^{-3} = |\text{cosec}^3 x|$. The minus sign in the exponent of the option is incorrect.

(D) $e^{3 \cot x}$. This would imply $P(x) = 3 \cot x$, which is not the case.

The correct integrating factor is $\text{cosec}^3 x$.

The general solution of a differential equation contains a number of arbitrary constants equal to the order of the differential equation.

The given general solution is:

$y = C_1 e^x + C_2 e^{2x}$

In this solution, we can identify the arbitrary constants:

$C_1$ and $C_2$.

There are exactly two arbitrary constants in the general solution.

Therefore, the order of the differential equation that has this general solution must be 2.

To illustrate, we can derive the differential equation:

Given $y = C_1 e^x + C_2 e^{2x}$

First derivative: $\frac{dy}{dx} = C_1 e^x + 2 C_2 e^{2x}$

Second derivative: $\frac{d^2y}{dx^2} = C_1 e^x + 4 C_2 e^{2x}$

We can eliminate $C_1$ and $C_2$. For example, let's try to form a second-order equation. We can see that the terms $C_1 e^x$ and $C_2 e^{2x}$ are involved. The characteristic equation for a second-order linear homogeneous differential equation with constant coefficients has roots corresponding to the exponents in the solution. If the roots are $r_1$ and $r_2$, the solution is $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}$. Here, the roots are 1 and 2.

The characteristic equation with roots 1 and 2 is $(r-1)(r-2) = 0$.

$r^2 - 3r + 2 = 0$

The corresponding differential equation is obtained by replacing $r^2$ with $\frac{d^2y}{dx^2}$, $r$ with $\frac{dy}{dx}$, and the constant term with $y$:

$\frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + 2y = 0$

This is a second-order differential equation, confirming that the number of arbitrary constants in the general solution corresponds to the order of the differential equation.

Since there are two arbitrary constants ($C_1$ and $C_2$), the order of the differential equation is 2.

The given differential equation is:

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$

To Find: The order and degree of the differential equation.

Solution:

The order of a differential equation is the order of the highest derivative present in the equation.

In the given equation, $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$, the highest derivative is $\frac{d^2y}{dx^2}$, which is the second derivative.

Therefore, the order of the differential equation is 2.

The degree of a differential equation is the highest power of the highest order derivative, provided that the differential equation is expressed as a polynomial in the derivatives.

The given equation is:

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + y = 0$

The highest order derivative is $\frac{d^2y}{dx^2}$. The power of this derivative is 1.

The equation is already a polynomial in its derivatives, so we can directly identify the degree.

Therefore, the degree of the differential equation is 1.

Conclusion:

The order of the differential equation is 2.

The degree of the differential equation is 1.

Given the differential equation: $$ \frac{d^2y}{dx^2} - y = 0 $$ And the proposed solution: $$ y = Ae^x + Be^{-x} $$ where A and B are arbitrary constants.

To Verify: We need to show that when we substitute $y$ and its second derivative into the differential equation, the equation holds true.

Step 1: Find the first derivative of y with respect to x ($\frac{dy}{dx}$)

Differentiating $y = Ae^x + Be^{-x}$ with respect to x:

$\frac{dy}{dx} = \frac{d}{dx}(Ae^x + Be^{-x})$

$\frac{dy}{dx} = A\frac{d}{dx}(e^x) + B\frac{d}{dx}(e^{-x})$

$\frac{dy}{dx} = Ae^x + B(-e^{-x})$

$$ \frac{dy}{dx} = Ae^x - Be^{-x} $$

Step 2: Find the second derivative of y with respect to x ($\frac{d^2y}{dx^2}$)

Differentiating $\frac{dy}{dx}$ with respect to x:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(Ae^x - Be^{-x})$

$\frac{d^2y}{dx^2} = A\frac{d}{dx}(e^x) - B\frac{d}{dx}(e^{-x})$

$\frac{d^2y}{dx^2} = Ae^x - B(-e^{-x})$

$$ \frac{d^2y}{dx^2} = Ae^x + Be^{-x} $$

Step 3: Substitute y and $\frac{d^2y}{dx^2}$ into the differential equation

The differential equation is $\frac{d^2y}{dx^2} - y = 0$.

Substitute the expressions for $\frac{d^2y}{dx^2}$ and $y$:

$(Ae^x + Be^{-x}) - (Ae^x + Be^{-x}) = 0$

$$ Ae^x + Be^{-x} - Ae^x - Be^{-x} = 0 $$

Step 4: Simplify the equation

The terms cancel out:

$(Ae^x - Ae^x) + (Be^{-x} - Be^{-x}) = 0$

$$ 0 + 0 = 0 $$

$$ 0 = 0 $$

Conclusion:

Since the substitution of $y = Ae^x + Be^{-x}$ and its second derivative into the given differential equation results in a true statement ($0 = 0$), we have verified that $y = Ae^x + Be^{-x}$ is indeed a solution to the differential equation $\frac{d^2y}{dx^2} - y = 0$.

Given the family of curves: $$ y = ax $$ where $a$ is an arbitrary constant.

Objective: To form a differential equation that represents this family of curves, we need to eliminate the arbitrary constant $a$.

Step 1: Differentiate the given equation with respect to $x$.

Differentiating $y = ax$ with respect to $x$:

$$ \frac{dy}{dx} = \frac{d}{dx}(ax) $$

$$ \frac{dy}{dx} = a $$

Step 2: Express the arbitrary constant $a$ in terms of $y$ and $x$.

From the differentiated equation, we have $a = \frac{dy}{dx}$.

Step 3: Substitute the expression for $a$ back into the original equation.

The original equation is $y = ax$. Substitute $a = \frac{dy}{dx}$ into this equation:

$$ y = \left(\frac{dy}{dx}\right)x $$

Step 4: Rearrange the equation to obtain the differential equation.

We can rewrite the equation as:

$$ y = x\frac{dy}{dx} $$

Or, arranging it in a standard form:

$$ x\frac{dy}{dx} - y = 0 $$

Final Answer:

The differential equation representing the family of curves $y = ax$ is: $$ x\frac{dy}{dx} - y = 0 $$

Given the differential equation: $$ \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} $$

This is a separable differential equation. We can separate the variables $x$ and $y$ to different sides of the equation.

Step 1: Separate the variables.

Multiply both sides by $dx$ and divide both sides by $(1 + y^2)$: $$ \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} $$

Step 2: Integrate both sides of the equation.

Integrate the left side with respect to $y$ and the right side with respect to $x$: $$ \int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2} $$

The integral of $\frac{1}{1+y^2}$ is $\arctan(y)$, and the integral of $\frac{1}{1+x^2}$ is $\arctan(x)$. So, we have:

$$ \arctan(y) = \arctan(x) + C $$

where $C$ is the constant of integration.

Step 3: Solve for $y$.

To solve for $y$, we can take the tangent of both sides:

$$ \tan(\arctan(y)) = \tan(\arctan(x) + C) $$

$$ y = \tan(\arctan(x) + C) $$

We can use the tangent addition formula, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Let $A = \arctan(x)$ and $B = C$. Then $\tan A = x$ and $\tan B = \tan C$. Let $\tan C = k$, where $k$ is a constant.

$$ y = \frac{\tan(\arctan(x)) + \tan(C)}{1 - \tan(\arctan(x))\tan(C)} $$

$$ y = \frac{x + k}{1 - xk} $$

Final Answer:

The general solution to the differential equation is: $$ y = \frac{x + k}{1 - xk} $$ or $$ \arctan(y) = \arctan(x) + C $$

Given the differential equation: $$ \frac{dy}{dx} = e^{x+y} $$

We can rewrite the right-hand side using the property of exponents $e^{a+b} = e^a \cdot e^b$: $$ \frac{dy}{dx} = e^x e^y $$

This is a separable differential equation. We can separate the variables $x$ and $y$ to different sides of the equation.

Step 1: Separate the variables.

Divide both sides by $e^y$ (or multiply by $e^{-y}$) and multiply both sides by $dx$: $$ e^{-y} dy = e^x dx $$

Step 2: Integrate both sides of the equation.

Integrate the left side with respect to $y$ and the right side with respect to $x$: $$ \int e^{-y} dy = \int e^x dx $$

The integral of $e^{-y}$ with respect to $y$ is $-e^{-y}$, and the integral of $e^x$ with respect to $x$ is $e^x$. So, we have:

$$ -e^{-y} = e^x + C $$

where $C$ is the constant of integration.

Step 3: Solve for $y$.

We can rearrange the equation to solve for $y$. First, multiply by -1:

$$ e^{-y} = -e^x - C $$

Let $K = -C$, where $K$ is also an arbitrary constant:

$$ e^{-y} = K - e^x $$

Now, take the natural logarithm of both sides:

$$ \ln(e^{-y}) = \ln(K - e^x) $$

$$ -y = \ln(K - e^x) $$

Finally, multiply by -1 to solve for $y$: $$ y = -\ln(K - e^x) $$

Final Answer:

The general solution to the differential equation is: $$ y = -\ln(K - e^x) $$ or $$ -e^{-y} = e^x + C $$

Given the differential equation: $$ (x-y) \frac{dy}{dx} = x + 2y $$

Part 1: Show that the differential equation is homogeneous.

First, rewrite the equation in the form $\frac{dy}{dx} = f(x, y)$: $$ \frac{dy}{dx} = \frac{x + 2y}{x - y} $$

A differential equation $\frac{dy}{dx} = f(x, y)$ is homogeneous if $f(tx, ty) = f(x, y)$ for any scalar $t$. Let's check this condition for our function $f(x, y) = \frac{x + 2y}{x - y}$.

Replace $x$ with $tx$ and $y$ with $ty$: $$ f(tx, ty) = \frac{(tx) + 2(ty)}{(tx) - (ty)} $$

Factor out $t$ from the numerator and the denominator: $$ f(tx, ty) = \frac{t(x + 2y)}{t(x - y)} $$

Cancel out $t$: $$ f(tx, ty) = \frac{x + 2y}{x - y} $$

Since $f(tx, ty) = f(x, y)$, the differential equation is homogeneous.

Part 2: Solve the differential equation.

For a homogeneous differential equation, we use the substitution $y = vx$. Then, $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x + 2y}{x - y}$: $$ v + x\frac{dv}{dx} = \frac{x + 2(vx)}{x - (vx)} $$

$$ v + x\frac{dv}{dx} = \frac{x(1 + 2v)}{x(1 - v)} $$

$$ v + x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} $$

Now, isolate $x\frac{dv}{dx}$: $$ x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v $$

Combine the terms on the right-hand side: $$ x\frac{dv}{dx} = \frac{1 + 2v - v(1 - v)}{1 - v} $$

$$ x\frac{dv}{dx} = \frac{1 + 2v - v + v^2}{1 - v} $$

$$ x\frac{dv}{dx} = \frac{1 + v + v^2}{1 - v} $$

Now, separate the variables $v$ and $x$: $$ \frac{1 - v}{1 + v + v^2} dv = \frac{1}{x} dx $$

Integrate both sides:

$$ \int \frac{1 - v}{1 + v + v^2} dv = \int \frac{1}{x} dx $$

For the left-hand integral, let's manipulate the numerator to relate it to the derivative of the denominator. The derivative of $1 + v + v^2$ is $1 + 2v$. We can write $1-v$ as $-\frac{1}{2}(1+2v) + \frac{3}{2}$.

$$ \int \frac{-\frac{1}{2}(1+2v) + \frac{3}{2}}{1 + v + v^2} dv = \int \frac{1}{x} dx $$

$$ -\frac{1}{2} \int \frac{1+2v}{1 + v + v^2} dv + \frac{3}{2} \int \frac{1}{1 + v + v^2} dv = \int \frac{1}{x} dx $$

The first integral on the left is of the form $\int \frac{f'(v)}{f(v)} dv = \ln|f(v)|$: $$ -\frac{1}{2} \ln|1 + v + v^2| $$

For the second integral, we complete the square for the denominator $1 + v + v^2$: $1 + v + v^2 = (v^2 + v + \frac{1}{4}) + 1 - \frac{1}{4} = (v + \frac{1}{2})^2 + \frac{3}{4}$. The integral becomes $\int \frac{1}{(v + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dv$. This is of the form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan(\frac{u}{a})$. Here, $u = v + \frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$.

$$ \frac{3}{2} \int \frac{1}{(v + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dv = \frac{3}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) $$

$$ = \frac{3}{\sqrt{3}} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) = \sqrt{3} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) $$

The integral on the right is $\int \frac{1}{x} dx = \ln|x| + C_1$.

Combining the integrated terms: $$ -\frac{1}{2} \ln|1 + v + v^2| + \sqrt{3} \arctan\left(\frac{2v + 1}{\sqrt{3}}\right) = \ln|x| + C_1 $$

Now, substitute back $v = \frac{y}{x}$: $$ -\frac{1}{2} \ln\left|1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2\right| + \sqrt{3} \arctan\left(\frac{2(\frac{y}{x}) + 1}{\sqrt{3}}\right) = \ln|x| + C_1 $$

$$ -\frac{1}{2} \ln\left|\frac{x^2 + xy + y^2}{x^2}\right| + \sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = \ln|x| + C_1 $$

$$ -\frac{1}{2} (\ln|x^2 + xy + y^2| - \ln|x^2|) + \sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = \ln|x| + C_1 $$

$$ -\frac{1}{2} \ln|x^2 + xy + y^2| + \frac{1}{2} \ln|x^2| + \sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = \ln|x| + C_1 $$

$$ -\frac{1}{2} \ln|x^2 + xy + y^2| + \ln|x| + \sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = \ln|x| + C_1 $$

The $\ln|x|$ terms cancel out:

$$ -\frac{1}{2} \ln|x^2 + xy + y^2| + \sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = C_1 $$

Multiply by -2 for a cleaner form and let $C = -2C_1$: $$ \ln|x^2 + xy + y^2| - 2\sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = C $$

Final Answer:

The differential equation is homogeneous. The solution is: $$ \ln|x^2 + xy + y^2| - 2\sqrt{3} \arctan\left(\frac{2y + x}{\sqrt{3}x}\right) = C $$

The given differential equation is: $$ \frac{dy}{dx} + \frac{y}{x} = x^2 $$

This is a first-order linear differential equation of the form: $$ \frac{dy}{dx} + P(x)y = Q(x) $$

By comparing the given equation with the standard form, we can identify $P(x)$ and $Q(x)$.

Here, $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.

The integrating factor (I.F.) of a linear differential equation is given by the formula: $$ \text{I.F.} = e^{\int P(x) dx} $$

Now, we need to find the integral of $P(x)$ with respect to $x$: $$ \int P(x) dx = \int \frac{1}{x} dx $$

$$ \int \frac{1}{x} dx = \ln|x| $$

Note that we do not need to add the constant of integration here as it will be incorporated into the final solution.

Now, substitute this integral back into the formula for the integrating factor: $$ \text{I.F.} = e^{\ln|x|} $$

Using the property $e^{\ln a} = a$: $$ \text{I.F.} = |x| $$

For practical purposes, and as $x$ often represents a domain where the solution is valid, we can consider the positive value. If the domain is restricted, we use $|x|$. If we consider $x>0$, then I.F. = $x$. If we consider $x<0$, then I.F. = $-x$. However, typically we can represent the integrating factor as $x$ for the purpose of solving the differential equation, assuming a suitable domain for $x$.

Final Answer:

The integrating factor of the given differential equation is $\mathbf{x}$.

The given differential equation is: $$ \frac{dy}{dx} + y \sec x = \tan x $$

This is a first-order linear differential equation of the form: $$ \frac{dy}{dx} + P(x)y = Q(x) $$

By comparing the given equation with the standard form, we identify: $P(x) = \sec x$ $Q(x) = \tan x$

The integrating factor (I.F.) is given by $e^{\int P(x) dx}$. First, calculate the integral of $P(x)$: $$ \int P(x) dx = \int \sec x dx $$

The integral of $\sec x$ is $\ln|\sec x + \tan x|$: $$ \int \sec x dx = \ln|\sec x + \tan x| $$

Now, find the integrating factor:

$$ \text{I.F.} = e^{\ln|\sec x + \tan x|} $$

$$ \text{I.F.} = |\sec x + \tan x| $$

For simplicity in solving, we can often drop the absolute value, assuming we are working in an interval where $\sec x + \tan x > 0$. Let's use $\sec x + \tan x$ as the integrating factor.

Multiply the entire differential equation by the integrating factor:

$$ (\sec x + \tan x) \frac{dy}{dx} + (\sec x + \tan x) y \sec x = (\sec x + \tan x) \tan x $$

The left side of the equation is the derivative of the product of $y$ and the integrating factor:

$$ \frac{d}{dx} [y (\sec x + \tan x)] = (\sec x + \tan x) \tan x $$

Expand the right side: $$ \frac{d}{dx} [y (\sec x + \tan x)] = \sec x \tan x + \tan^2 x $$

Now, integrate both sides with respect to $x$: $$ \int \frac{d}{dx} [y (\sec x + \tan x)] dx = \int (\sec x \tan x + \tan^2 x) dx $$

The left side integrates to $y (\sec x + \tan x)$.

For the right side, we integrate term by term:

$$ \int \sec x \tan x dx = \sec x $$

For $\int \tan^2 x dx$, we use the identity $\tan^2 x = \sec^2 x - 1$: $$ \int \tan^2 x dx = \int (\sec^2 x - 1) dx = \int \sec^2 x dx - \int 1 dx = \tan x - x $$

So, the integral of the right side is: $$ \int (\sec x \tan x + \tan^2 x) dx = \sec x + \tan x - x + C $$

Equating the integrated left and right sides:

$$ y (\sec x + \tan x) = \sec x + \tan x - x + C $$

Finally, solve for $y$ by dividing by $(\sec x + \tan x)$: $$ y = \frac{\sec x + \tan x - x + C}{\sec x + \tan x} $$

We can simplify this by dividing each term in the numerator by the denominator:

$$ y = \frac{\sec x + \tan x}{\sec x + \tan x} - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x} $$

$$ y = 1 - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x} $$

We can also write $\frac{1}{\sec x + \tan x} = \frac{\cos x}{1 + \sin x}$.

$$ y = 1 - \frac{x \cos x}{1 + \sin x} + \frac{C \cos x}{1 + \sin x} $$

Final Answer:

The solution to the differential equation is: $$ y = 1 + \frac{C - x}{\sec x + \tan x} $$ or $$ y = 1 + \frac{(C - x)\cos x}{1 + \sin x} $$

The given differential equation is: $$ \left(\frac{d^3y}{dx^3}\right)^2 + x \left(\frac{dy}{dx}\right)^3 + y^4 = 0 $$

To find the order:

The order of a differential equation is the order of the highest derivative present in the equation.

In the given equation, the derivatives present are:

• $\frac{d^3y}{dx^3}$ (third derivative)
• $\frac{dy}{dx}$ (first derivative)

The highest order of derivative is 3 (from $\frac{d^3y}{dx^3}$).

Therefore, the order of the differential equation is 3.

To find the degree:

The degree of a differential equation is the power of the highest order derivative, after the equation has been cleared of radicals and fractions so that derivatives are polynomials.

In the given equation, the highest order derivative is $\frac{d^3y}{dx^3}$. The power of this derivative is 2, as it is raised to the power of 2: $\left(\frac{d^3y}{dx^3}\right)^2$.

The equation is already in a polynomial form with respect to the derivatives (no radicals or fractions involving derivatives).

Therefore, the degree of the differential equation is 2.

Summary:

• Order: 3
• Degree: 2

Given the differential equation: $$ xy' = y + x \sqrt{x^2 - y^2} $$ and the proposed solution: $$ y = x \sin x $$ where $x > 0$ and $-x < y < x$.

To Verify: We need to substitute $y$ and its derivative $y'$ into the differential equation and check if the equation holds true.

Step 1: Find the derivative of $y$ with respect to $x$ ($y'$).

Using the product rule for differentiation on $y = x \sin x$: $$ y' = \frac{d}{dx}(x \sin x) $$ $$ y' = (1) \sin x + x (\cos x) $$ $$ y' = \sin x + x \cos x $$

Step 2: Substitute $y$ and $y'$ into the left side of the differential equation ($xy'$).

Left Side = $xy'$ $$ xy' = x(\sin x + x \cos x) $$ $$ xy' = x \sin x + x^2 \cos x $$

Step 3: Substitute $y$ into the right side of the differential equation ($y + x \sqrt{x^2 - y^2}$).

Right Side = $y + x \sqrt{x^2 - y^2}$ Substitute $y = x \sin x$: $$ y + x \sqrt{x^2 - y^2} = (x \sin x) + x \sqrt{x^2 - (x \sin x)^2} $$ $$ = x \sin x + x \sqrt{x^2 - x^2 \sin^2 x} $$ Factor out $x^2$ from the square root: $$ = x \sin x + x \sqrt{x^2(1 - \sin^2 x)} $$ Using the trigonometric identity $\cos^2 x = 1 - \sin^2 x$: $$ = x \sin x + x \sqrt{x^2 \cos^2 x} $$ Since $x > 0$, $\sqrt{x^2} = x$. We are given $-x < y < x$, which implies $-x < x \sin x < x$. For $x > 0$, this means $-1 < \sin x < 1$, which is always true. However, the term $\sqrt{x^2 \cos^2 x}$ depends on the sign of $\cos x$. We have $\sqrt{x^2 \cos^2 x} = |x \cos x|$. Since $x > 0$, this is $|x| |\cos x| = x |\cos x|$. The condition $-x < y < x$ means $-x < x \sin x < x$. If $x>0$, then $-1 < \sin x < 1$. The term $\sqrt{x^2-y^2}$ implies that $x^2-y^2 \ge 0$, so $y^2 \le x^2$, which means $|y| \le |x|$. Since $x>0$, this means $|y| \le x$, or $-x \le y \le x$. The problem statement gives $-x < y < x$, which excludes the boundaries. If we assume we are in an interval where $\cos x > 0$ (e.g., $0 < x < \frac{\pi}{2}$), then $\sqrt{x^2 \cos^2 x} = x \cos x$. $$ \text{Right Side} = x \sin x + x (x \cos x) $$ $$ \text{Right Side} = x \sin x + x^2 \cos x $$

Step 4: Compare the left and right sides.

Left Side = $x \sin x + x^2 \cos x$ Right Side = $x \sin x + x^2 \cos x$

The left side equals the right side. This holds true provided that $\cos x > 0$, which is consistent with the constraint $-x < y < x$ for suitable intervals of $x$ (e.g., when $\sin x$ is not 1 or -1). More formally, the term $\sqrt{x^2 - y^2}$ requires $x^2 - y^2 \ge 0$, which means $|y| \le |x|$. Given $x>0$, this is $|y| \le x$, or $-x \le y \le x$. The problem statement specifies $-x < y < x$. The term $\sqrt{x^2 \cos^2 x}$ must be $x \cos x$ for the equality to hold. This implies $\cos x \ge 0$. If $\cos x = 0$, then $y = x \sin x$ would be $y = x(\pm 1)$ leading to $|y|=x$, which is excluded by $-x < y < x$. So, $\cos x > 0$ is implicitly required for the term $\sqrt{x^2-y^2}$ in this form.

Conclusion:

Since substituting $y = x \sin x$ and its derivative $y' = \sin x + x \cos x$ into the differential equation $xy' = y + x \sqrt{x^2 - y^2}$ results in an identity ($x \sin x + x^2 \cos x = x \sin x + x^2 \cos x$), the proposed solution $y = x \sin x$ is verified to be a solution of the given differential equation under the specified conditions.

A circle touching the x-axis at the origin must have its center on the y-axis. Let the radius of such a circle be $a$. Since it touches the x-axis at the origin (0,0), the center of the circle will be at $(0, a)$ or $(0, -a)$.

The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

Case 1: Center at $(0, a)$ and radius $a$.

The equation of the circle is $(x-0)^2 + (y-a)^2 = a^2$. $$ x^2 + (y-a)^2 = a^2 $$ $$ x^2 + y^2 - 2ay + a^2 = a^2 $$ $$ x^2 + y^2 - 2ay = 0 $$

Case 2: Center at $(0, -a)$ and radius $a$.

The equation of the circle is $(x-0)^2 + (y-(-a))^2 = a^2$. $$ x^2 + (y+a)^2 = a^2 $$ $$ x^2 + y^2 + 2ay + a^2 = a^2 $$ $$ x^2 + y^2 + 2ay = 0 $$

We can combine both cases into a single equation by replacing $a$ with $\pm a$. From $x^2 + y^2 - 2ay = 0$, we get $2ay = x^2 + y^2$. From $x^2 + y^2 + 2ay = 0$, we get $-2ay = x^2 + y^2$. Both can be represented as $y^2 + x^2 = \pm 2ay$. Let's consider the equation $x^2 + y^2 - 2ay = 0$. We need to eliminate the arbitrary constant $a$.

Step 1: Differentiate the equation with respect to $x$.

Consider the family of circles $x^2 + y^2 - 2ay = 0$. Differentiate implicitly with respect to $x$: $$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ay) = \frac{d}{dx}(0) $$ $$ 2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0 $$ $$ 2x + (2y - 2a) \frac{dy}{dx} = 0 $$

Step 2: Eliminate the arbitrary constant $a$.

From the original equation $x^2 + y^2 - 2ay = 0$, we can express $a$ in terms of $x$ and $y$: $$ 2ay = x^2 + y^2 $$ $$ a = \frac{x^2 + y^2}{2y} $$ (Note: This assumes $y \neq 0$. If $y=0$, then from the original equation $x^2=0$, so $x=0$. This is the point of tangency, and the derivative is undefined there). Substitute this expression for $a$ into the differentiated equation: $$ 2x + (2y - 2 \left(\frac{x^2 + y^2}{2y}\right)) \frac{dy}{dx} = 0 $$ $$ 2x + \left(2y - \frac{x^2 + y^2}{y}\right) \frac{dy}{dx} = 0 $$ $$ 2x + \left(\frac{2y^2 - (x^2 + y^2)}{y}\right) \frac{dy}{dx} = 0 $$ $$ 2x + \left(\frac{2y^2 - x^2 - y^2}{y}\right) \frac{dy}{dx} = 0 $$ $$ 2x + \left(\frac{y^2 - x^2}{y}\right) \frac{dy}{dx} = 0 $$

Step 3: Rearrange to obtain the differential equation.

Multiply the entire equation by $y$: $$ 2xy + (y^2 - x^2) \frac{dy}{dx} = 0 $$ This can be written as: $$ (y^2 - x^2) \frac{dy}{dx} + 2xy = 0 $$

Final Answer:

The differential equation representing the family of circles touching the x-axis at the origin is: $$ (y^2 - x^2) \frac{dy}{dx} + 2xy = 0 $$

Given the differential equation: $$ \frac{dy}{dx} = (1+x^2)(1+y^2) $$

This is a separable differential equation. We can separate the variables $x$ and $y$ to different sides of the equation.

Step 1: Separate the variables.

Divide both sides by $(1+y^2)$ and multiply both sides by $dx$: $$ \frac{dy}{1+y^2} = (1+x^2) dx $$

Step 2: Integrate both sides of the equation.

Integrate the left side with respect to $y$ and the right side with respect to $x$: $$ \int \frac{dy}{1+y^2} = \int (1+x^2) dx $$

The integral of $\frac{1}{1+y^2}$ is $\arctan(y)$. The integral of $(1+x^2)$ is $x + \frac{x^3}{3}$.

So, we have:

$$ \arctan(y) = x + \frac{x^3}{3} + C $$

where $C$ is the constant of integration.

Step 3: Solve for $y$.

To solve for $y$, take the tangent of both sides:

$$ \tan(\arctan(y)) = \tan\left(x + \frac{x^3}{3} + C\right) $$

$$ y = \tan\left(x + \frac{x^3}{3} + C\right) $$

Final Answer:

The general solution to the differential equation is: $$ y = \tan\left(x + \frac{x^3}{3} + C\right) $$

Given the differential equation: $$ \frac{dy}{dx} = \frac{x+y}{x} $$

We can rewrite the equation as: $$ \frac{dy}{dx} = \frac{x}{x} + \frac{y}{x} $$ $$ \frac{dy}{dx} = 1 + \frac{y}{x} $$

This is a homogeneous differential equation because we can write it in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$. Here, $f\left(\frac{y}{x}\right) = 1 + \frac{y}{x}$.

To solve a homogeneous differential equation, we use the substitution $y = vx$. Differentiating 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:

$$ v + x\frac{dv}{dx} = 1 + v $$

Now, we can separate the variables $v$ and $x$. Subtract $v$ from both sides: $$ x\frac{dv}{dx} = 1 $$

Separate the variables: $$ dv = \frac{1}{x} dx $$

Integrate both sides: $$ \int dv = \int \frac{1}{x} dx $$

$$ v = \ln|x| + C $$

where $C$ is the constant of integration.

Now, substitute back $v = \frac{y}{x}$: $$ \frac{y}{x} = \ln|x| + C $$

Solve for $y$: $$ y = x(\ln|x| + C) $$

Final Answer:

The solution to the differential equation is: $$ y = x(\ln|x| + C) $$

The given differential equation is: $$ \frac{dy}{dx} + y \cot x = 2x + x^2 \cot x $$

This is a first-order linear differential equation of the form: $$ \frac{dy}{dx} + P(x)y = Q(x) $$

By comparing the given equation with the standard form, we identify:

$P(x) = \cot x$

$Q(x) = 2x + x^2 \cot x$

The integrating factor (I.F.) is given by the formula: $$ \text{I.F.} = e^{\int P(x) dx} $$

Now, we need to find the integral of $P(x)$ with respect to $x$: $$ \int P(x) dx = \int \cot x dx $$

The integral of $\cot x$ is $\ln|\sin x|$: $$ \int \cot x dx = \ln|\sin x| $$

Note that we do not need to add the constant of integration here as it will be incorporated into the final solution.

Now, substitute this integral back into the formula for the integrating factor: $$ \text{I.F.} = e^{\ln|\sin x|} $$

Using the property $e^{\ln a} = a$: $$ \text{I.F.} = |\sin x| $$

For practical purposes, and assuming a domain where $\sin x > 0$ (e.g., $0 < x < \pi$), we can use $\sin x$ as the integrating factor.

Final Answer:

The integrating factor of the given differential equation is $\mathbf{\sin x}$.

The given differential equation is: $$ x \frac{dy}{dx} + 2y = x^2 \log x $$

To solve this, we first need to convert it into the standard form of a first-order linear differential equation, which is $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide the entire equation by $x$ (assuming $x \neq 0$): $$ \frac{dy}{dx} + \frac{2}{x} y = \frac{x^2 \log x}{x} $$ $$ \frac{dy}{dx} + \frac{2}{x} y = x \log x $$

Now, we can identify $P(x)$ and $Q(x)$: $P(x) = \frac{2}{x}$ $Q(x) = x \log x$

The integrating factor (I.F.) is given by $e^{\int P(x) dx}$.

Calculate the integral of $P(x)$: $$ \int P(x) dx = \int \frac{2}{x} dx $$ $$ = 2 \int \frac{1}{x} dx $$ $$ = 2 \ln|x| $$ $$ = \ln|x^2| $$

Now, find the integrating factor: $$ \text{I.F.} = e^{\ln|x^2|} $$ $$ \text{I.F.} = |x^2| $$ Since $x^2$ is always non-negative, and we typically consider a domain where $x \neq 0$, we can use $x^2$ as the integrating factor.

Multiply the standard form of the differential equation by the integrating factor $x^2$: $$ x^2 \left(\frac{dy}{dx} + \frac{2}{x} y\right) = x^2 (x \log x) $$ $$ x^2 \frac{dy}{dx} + 2xy = x^3 \log x $$

The left side of the equation is the derivative of the product of $y$ and the integrating factor ($y \cdot \text{I.F.}$): $$ \frac{d}{dx} (y \cdot x^2) = x^3 \log x $$

Now, integrate both sides with respect to $x$: $$ \int \frac{d}{dx} (y x^2) dx = \int x^3 \log x dx $$ $$ y x^2 = \int x^3 \log x dx $$

To solve the integral $\int x^3 \log x dx$, we use integration by parts. Let $u = \log x$ and $dv = x^3 dx$. Then $du = \frac{1}{x} dx$ and $v = \int x^3 dx = \frac{x^4}{4}$.

Using the integration by parts formula $\int u dv = uv - \int v du$: $$ \int x^3 \log x dx = (\log x) \left(\frac{x^4}{4}\right) - \int \left(\frac{x^4}{4}\right) \left(\frac{1}{x}\right) dx $$ $$ = \frac{x^4}{4} \log x - \int \frac{x^3}{4} dx $$ $$ = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx $$ $$ = \frac{x^4}{4} \log x - \frac{1}{4} \left(\frac{x^4}{4}\right) $$ $$ = \frac{x^4}{4} \log x - \frac{x^4}{16} $$

So, we have: $$ y x^2 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C $$ where $C$ is the constant of integration.

Finally, solve for $y$ by dividing by $x^2$: $$ y = \frac{1}{x^2} \left(\frac{x^4}{4} \log x - \frac{x^4}{16} + C\right) $$ $$ y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2} $$

Final Answer:

The solution to the differential equation is: $$ y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2} $$

Given the family of curves: $$ y = A \cos(x+B) $$ where $A$ and $B$ are arbitrary constants.

To form a differential equation, we need to eliminate the two arbitrary constants, $A$ and $B$. This means we need to differentiate the given equation twice with respect to $x$ and then combine the equations to eliminate $A$ and $B$.

Step 1: Find the first derivative ($y'$).

Differentiate $y = A \cos(x+B)$ with respect to $x$: $$ y' = \frac{d}{dx} [A \cos(x+B)] $$ $$ y' = A (-\sin(x+B)) \cdot \frac{d}{dx}(x+B) $$ Since $B$ is a constant, $\frac{d}{dx}(x+B) = 1$. $$ y' = -A \sin(x+B) $$

Step 2: Find the second derivative ($y''$).

Differentiate $y' = -A \sin(x+B)$ with respect to $x$: $$ y'' = \frac{d}{dx} [-A \sin(x+B)] $$ $$ y'' = -A (\cos(x+B)) \cdot \frac{d}{dx}(x+B) $$ $$ y'' = -A \cos(x+B) $$

Step 3: Eliminate the constants $A$ and $B$.

We have the following equations: 1) $y = A \cos(x+B)$ 2) $y' = -A \sin(x+B)$ 3) $y'' = -A \cos(x+B)$

From equation (1), we can see that $y'' = -y$.

Alternatively, we can express $A$ and use it to eliminate $B$. From (1), $A = \frac{y}{\cos(x+B)}$. From (2), $A = \frac{-y'}{\sin(x+B)}$. Equating these: $\frac{y}{\cos(x+B)} = \frac{-y'}{\sin(x+B)}$ $\frac{y}{-y'} = \frac{\cos(x+B)}{\sin(x+B)} = \cot(x+B)$. This doesn't immediately eliminate $A$ and $B$ cleanly without further manipulation. Let's use equation (3) which is $y'' = -A \cos(x+B)$. From equation (1), we know $y = A \cos(x+B)$. Substituting equation (1) into equation (3): $$ y'' = -(A \cos(x+B)) $$ $$ y'' = -y $$

Rearranging this equation gives us the differential equation:

$$ y'' + y = 0 $$

Final Answer:

The differential equation representing the family of curves $y = A \cos(x+B)$ is: $$ \frac{d^2y}{dx^2} + y = 0 $$

Given the differential equation: $$ (e^x + 1) y dy = e^x (y^2 + 1) dx $$

This is a separable differential equation. We need to group terms involving $y$ with $dy$ and terms involving $x$ with $dx$.

Step 1: Separate the variables.

Divide both sides by $(e^x + 1)(y^2 + 1)$: $$ \frac{y}{y^2 + 1} dy = \frac{e^x}{e^x + 1} dx $$

Step 2: Integrate both sides of the equation.

Integrate the left side with respect to $y$ and the right side with respect to $x$: $$ \int \frac{y}{y^2 + 1} dy = \int \frac{e^x}{e^x + 1} dx $$

For the left side integral, $\int \frac{y}{y^2 + 1} dy$: Let $u = y^2 + 1$. Then $du = 2y dy$, so $y dy = \frac{1}{2} du$. The integral becomes $\int \frac{1}{u} \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(y^2 + 1)$. (Since $y^2+1 > 0$, the absolute value is not needed).

For the right side integral, $\int \frac{e^x}{e^x + 1} dx$: Let $v = e^x + 1$. Then $dv = e^x dx$. The integral becomes $\int \frac{1}{v} dv = \ln|v| = \ln(e^x + 1)$. (Since $e^x+1 > 0$, the absolute value is not needed).

Equating the results of the integration: $$ \frac{1}{2} \ln(y^2 + 1) = \ln(e^x + 1) + C $$ where $C$ is the constant of integration.

Step 3: Solve for $y$.

To simplify, we can multiply the entire equation by 2: $$ \ln(y^2 + 1) = 2 \ln(e^x + 1) + 2C $$ Let $K = 2C$, which is still an arbitrary constant: $$ \ln(y^2 + 1) = 2 \ln(e^x + 1) + K $$ Using the logarithm property $a \ln b = \ln b^a$: $$ \ln(y^2 + 1) = \ln((e^x + 1)^2) + K $$ Exponentiate both sides: $$ e^{\ln(y^2 + 1)} = e^{\ln((e^x + 1)^2) + K} $$ $$ y^2 + 1 = e^{\ln((e^x + 1)^2)} \cdot e^K $$ $$ y^2 + 1 = (e^x + 1)^2 \cdot e^K $$ Let $A = e^K$, where $A$ is a positive constant: $$ y^2 + 1 = A (e^x + 1)^2 $$ $$ y^2 = A (e^x + 1)^2 - 1 $$ $$ y = \pm \sqrt{A (e^x + 1)^2 - 1} $$

Final Answer:

The general solution to the differential equation is: $$ \ln(y^2 + 1) = 2 \ln(e^x + 1) + C $$ or equivalently $$ y^2 + 1 = A (e^x + 1)^2 $$ where $A = e^C$ is a positive constant.

The given differential equation is: $$ \frac{dy}{dx} + y = e^{-x} $$

This is a first-order linear differential equation of the form: $$ \frac{dy}{dx} + P(x)y = Q(x) $$

By comparing the given equation with the standard form, we identify: $P(x) = 1$ $Q(x) = e^{-x}$

The integrating factor (I.F.) is given by $e^{\int P(x) dx}$.

Calculate the integral of $P(x)$: $$ \int P(x) dx = \int 1 dx $$ $$ = x $$

Now, find the integrating factor: $$ \text{I.F.} = e^{x} $$

Multiply the differential equation by the integrating factor $e^x$: $$ e^x \left(\frac{dy}{dx} + y\right) = e^x (e^{-x}) $$ $$ e^x \frac{dy}{dx} + e^x y = e^0 $$ $$ e^x \frac{dy}{dx} + e^x y = 1 $$

The left side of the equation is the derivative of the product of $y$ and the integrating factor ($y \cdot \text{I.F.}$): $$ \frac{d}{dx} (y e^x) = 1 $$

Now, integrate both sides with respect to $x$: $$ \int \frac{d}{dx} (y e^x) dx = \int 1 dx $$ $$ y e^x = x + C $$

where $C$ is the constant of integration.

Finally, solve for $y$ by dividing by $e^x$: $$ y = \frac{x + C}{e^x} $$ $$ y = (x + C)e^{-x} $$

Final Answer:

The solution to the differential equation is: $$ y = (x + C)e^{-x} $$

Given the differential equation: $$ y' = \frac{x+y}{x} $$

Part 1: Show that the differential equation is homogeneous.

A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(tx, ty) = f(x,y)$ for any scalar $t$.

Let $f(x,y) = \frac{x+y}{x}$.

Replace $x$ with $tx$ and $y$ with $ty$: $$ f(tx, ty) = \frac{tx + ty}{tx} $$

Factor out $t$ from the numerator and the denominator: $$ f(tx, ty) = \frac{t(x + y)}{t(x)} $$

Cancel out $t$: $$ f(tx, ty) = \frac{x + y}{x} $$

Since $f(tx, ty) = f(x,y)$, the differential equation is homogeneous.

Part 2: Find the general solution.

To solve a homogeneous differential equation, we use the substitution $y = vx$. Differentiating 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 $\frac{dy}{dx} = \frac{x+y}{x}$: $$ v + x\frac{dv}{dx} = \frac{x + vx}{x} $$

Simplify the right side: $$ v + x\frac{dv}{dx} = \frac{x(1 + v)}{x} $$ $$ v + x\frac{dv}{dx} = 1 + v $$

Now, we can separate the variables $v$ and $x$. Subtract $v$ from both sides: $$ x\frac{dv}{dx} = 1 $$

Separate the variables: $$ dv = \frac{1}{x} dx $$

Integrate both sides: $$ \int dv = \int \frac{1}{x} dx $$

$$ v = \ln|x| + C $$

where $C$ is the constant of integration.

Now, substitute back $v = \frac{y}{x}$: $$ \frac{y}{x} = \ln|x| + C $$

Solve for $y$: $$ y = x(\ln|x| + C) $$

Final Answer:

The differential equation is homogeneous. The general solution is: $$ y = x(\ln|x| + C) $$

Given the differential equation: $$ \frac{dy}{dx} + 2xy = 0 $$ And the proposed solution: $$ y = c e^{-x^2} $$ where $c$ is an arbitrary constant.

To Verify: We need to substitute $y$ and its derivative $\frac{dy}{dx}$ into the differential equation and check if the equation holds true.

Step 1: Find the derivative of $y$ with respect to $x$.

Differentiate $y = c e^{-x^2}$ with respect to $x$ using the chain rule:

$$ \frac{dy}{dx} = \frac{d}{dx} (c e^{-x^2}) $$

$$ \frac{dy}{dx} = c \cdot e^{-x^2} \cdot \frac{d}{dx}(-x^2) $$

$$ \frac{dy}{dx} = c \cdot e^{-x^2} \cdot (-2x) $$

$$ \frac{dy}{dx} = -2cx e^{-x^2} $$

Step 2: Substitute $y$ and $\frac{dy}{dx}$ into the differential equation.

The differential equation is $\frac{dy}{dx} + 2xy = 0$.

Substitute the expressions for $\frac{dy}{dx}$ and $y$: $$ (-2cx e^{-x^2}) + 2x (c e^{-x^2}) = 0 $$

Step 3: Simplify the equation.

$$ -2cx e^{-x^2} + 2cx e^{-x^2} = 0 $$

The terms cancel each other out: $$ 0 = 0 $$

Conclusion:

Since substituting $y = c e^{-x^2}$ and its derivative into the differential equation results in an identity ($0 = 0$), we have verified that $y = c e^{-x^2}$ is indeed a solution to the differential equation $\frac{dy}{dx} + 2xy = 0$.

The given differential equation is: $$ (x + 2y^3) \frac{dy}{dx} = y $$

First, let's rewrite the equation in the form $\frac{dy}{dx} = f(x,y)$: $$ \frac{dy}{dx} = \frac{y}{x + 2y^3} $$

This does not immediately look like a linear equation in $y$. Let's try rewriting it in the form $\frac{dx}{dy} = g(x,y)$.

$$ \frac{dx}{dy} = \frac{x + 2y^3}{y} $$ $$ \frac{dx}{dy} = \frac{x}{y} + \frac{2y^3}{y} $$ $$ \frac{dx}{dy} = \frac{x}{y} + 2y^2 $$

Now, rearrange this into the standard form of a linear differential equation in terms of $x$ as a function of $y$: $$ \frac{dx}{dy} - \frac{1}{y} x = 2y^2 $$

This is a first-order linear differential equation of the form: $$ \frac{dx}{dy} + P(y)x = Q(y) $$

By comparing the equation with the standard form, we identify:

$P(y) = -\frac{1}{y}$

$Q(y) = 2y^2$

The integrating factor (I.F.) for an equation of this form is given by $e^{\int P(y) dy}$.

Calculate the integral of $P(y)$: $$ \int P(y) dy = \int -\frac{1}{y} dy $$ $$ = -\int \frac{1}{y} dy $$ $$ = -\ln|y| $$

Now, find the integrating factor: $$ \text{I.F.} = e^{-\ln|y|} $$ Using the property $a \ln b = \ln b^a$: $$ \text{I.F.} = e^{\ln|y|^{-1}} $$ $$ \text{I.F.} = |y|^{-1} $$ $$ \text{I.F.} = \frac{1}{|y|} $$

For simplicity, we can consider the case where $y > 0$, in which case the integrating factor is $\frac{1}{y}$. If $y < 0$, the integrating factor is $-\frac{1}{y}$. However, $\frac{1}{y}$ is generally used.

Final Answer:

The integrating factor of the differential equation is $\mathbf{\frac{1}{y}}$.

The given differential equation is: $$ (1 + x^2) \frac{dy}{dx} + y = \tan^{-1}x $$

To solve this, we first convert it into the standard form of a first-order linear differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide the entire equation by $(1 + x^2)$: $$ \frac{dy}{dx} + \frac{1}{1 + x^2} y = \frac{\tan^{-1}x}{1 + x^2} $$

Now, we can identify $P(x)$ and $Q(x)$: $P(x) = \frac{1}{1 + x^2}$ $Q(x) = \frac{\tan^{-1}x}{1 + x^2}$

The integrating factor (I.F.) is given by $e^{\int P(x) dx}$.

Calculate the integral of $P(x)$: $$ \int P(x) dx = \int \frac{1}{1 + x^2} dx $$ $$ = \tan^{-1}x $$

Now, find the integrating factor: $$ \text{I.F.} = e^{\tan^{-1}x} $$

Multiply the standard form of the differential equation by the integrating factor $e^{\tan^{-1}x}$: $$ e^{\tan^{-1}x} \left(\frac{dy}{dx} + \frac{1}{1 + x^2} y\right) = e^{\tan^{-1}x} \left(\frac{\tan^{-1}x}{1 + x^2}\right) $$ The left side is the derivative of the product of $y$ and the integrating factor ($y \cdot \text{I.F.}$): $$ \frac{d}{dx} \left(y e^{\tan^{-1}x}\right) = \frac{\tan^{-1}x \cdot e^{\tan^{-1}x}}{1 + x^2} $$

Now, integrate both sides with respect to $x$: $$ \int \frac{d}{dx} \left(y e^{\tan^{-1}x}\right) dx = \int \frac{\tan^{-1}x \cdot e^{\tan^{-1}x}}{1 + x^2} dx $$ $$ y e^{\tan^{-1}x} = \int \frac{\tan^{-1}x \cdot e^{\tan^{-1}x}}{1 + x^2} dx $$

To solve the integral on the right side, let $u = \tan^{-1}x$. Then $du = \frac{1}{1 + x^2} dx$.

The integral becomes: $$ \int u e^u du $$

We use integration by parts: let $f = u$ and $dg = e^u du$. Then $df = du$ and $g = e^u$.

$$ \int u e^u du = u e^u - \int e^u du $$ $$ = u e^u - e^u $$ Substitute back $u = \tan^{-1}x$: $$ = \tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} $$

So, we have: $$ y e^{\tan^{-1}x} = \tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} + C $$ where $C$ is the constant of integration.

Finally, solve for $y$ by dividing by $e^{\tan^{-1}x}$: $$ y = \frac{\tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} + C}{e^{\tan^{-1}x}} $$ $$ y = \tan^{-1}x - 1 + C e^{-\tan^{-1}x} $$

Final Answer:

The solution to the differential equation is: $$ y = \tan^{-1}x - 1 + C e^{-\tan^{-1}x} $$

Rewrite the equation as $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$. This is a homogeneous differential equation.

Use the substitution $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

$$ v + x\frac{dv}{dx} = \frac{vx + \sqrt{x^2 + (vx)^2}}{x} = v + \sqrt{1+v^2} $$

This simplifies to $x\frac{dv}{dx} = \sqrt{1+v^2}$.

Separating variables: $\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$.

Integrating both sides: $\int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x}$.

$$ \ln|v + \sqrt{1+v^2}| = \ln|x| + C_1 $$

Exponentiating both sides: $v + \sqrt{1+v^2} = C|x|$, where $C = \pm e^{C_1}$.

Substitute $v = y/x$: $\frac{y}{x} + \frac{\sqrt{x^2+y^2}}{|x|} = C|x|$.

Multiply by $x$: $y + \frac{x}{|x|}\sqrt{x^2+y^2} = Cx|x|$.

If $x > 0$, $y + \sqrt{x^2+y^2} = Cx^2$. If $x < 0$, $y - \sqrt{x^2+y^2} = Cx^2$.

From $y + \sqrt{x^2+y^2} = Cx^2$, we get $\sqrt{x^2+y^2} = Cx^2 - y$. Squaring both sides gives $x^2+y^2 = C^2x^4 - 2Cx^2y + y^2$, which simplifies to $x^2 = C^2x^4 - 2Cx^2y$.

Solving for $y$: $2Cx^2y = C^2x^4 - x^2 \implies y = \frac{C^2x^2 - 1}{2C}$.

Let $a = C/2$. Then $y = ax^2 - \frac{1}{4a}$.

Final Answer:

The solution is $y = ax^2 - \frac{1}{4a}$, where $a$ is an arbitrary non-zero constant.

The given differential equation is: $$ (1 + e^{x/y}) dx + e^{x/y} \left(1 - \frac{x}{y}\right) dy = 0 $$

Let's check if this is a homogeneous differential equation. We can rewrite it in the form $\frac{dy}{dx} = -\frac{M(x,y)}{N(x,y)}$, where $M(x,y) = 1 + e^{x/y}$ and $N(x,y) = e^{x/y} (1 - \frac{x}{y})$.

Consider the ratio of terms involving $x/y$. Let $t$ be a scalar. For $M(tx, ty) = 1 + e^{(tx)/(ty)} = 1 + e^{x/y} = M(x,y)$. For $N(tx, ty) = e^{(tx)/(ty)} (1 - \frac{tx}{ty}) = e^{x/y} (1 - \frac{x}{y}) = N(x,y)$. Since both $M$ and $N$ are homogeneous functions of degree 0, the differential equation is homogeneous.

For homogeneous equations of the form $M dx + N dy = 0$, it is often useful to consider the substitution $x = vy$. Then $dx = v dy + y dv$.

Substitute $x = vy$ and $dx = v dy + y dv$ into the given differential equation:

$$ (1 + e^{vy/y}) (v dy + y dv) + e^{vy/y} \left(1 - \frac{vy}{y}\right) dy = 0 $$

$$ (1 + e^v) (v dy + y dv) + e^v (1 - v) dy = 0 $$

Expand the terms:

$$ (1 + e^v) v dy + (1 + e^v) y dv + e^v (1 - v) dy = 0 $$

Group terms with $dy$ and $dv$: $$ [v(1 + e^v) + e^v(1 - v)] dy + y(1 + e^v) dv = 0 $$

Simplify the coefficient of $dy$: $$ [v + ve^v + e^v - ve^v] dy + y(1 + e^v) dv = 0 $$ $$ [v + e^v] dy + y(1 + e^v) dv = 0 $$

Now, separate the variables $y$ and $v$. Divide by $y(1+e^v)$: $$ \frac{v + e^v}{1 + e^v} \frac{dy}{y} + dv = 0 $$

Rearrange: $$ dv = -\frac{v + e^v}{1 + e^v} \frac{dy}{y} $$ $$ \frac{1+e^v}{v+e^v} dv = -\frac{dy}{y} $$

Integrate both sides:

$$ \int \frac{1+e^v}{v+e^v} dv = -\int \frac{dy}{y} $$

For the left side integral, let $u = v+e^v$. Then $du = (1+e^v) dv$. The integral becomes $\int \frac{1}{u} du = \ln|u| = \ln|v+e^v|$.

For the right side integral, $-\int \frac{dy}{y} = -\ln|y| = \ln|y|^{-1}$.

So, we have: $$ \ln|v+e^v| = \ln|y|^{-1} + C_1 $$ where $C_1$ is the constant of integration.

Exponentiate both sides: $$ |v+e^v| = e^{\ln|y|^{-1} + C_1} = e^{C_1} |y|^{-1} $$ Let $C = \pm e^{C_1}$ (a non-zero constant): $$ v+e^v = Cy^{-1} = \frac{C}{y} $$

Substitute back $v = \frac{x}{y}$: $$ \frac{x}{y} + e^{x/y} = \frac{C}{y} $$

Multiply by $y$: $$ x + y e^{x/y} = C $$

Final Answer:

The solution to the differential equation is: $$ x + y e^{x/y} = C $$

The given differential equation is: $$ (1 + y^2) dx + (x - e^{\tan^{-1}y}) dy = 0 $$

Let's rewrite the equation to identify its form. It's often easier to work with if we express $\frac{dx}{dy}$ or $\frac{dy}{dx}$.

Let's try to express it as a linear equation in $x$ as a function of $y$: $$ (1 + y^2) dx = -(x - e^{\tan^{-1}y}) dy $$ $$ \frac{dx}{dy} = -\frac{x - e^{\tan^{-1}y}}{1 + y^2} $$ $$ \frac{dx}{dy} = \frac{e^{\tan^{-1}y} - x}{1 + y^2} $$ $$ \frac{dx}{dy} = \frac{e^{\tan^{-1}y}}{1 + y^2} - \frac{x}{1 + y^2} $$

Rearranging into the standard form for a linear differential equation in $x$ with respect to $y$: $$ \frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{e^{\tan^{-1}y}}{1 + y^2} $$

This is a first-order linear differential equation of the form: $$ \frac{dx}{dy} + P(y)x = Q(y) $$

By comparing, we have: $P(y) = \frac{1}{1 + y^2}$ $Q(y) = \frac{e^{\tan^{-1}y}}{1 + y^2}$

The integrating factor (I.F.) is given by $e^{\int P(y) dy}$.

Calculate the integral of $P(y)$: $$ \int P(y) dy = \int \frac{1}{1 + y^2} dy = \tan^{-1}y $$

Now, find the integrating factor: $$ \text{I.F.} = e^{\tan^{-1}y} $$

Multiply the standard form of the differential equation by the integrating factor $e^{\tan^{-1}y}$: $$ e^{\tan^{-1}y} \left(\frac{dx}{dy} + \frac{1}{1 + y^2} x\right) = e^{\tan^{-1}y} \left(\frac{e^{\tan^{-1}y}}{1 + y^2}\right) $$ The left side is the derivative of the product of $x$ and the integrating factor ($x \cdot \text{I.F.}$): $$ \frac{d}{dy} (x e^{\tan^{-1}y}) = \frac{(e^{\tan^{-1}y})^2}{1 + y^2} $$ $$ \frac{d}{dy} (x e^{\tan^{-1}y}) = \frac{e^{2 \tan^{-1}y}}{1 + y^2} $$

Now, integrate both sides with respect to $y$: $$ \int \frac{d}{dy} (x e^{\tan^{-1}y}) dy = \int \frac{e^{2 \tan^{-1}y}}{1 + y^2} dy $$ $$ x e^{\tan^{-1}y} = \int \frac{e^{2 \tan^{-1}y}}{1 + y^2} dy $$

To solve the integral on the right side, let $u = 2 \tan^{-1}y$. Then $du = \frac{2}{1 + y^2} dy$, so $\frac{1}{1 + y^2} dy = \frac{1}{2} du$. The integral becomes: $$ \int e^u \frac{1}{2} du = \frac{1}{2} \int e^u du = \frac{1}{2} e^u $$

Substitute back $u = 2 \tan^{-1}y$: $$ \int \frac{e^{2 \tan^{-1}y}}{1 + y^2} dy = \frac{1}{2} e^{2 \tan^{-1}y} $$

So, we have: $$ x e^{\tan^{-1}y} = \frac{1}{2} e^{2 \tan^{-1}y} + C $$ where $C$ is the constant of integration.

Now, we need to find the particular solution using the initial condition $y = 0$ when $x = 1$.

Substitute $x=1$ and $y=0$ into the general solution: $$ 1 \cdot e^{\tan^{-1}0} = \frac{1}{2} e^{2 \tan^{-1}0} + C $$ Since $\tan^{-1}0 = 0$, we have $e^0 = 1$. $$ 1 \cdot 1 = \frac{1}{2} \cdot 1 + C $$ $$ 1 = \frac{1}{2} + C $$ $$ C = 1 - \frac{1}{2} = \frac{1}{2} $$

Substitute the value of $C$ back into the general solution:

$$ x e^{\tan^{-1}y} = \frac{1}{2} e^{2 \tan^{-1}y} + \frac{1}{2} $$

We can solve for $x$: $$ x = \frac{\frac{1}{2} e^{2 \tan^{-1}y} + \frac{1}{2}}{e^{\tan^{-1}y}} $$ $$ x = \frac{1}{2} \frac{e^{2 \tan^{-1}y}}{e^{\tan^{-1}y}} + \frac{1}{2} \frac{1}{e^{\tan^{-1}y}} $$ $$ x = \frac{1}{2} e^{\tan^{-1}y} + \frac{1}{2} e^{-\tan^{-1}y} $$

Final Answer:

The particular solution is: $$ x = \frac{1}{2} \left(e^{\tan^{-1}y} + e^{-\tan^{-1}y}\right) $$ This can also be written using the hyperbolic cosine function: $$ x = \cosh(\tan^{-1}y) $$