Solution

$\Rightarrow y^{\prime \prime \prime \prime}+\sin (y^{\prime \prime \prime})=0$

The highest order derivative present in the differential equation is $y^{\prime \prime \prime \prime}$. Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Solution

The given differential equation is:

$y^{\prime}+5 y=0$

The highest order derivative present in the differential equation is $y^{\prime}$. Therefore, its order is one.

It is a polynomial equation in $y^{\prime}$. The highest power raised to $y^{\prime}$ is 1 . Hence, its degree is one.

Solution

$(\frac{d s}{d t})^{4}+3 \frac{d^{2} s}{d t^{2}}=0$

The highest order derivative present in the given differential equation is $\frac{d^{2} s}{d t^{2}}$. Therefore, its order is two.

It is a polynomial equation in $\frac{d^{2} s}{d t^{2}}$ and $\frac{d s}{d t}$. The power raised to $\frac{d^{2} s}{d t^{2}}$ is 1 . Hence, its degree is one.

Solution

$(\frac{d^{2} y}{d x^{2}})^{2}+\cos (\frac{d y}{d x})=0$

The highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$. Therefore, its order is 2.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Solution

$\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x$

$\Rightarrow \frac{d^{2} y}{d x^{2}}-\cos 3 x-\sin 3 x=0$

The highest order derivative present in the differential equation is $\frac{d^{2} y}{d x^{2}}$. Therefore, its order is two.

It is a polynomial equation in $\frac{d^{2} y}{d x^{2}}$ and the power raised to $\frac{d^{2} y}{d x^{2}}$ is 1 .

Hence, its degree is one.

Solution

$(y^{\prime \prime \prime})^{2}+(y^{\prime \prime})^{3}+(y^{\prime})+y^{5}=0$

The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$. Therefore, its order is three.

The given differential equation is a polynomial equation in $y^{\prime \prime \prime}, y^{\prime \prime}$, and $y^{\prime}$.

The highest power raised to $y^{\prime \prime \prime}$ is 2 . Hence, its degree is 2 .

Solution

$y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$

The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$. Therefore, its order is three.

It is a polynomial equation in $y^{\prime \prime \prime}, y^{\prime \prime}$ and $y^{\prime}$. The highest power raised to $y^{\prime \prime \prime}$ is 1 . Hence, its degree is 1 .

Solution

$ \begin{aligned} & y^{\prime}+y=e^{x} \\ & \Rightarrow y^{\prime}+y-e^{x}=0 \end{aligned} $

The highest order derivative present in the differential equation is $y^{\prime}$. Therefore, its order is one.

The given differential equation is a polynomial equation in $y^{\prime}$ and the highest power raised to $y^{\prime}$ is one. Hence, its degree is one.

Solution

$y^{\prime \prime}+(y^{\prime})^{2}+2 y=0$

The highest order derivative present in the differential equation is $y^{\prime \prime}$. Therefore, its order is two.

The given differential equation is a polynomial equation in $y^{\prime \prime}$ and $y^{\prime}$ and the highest power raised to $y^{\prime \prime}$ is one.

Hence, its degree is one.

Solution

$y^{\prime \prime}+2 y^{\prime}+\sin y=0$

The highest order derivative present in the differential equation is $y^{\prime \prime}$. Therefore, its order is two.

This is a polynomial equation in $y^{\prime \prime}$ and $y^{\prime}$ and the highest power raised to $y^{\prime \prime}$ is one. Hence, its degree is one.

Solution

$(\frac{d^{2} y}{d x^{2}})^{3}+(\frac{d y}{d x})^{2}+\sin (\frac{d y}{d x})+1=0$

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

Solution

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

The highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$. Therefore, its order is two.

Hence, the correct answer is A.

Solution

$y=e^{x}+1$

Differentiating both sides of this equation with respect to $x$, we get:

$\frac{d y}{d x}=\frac{d}{d x}(e^{x}+1)$

$\Rightarrow y^{\prime}=e^{x}$

Now, differentiating equation (1) with respect to $x$, we get:

$\frac{d}{d x}(y^{\prime})=\frac{d}{d x}(e^{x})$

$\Rightarrow y^{\prime \prime}=e^{x}$

Substituting the values of $y^{\prime}$ and $y^{\prime \prime}$ in the given differential equation, we get the L.H.S. as:

$y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=0=$ R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

Solution

$y=x^{2}+2 x+C$

Differentiating both sides of this equation with respect to $x$, we get:

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

$\Rightarrow y^{\prime}=2 x+2$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{\prime}-2 x-2=2 x+2-2 x-2=0=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=\cos x+C$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(\cos x+C)$

$\Rightarrow y^{\prime}=-\sin x$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{\prime}+\sin x=-\sin x+\sin x=0=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=\sqrt{1+x^{2}}$

Differentiating both sides of the equation with respect to $x$, we get: $y^{\prime}=\frac{d}{d x}(\sqrt{1+x^{2}})$

$y^{\prime}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot \frac{d}{d x}(1+x^{2})$

$y^{\prime}=\frac{2 x}{2 \sqrt{1+x^{2}}}$

$y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \times \sqrt{1+x^{2}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \cdot y$

$\Rightarrow y^{\prime}=\frac{x y}{1+x^{2}}$

$\therefore$ L.H.S. $=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=A x$

Differentiating both sides with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(A x)$

$\Rightarrow y^{\prime}=A$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=x y^{\prime}=x \cdot A=A x=y=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=x \sin x$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(x \sin x)$

$\Rightarrow y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow y^{\prime}=\sin x+x \cos x$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

$ \begin{aligned} \text{ L.H.S. }=x y^{\prime} & =x(\sin x+x \cos x) \\ & =x \sin x+x^{2} \cos x \\ & =y+x^{2} \cdot \sqrt{1-\sin ^{2}} x \\ & =y+x^{2} \sqrt{1-(\frac{y}{x})^{2}} \\ & =y+x \sqrt{y^{2}-x^{2}} \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

$x y=\log y+C$

Differentiating both sides of this equation with respect to $x$, we get: $\frac{d}{d x}(x y)=\frac{d}{d x}(\log y)$

$\Rightarrow y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}$

$\Rightarrow y+x y^{\prime}=\frac{1}{y} y^{\prime}$

$\Rightarrow y^{2}+x y y^{\prime}=y^{\prime}$

$\Rightarrow(x y-1) y^{\prime}=-y^{2}$

$\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y}$

$\therefore$ L.H.S. $=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y-\cos y=x$

Differentiating both sides of the equation with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}-\frac{d}{d x}(\cos y)=\frac{d}{d x}(x) \\ & \Rightarrow y^{\prime}+\sin y \cdot y^{\prime}=1 \\ & \Rightarrow y^{\prime}(1+\sin y)=1 \\ & \Rightarrow y^{\prime}=\frac{1}{1+\sin y} \end{aligned} $

Substituting the value of $y^{\prime}$ in equation (1), we get:

L.H.S. $=(y \sin y+\cos y+x) y^{\prime}$

$ \begin{aligned} & =(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y} \\ & =y(1+\sin y) \cdot \frac{1}{1+\sin y} \\ & =y \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

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

Differentiating both sides of this equation with respect to $x$, we get:

$ \frac{d}{d x}(x+y)=\frac{d}{d x}(\tan ^{-1} y) $

$ \Rightarrow 1+y^{\prime}=[\frac{1}{1+y^{2}}] y^{\prime} $

$\Rightarrow y^{\prime}[\frac{1}{1+y^{2}}-1]=1$

$\Rightarrow y^{\prime}[\frac{1-(1+y^{2})}{1+y^{2}}]=1$

$\Rightarrow y^{\prime}[\frac{-y^{2}}{1+y^{2}}]=1$

$\Rightarrow y^{\prime}=\frac{-(1+y^{2})}{y^{2}}$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{2} y^{\prime}+y^{2}+1=y^{2}[\frac{-(1+y^{2})}{y^{2}}]+y^{2}+1$

$ \begin{aligned} & =-1-y^{2}+y^{2}+1 \\ & =0 \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=\sqrt{a^{2}-x^{2}}$

Differentiating both sides of this equation with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(\sqrt{a^{2}-x^{2}}) \\ & \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}(a^{2}-x^{2}) \\ & =\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x) \\ & =\frac{-x}{\sqrt{a^{2}-x^{2}}} \end{aligned} $

Substituting the value of $\frac{d y}{d x}$ in the given differential equation, we get:

L.H.S. $=x+y \frac{d y}{d x}=x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}}$

$ \begin{aligned} & =x-x \\ & =0 \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

We know that the number of constants in the general solution of a differential equation of order $n$ is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is D.

Solution

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is D.

Solution

The given differential equation is:

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

$\Rightarrow \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2}$

$\Rightarrow \frac{d y}{d x}=(\sec ^{2} \frac{x}{2}-1)$

Separating the variables,we get:

$d y=(\sec ^{2} \frac{x}{2}-1) d x$

Now, integrating both sides of this equation, we get:

$ \begin{aligned} & \int d y=\int(\sec ^{2} \frac{x}{2}-1) d x=\int \sec ^{2} \frac{x}{2} d x-\int d x \\ & \Rightarrow y=2 \tan \frac{x}{2}-x+C \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is: $\frac{d y}{d x}=\sqrt{4-y^{2}}$

Separating the variables, we get:

$\Rightarrow \frac{d y}{\sqrt{4-y^{2}}}=d x$

Now, integrating both sides of this equation, we get:

$ \begin{aligned} & \int \frac{d y}{\sqrt{4-y^{2}}}=\int d x \\ & \Rightarrow \sin ^{-1} \frac{y}{2}=x+C \\ & \Rightarrow \frac{y}{2}=\sin (x+C) \\ & \Rightarrow y=2 \sin (x+C) \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$ \begin{aligned} & \frac{d y}{d x}+y=1 \\ & \Rightarrow d y+y d x=d x \\ & \Rightarrow d y=(1-y) d x \end{aligned} $

Separating the variables, we get:

$\Rightarrow \frac{d y}{1-y}=d x$

Now, integrating both sides, we get: $\int \frac{d y}{1-y}=\int d x$

$\Rightarrow -\log (1-y)=x+\log (C)$

$\Rightarrow-\log C-\log (1-y)=x$

$\Rightarrow \log C(1-y)=-x$

$\Rightarrow C(1-y)=e^{-x}$

$\Rightarrow 1-y=\frac{1}{C} e^{-x}$

$\Rightarrow y=1-\frac{1}{C} e^{-x}$

$\Rightarrow y=1+A e^{-x}$ (where $A=-\frac{1}{C}$ )

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

$\Rightarrow \frac{\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y}{\tan x \tan y}=0$

$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0$

$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$

Integrating both sides of this equation, we get:

$$ \begin{equation*} \int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y \tag{1} \end{equation*} $$

Let $\tan x=t$.

$\therefore \frac{d}{d x}(\tan x)=\frac{d t}{d x}$

$\Rightarrow \sec ^{2} x=\frac{d t}{d x}$

$\Rightarrow \sec ^{2} x d x=d t$

Now, $\int \frac{\sec ^{2} x}{\tan x} d x=\int_t^1 d t$.

$ \begin{aligned} & =\log t \\ & =\log (\tan x) \end{aligned} $

Similarly, $\int \frac{\sec ^{2} x}{\tan x} d y=\log (\tan y)$.

Substituting these values in equation (1), we get:

$ \begin{aligned} & \log (\tan x)=-\log (\tan y)+\log C \\ & \Rightarrow \log (\tan x)=\log (\frac{C}{\tan y}) \\ & \Rightarrow \tan x=\frac{C}{\tan y} \\ & \Rightarrow \tan x \tan y=C \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$ \begin{aligned} & (e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0 \\ & \Rightarrow(e^{x}+e^{-x}) d y=(e^{x}-e^{-x}) d x \\ & \Rightarrow d y=[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}] d x \end{aligned} $

Integrating both sides of this equation, we get: $\int d y=\int[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}] d x+C$

$\Rightarrow y=\int[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}] d x+C$

Let $(e^{x}+e^{-x})=t$.

Differentiating both sides with respect to $x$, we get:

$\frac{d}{d x}(e^{x}+e^{-x})=\frac{d t}{d x}$

$\Rightarrow e^{x}-e^{-x}=\frac{d t}{d t}$

$\Rightarrow(e^{x}-e^{-x}) d x=d t$

Substituting this value in equation (1), we get:

$ \begin{aligned} & y=\int \frac{1}{t} d t+C \\ & \Rightarrow y=\log (t)+C \\ & \Rightarrow y=\log (e^{x}+e^{-x})+C \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$ \begin{aligned} & \frac{d y}{d x}=(1+x^{2})(1+y^{2}) \\ & \Rightarrow \frac{d y}{1+y^{2}}=(1+x^{2}) d x \end{aligned} $

Integrating both sides of this equation, we get: $\int \frac{d y}{1+y^{2}}=\int(1+x^{2}) d x$

$\Rightarrow \tan ^{-1} y=\int d x+\int x^{2} d x$

$\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+C$

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$y \log y d x-x d y=0$

$\Rightarrow y \log y d x=x d y$

$\Rightarrow \frac{d y}{y \log y}=\frac{d x}{x}$

Integrating both sides, we get:

$\int \frac{d y}{y \log y}=\int \frac{d x}{x}$

Let $\log y=t$.

$\therefore \frac{d}{d y}(\log y)=\frac{d t}{d y}$

$\Rightarrow \frac{1}{y}=\frac{d t}{d y}$

$\Rightarrow \frac{1}{y} d y=d t$

Substituting this value in equation (1), we get:

$ \begin{aligned} & \int \frac{d t}{t}=\int \frac{d x}{x} \\ & \Rightarrow \log t=\log x+\log C \\ & \Rightarrow \log (\log y)=\log C x \\ & \Rightarrow \log y=C x \\ & \Rightarrow y=e^{C x} \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$x^{5} \frac{d y}{d x}=-y^{5}$

$\Rightarrow \frac{d y}{y^{5}}=-\frac{d x}{x^{5}}$

$\Rightarrow \frac{d x}{x^{5}}+\frac{d y}{y^{5}}=0$

Integrating both sides, we get:

$ \begin{aligned} & \int \frac{d x}{x^{5}}+\int \frac{d y}{y^{5}}=k \quad \text{ (where } k \text{ is any constant) } \\ & \Rightarrow \int x^{-5} d x+\int y^{-5} d y=k \\ & \Rightarrow \frac{x^{-4}}{-4}+\frac{y^{-4}}{-4}=k \\ & \Rightarrow x^{-4}+y^{-4}=-4 k \\ & \Rightarrow x^{-4}+y^{-4}=C \quad(C=-4 k) \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$\frac{d y}{d x}=\sin ^{-1} x$

$\Rightarrow d y=\sin ^{-1} x d x$

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\int \sin ^{-1} x d x \\ & \Rightarrow y=\int(\sin ^{-1} x \cdot 1) d x \\ & \Rightarrow y=\sin ^{-1} x \cdot \int(1) d x-\int[(\frac{d}{d x}(\sin ^{-1} x) \cdot \int(1) d x)] d x \\ & \Rightarrow y=\sin ^{-1} x \cdot x-\int(\frac{1}{\sqrt{1-x^{2}}} \cdot x) d x \\ & \Rightarrow y=x \sin ^{-1} x+\int \frac{-x}{\sqrt{1-x^{2}}} d x \tag{1} \end{align*} $$

Let $1-x^{2}=t$.

$\Rightarrow \frac{d}{d x}(1-x^{2})=\frac{d t}{d x}$

$\Rightarrow-2 x=\frac{d t}{d x}$

$\Rightarrow x d x=-\frac{1}{2} d t$

Substituting this value in equation (1), we get:

$ \begin{aligned} & y=x \sin ^{-1} x+\int \frac{1}{2 \sqrt{t}} d t \\ & \Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \int(t)^{-\frac{1}{2}} d t \\ & \Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\ & \Rightarrow y=x \sin ^{-1} x+\sqrt{t}+C \\ & \Rightarrow y=x \sin ^{-1} x+\sqrt{1-x^{2}}+C \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$e^{x} \tan y d x+(1-e^{x}) \sec ^{2} y d y=0$

$(1-e^{x}) \sec ^{2} y d y=-e^{x} \tan y d x$

Separating the variables, we get:

$ \frac{\sec ^{2} y}{\tan y} d y=\frac{-e^{x}}{1-e^{x}} d x $

Integrating both sides, we get:

$$ \begin{equation*} \int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{-e^{x}}{1-e^{x}} d x \tag{1} \end{equation*} $$

Let $\tan y=u$.

$ \begin{aligned} & \Rightarrow \frac{d}{d y}(\tan y)=\frac{d u}{d y} \\ & \Rightarrow \sec ^{2} y=\frac{d u}{d y} \\ & \Rightarrow \sec ^{2} y d y=d u \\ & \therefore \int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{d u}{u}=\log u=\log (\tan y) \end{aligned} $

Now, let $1-e^{x}=t$.

$\therefore \frac{d}{d x}(1-e^{x})=\frac{d t}{d x}$

$\Rightarrow-e^{x}=\frac{d t}{d x}$

$\Rightarrow-e^{x} d x=d t$

$\Rightarrow \int \frac{-e^{x}}{1-e^{x}} d x=\int \frac{d t}{t}=\log t=\log (1-e^{x})$

Substituting the values of $\int \frac{\sec ^{2} y}{\tan y} d y$ and $\int \frac{-e^{x}}{1-e^{x}} d x$ in equation (1), we get:

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

$\Rightarrow \log (\tan y)=\log [C(1-e^{x})]$

$\Rightarrow \tan y=C(1-e^{x})$

This is the required general solution of the given differential equation.

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

Solution

The given differential equation is:

$ \begin{aligned} & (x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x \\ & \Rightarrow \frac{d y}{d x}=\frac{2 x^{2}+x}{(x^{3}+x^{2}+x+1)} \\ & \Rightarrow d y=\frac{2 x^{2}+x}{(x+1)(x^{2}+1)} d x \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\int \frac{2 x^{2}+x}{(x+1)(x^{2}+1)} d x \tag{1}\\ & \text{ Let } \frac{2 x^{2}+x}{(x+1)(x^{2}+1)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1} . \tag{2}\\ & \Rightarrow \frac{2 x^{2}+x}{(x+1)(x^{2}+1)}=\frac{A x^{2}+A+(B x+C)(x+1)}{(x+1)(x^{2}+1)} \\ & \Rightarrow 2 x^{2}+x=A x^{2}+A+B x^{2}+B x+C x+C \\ & \Rightarrow 2 x^{2}+x=(A+B) x^{2}+(B+C) x+(A+C) \end{align*} $$

Comparing the coefficients of $x^{2}$ and $x$, we get:

$A+B=2$

$B+C=1$

$A+C=0$

Solving these equations, we get:

$ A=\frac{1}{2}, B=\frac{3}{2} \text{ and } C=\frac{-1}{2} $

Substituting the values of $A, B$, and $C$ in equation (2), we get:

$ \frac{2 x^{2}+x}{(x+1)(x^{2}+1)}=\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \frac{(3 x-1)}{(x^{2}+1)} $

Therefore, equation (1) becomes:

$$ \begin{align*} & \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^{2}+1} d x \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{1}{x^{2}+1} d x \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \cdot \int \frac{2 x}{x^{2}+1} d x-\frac{1}{2} \tan ^{-1} x+C \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log (x^{2}+1)-\frac{1}{2} \tan ^{-1} x+C \\ & \Rightarrow y=\frac{1}{4}[2 \log (x+1)+3 \log (x^{2}+1)]-\frac{1}{2} \tan ^{-1} x+C \\ & \Rightarrow y=\frac{1}{4}\log (x+1)^{2}(x^{2}+1)^{3}]-\frac{1}{2} \tan ^{-1} x+C \tag{3} \end{align*} $$

Now, $y=1$ when $x=0$.

$\Rightarrow 1=\frac{1}{4} \log (1)-\frac{1}{2} \tan ^{-1} 0+C$

$\Rightarrow 1=\frac{1}{4} \times 0-\frac{1}{2} \times 0+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (3), we get:

$y=\frac{1}{4}[\log (x+1)^{2}(x^{2}+1)^{3}]-\frac{1}{2} \tan ^{-1} x+1$

Solution

$ \begin{aligned} & x(x^{2}-1) \frac{d y}{d x}=1 \\ & \Rightarrow d y=\frac{d x}{x(x^{2}-1)} \\ & \Rightarrow d y=\frac{1}{x(x-1)(x+1)} d x \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\int \frac{1}{x(x-1)(x+1)} d x \tag{1}\\ & \text{ Let } \frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} . \tag{2}\\ & \Rightarrow \frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)} \\ & =\frac{(A+B+C) x^{2}+(B-C) x-A}{x(x-1)(x+1)} \end{align*} $$

Comparing the coefficients of $x^{2}, x$, and constant, we get:

$A=-1$

$B-C=0$

$A+B+C=0$

Solving these equations, we get $B=\frac{1}{2}$ and $C=\frac{1}{2}$.

Substituting the values of $A, B$, and $C$ in equation (2), we get:

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

Therefore, equation (1) becomes:

$$ \begin{align*} & \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x+1} d x \\ & \Rightarrow y=-\log x+\frac{1}{2} \log (x-1)+\frac{1}{2} \log (x+1)+\log k \\ & \Rightarrow y=\frac{1}{2} \log [\frac{k^{2}(x-1)(x+1)}{x^{2}}] \tag{3} \end{align*} $$

Now, $y=0$ when $x=2$.

$\Rightarrow 0=\frac{1}{2} \log [\frac{k^{2}(2-1)(2+1)}{4}]$

$\Rightarrow \log (\frac{3 k^{2}}{4})=0$

$\Rightarrow \frac{3 k^{2}}{4}=1$

$\Rightarrow 3 k^{2}=4$

$\Rightarrow k^{2}=\frac{4}{3}$

Substituting the value of $k^{2}$ in equation (3), we get:

$y=\frac{1}{2} \log [\frac{4(x-1)(x+1)}{3 x^{2}}]$

$y=\frac{1}{2} \log [\frac{4(x^{2}-1)}{3 x^{2}}]$

Solution

$\cos (\frac{d y}{d x})=a$

$\Rightarrow \frac{d y}{d x}=\cos ^{-1} a$

$\Rightarrow d y=\cos ^{-1} a d x$

Integrating both sides, we get:

$$ \begin{align*} & \int d y=\cos ^{-1} a \int d x \\ & \Rightarrow y=\cos ^{-1} a \cdot x+C \\ & \Rightarrow y=x \cos ^{-1} a+C \tag{1} \end{align*} $$

Now, $y=1$ when $x=0$.

$\Rightarrow 1=0 \cdot \cos ^{-1} a+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (1), we get:

$y=x \cos ^{-1} a+1$ $\Rightarrow \frac{y-1}{x}=\cos ^{-1} a$ $\Rightarrow \cos (\frac{y-1}{x})=a$

Solution

$\frac{d y}{d x}=y \tan x$

$\Rightarrow \frac{d y}{y}=\tan x d x$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{d y}{y}=\int \tan x d x \\ & \Rightarrow \log y=\log (\sec x)+\log C \\ & \Rightarrow \log y=\log (C \sec x) \\ & \Rightarrow y=C \sec x \tag{1} \end{align*} $$

Now, $y=1$ when $x=0$.

$\Rightarrow 1=C \times \sec 0$

$\Rightarrow 1=C \times 1$

$\Rightarrow C=1$

Substituting $C=1$ in equation (1), we get:

$y=\sec x$

Solution

The differential equation of the curve is:

$$ \begin{aligned} & y^{\prime}=e^{x} \sin x \\ & \Rightarrow \frac{d y}{d x}=e^{x} \sin x \\ & \Rightarrow d y=e^{x} \sin x \end{aligned} $$

Integrating both sides, we get:

$$ \begin{equation*} \int d y=\int e^{x} \sin x d x \tag{1} \end{equation*} $$

Let $I=\int e^{x} \sin x d x$.

$\Rightarrow I=\sin x \int e^{x} d x-\int(\frac{d}{d x}(\sin x) \cdot \int e^{x} d x) d x$ $\Rightarrow I=\sin x \cdot e^{x}-\int \cos x \cdot e^{x} d x$

$\Rightarrow I=\sin x \cdot e^{x}-[\cos x \cdot \int e^{x} d x-\int(\frac{d}{d x}(\cos x) \cdot \int e^{x} d x) d x]$

$\Rightarrow I=\sin x \cdot e^{x}-[\cos x \cdot e^{x}-\int(-\sin x) \cdot e^{x} d x]$

$\Rightarrow I=e^{x} \sin x-e^{x} \cos x-I$

$\Rightarrow 2 I=e^{x}(\sin x-\cos x)$

$\Rightarrow I=\frac{e^{x}(\sin x-\cos x)}{2}$

Substituting this value in equation (1), we get:

$y=\frac{e^{x}(\sin x-\cos x)}{2}+C$

Now, the curve passes through point $(0,0)$.

$\therefore 0=\frac{e^{0}(\sin 0-\cos 0)}{2}+C$

$\Rightarrow 0=\frac{1(0-1)}{2}+C$

$\Rightarrow C=\frac{1}{2}$

Substituting $C=\frac{1}{2}$ in equation (2), we get:

$y=\frac{e^{x}(\sin x-\cos x)}{2}+\frac{1}{2}$

$\Rightarrow 2 y=e^{x}(\sin x-\cos x)+1$

$\Rightarrow 2 y-1=e^{x}(\sin x-\cos x)$

Hence, the required equation of the curve is $2 y-1=e^{x}(\sin x-\cos x)$.

Solution

The differential equation of the given curve is:

$x y \frac{d y}{d x}=(x+2)(y+2)$

$\Rightarrow(\frac{y}{y+2}) d y=(\frac{x+2}{x}) d x$

$\Rightarrow(1-\frac{2}{y+2}) d y=(1+\frac{2}{x}) d x$

Integrating both sides, we get:

$$ \begin{align*} & \int(1-\frac{2}{y+2}) d y=\int(1+\frac{2}{x}) d x \\ & \Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x \\ & \Rightarrow y-2 \log (y+2)=x+2 \log x+C \\ & \Rightarrow y-x-C=\log x^{2}+\log (y+2)^{2} \\ & \Rightarrow y-x-C=\log [x^{2}(y+2)^{2}] \tag{1} \end{align*} $$

Now, the curve passes through point $(1,-1)$.

$ \begin{aligned} & \Rightarrow-1-1-C=\log [(1)^{2}(-1+2)^{2}] \\ & \Rightarrow-2-C=\log 1=0 \\ & \Rightarrow C=-2 \end{aligned} $

Substituting $C=-2$ in equation (1), we get:

$ y-x+2=\log [x^{2}(y+2)^{2}] $

This is the required solution of the given curve.

Solution

Let $x$ and $y$ be the $x$-coordinate and $y$-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the $\frac{d y}{d x}$

According to the given information, we get:

$y \cdot \frac{d y}{d x}=x$

$\Rightarrow y d y=x d x$

Integrating both sides, we get:

$\int y d y=\int x d x$

$\Rightarrow \frac{y^{2}}{2}=\frac{x^{2}}{2}+C$

$\Rightarrow y^{2}-x^{2}=2 C$

Now, the curve passes through point $(0,-2)$.

$\therefore(-2)^{2}-0^{2}=2 C$

$\Rightarrow 2 C=4$

Substituting $2 C=4$ in equation (1), we get:

$y^{2}-x^{2}=4$

This is the required equation of the curve.

Solution

It is given that $(x, y)$ is the point of contact of the curve and its tangent.

The slope $(m_1)$ of the line segment joining $(x, y)$ and $(-4,-3)$ is $\frac{y+3}{x+4}$.

We know that the slope of the tangent to the curve is given by the relation, $\frac{d y}{d x}$

$\therefore$ Slope $(m_2)$ of the tangent $=\frac{d y}{d x}$

According to the given information:

$m_2=2 m_1$

$\Rightarrow \frac{d y}{d x}=\frac{2(y+3)}{x+4}$

$\Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{d y}{y+3}=2 \int \frac{d x}{x+4} \\ & \Rightarrow \log (y+3)=2 \log (x+4)+\log C \\ & \Rightarrow \log (y+3) \log C(x+4)^{2} \\ & \Rightarrow y+3=C(x+4)^{2} \tag{1} \end{align*} $$

This is the general equation of the curve.

It is given that it passes through point $(-2,1)$.

$ \begin{aligned} & \Rightarrow 1+3=C(-2+4)^{2} \\ & \Rightarrow 4=4 C \\ & \Rightarrow C=1 \end{aligned} $

Substituting $C=1$ in equation (1), we get:

$y+3=(x+4)^{2}$

This is the required equation of the curve.

Solution

Let the rate of change of the volume of the balloon be $k$ (where $k$ is a constant).

$\Rightarrow \frac{d v}{d t}=k$

$\Rightarrow \frac{d}{d t}(\frac{4}{3} \pi r^{3})=k$

$[.$ Volume of sphere $.=\frac{4}{3} \pi r^{3}]$

$\Rightarrow \frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=k$

$\Rightarrow 4 \pi r^{2} d r=k d t$

Integrating both sides, we get:

$4 \pi \int r^{2} d r=k \int d t$

$\Rightarrow 4 \pi \cdot \frac{r^{3}}{3}=k t+C$

$\Rightarrow 4 \pi r^{3}=3(k t+C)$

Now, at $t=0, r=3$ :

$\Rightarrow 4 \pi \times 3^{3}=3(k \times 0+C)$

$\Rightarrow 108 \pi=3 C$

$\Rightarrow C=36 \Pi$

At $t=3, r=6:$

$\Rightarrow 4 \pi \times 6^{3}=3(k \times 3+C)$

$\Rightarrow 864 \pi=3(3 k+36 \pi)$

$\Rightarrow 3 k=-288 \pi-36 \pi=252 \pi$

$\Rightarrow k=84 \pi$

Substituting the values of $k$ and $C$ in equation (1), we get:

$ \begin{aligned} & 4 \pi r^{3}=3[84 \pi t+36 \pi] \\ & \Rightarrow 4 \pi r^{3}=4 \pi(63 t+27) \\ & \Rightarrow r^{3}=63 t+27 \\ & \Rightarrow r=(63 t+27)^{\frac{1}{3}} \end{aligned} $

Thus, the radius of the balloon after $t$ seconds is $(63 t+27)^{\frac{1}{3}}$.

Solution

Let $p, t$, and $r$ represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of $r %$ per year.

$\Rightarrow \frac{d p}{d t}=(\frac{r}{100}) p$

$\Rightarrow \frac{d p}{p}=(\frac{r}{100}) d t$

Integrating both sides, we get:

$\int \frac{d p}{p}=\frac{r}{100} \int d t$

$\Rightarrow \log p=\frac{r t}{100}+k$

$\Rightarrow p=e^{\frac{r t}{100}+k}$

It is given that when $t=0, p=100$.

$\Rightarrow 100=e^{k}$

Now, if $t=10$, then $p=2 \times 100=200$.

Therefore, equation (1) becomes:

$ \begin{aligned} & 200=e^{\frac{r}{10}+k} \\ & \Rightarrow 200=e^{\frac{r}{10}} \cdot e^{k} \\ & \Rightarrow 200=e^{\frac{r}{10}} \cdot 100 \\ & \Rightarrow e^{\frac{r}{10}}=2 \\ & \Rightarrow \frac{r}{10}=\log _{e} 2 \\ & \Rightarrow \frac{r}{10}=0.6931 \\ & \Rightarrow r=6.931 \end{aligned} $

Hence, the value of $r$ is $6.93 %$.

Solution

Let $p$ and $t$ be the principal and time respectively.

It is given that the principal increases continuously at the rate of $5 %$ per year.

$\Rightarrow \frac{d p}{d t}=(\frac{5}{100}) p$

$\Rightarrow \frac{d p}{d t}=\frac{p}{20}$

$\Rightarrow \frac{d p}{p}=\frac{d t}{20}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{d p}{p}=\frac{1}{20} \int d t \\ & \Rightarrow \log p=\frac{t}{20}+C \\ & \Rightarrow p=e^{\frac{1}{20}+C} \tag{1} \end{align*} $$

Now, when $t=0, p=1000$.

$\Rightarrow 1000=e^{C}$

At $t=10$, equation (1) becomes:

$ \begin{aligned} & p=e^{\frac{1}{2}+C} \\ & \Rightarrow p=e^{0.5} \times e^{C} \\ & \Rightarrow p=1.648 \times 1000 \\ & \Rightarrow p=1648 \end{aligned} $

Hence, after 10 years the amount will worth Rs 1648.

Solution

Let $y$ be the number of bacteria at any instant $t$.

It is given that the rate of growth of the bacteria is proportional to the number present.

$\therefore \frac{d y}{d t} \propto y$

$\Rightarrow \frac{d y}{d t}=k y$ (where $k$ is a constant)

$\Rightarrow \frac{d y}{y}=k d t$

Integrating both sides, we get:

$\int \frac{d y}{y}=k \int d t$

$\Rightarrow \log y=k t+C$

Let $y_0$ be the number of bacteria at $t=0$.

$\Rightarrow \log y_0=C$

Substituting the value of $C$ in equation (1), we get:

$$ \begin{align*} & \log y=k t+\log y_0 \\ & \Rightarrow \log y-\log y_0=k t \\ & \Rightarrow \log (\frac{y}{y_0})=k t \\ & \Rightarrow k t=\log (\frac{y}{y_0}) \tag{2} \end{align*} $$

Also, it is given that the number of bacteria increases by $10 %$ in 2 hours. $\Rightarrow y=\frac{110}{100} y_0$

$\Rightarrow \frac{y}{y_0}=\frac{11}{10}$

Substituting this value in equation (2), we get:

$$ \begin{aligned} & k \cdot 2=\log (\frac{11}{10}) \\ & \Rightarrow k=\frac{1}{2} \log (\frac{11}{10}) \end{aligned} $$

Therefore, equation (2) becomes:

$$ \begin{align*} & \frac{1}{2} \log (\frac{11}{10}) \cdot t=\log (\frac{y}{y_0}) \\ & \Rightarrow t=\frac{2 \log (\frac{y}{y_0})}{\log (\frac{11}{10})} \tag{4} \end{align*} $$

Now, let the time when the number of bacteria increases from 100000 to 200000 be $t_1$.

$\Rightarrow y=2 y_0$ at $t=t_1$

From equation (4), we get:

$t_1=\frac{2 \log (\frac{y}{y_0})}{\log (\frac{11}{10})}=\frac{2 \log 2}{\log (\frac{11}{10})}$

Hence, in $\frac{2 \log 2}{\log (\frac{11}{10})}$ hours the number of bacteria increases from 100000 to 200000.

Solution

$ \begin{aligned} & \frac{d y}{d x}=e^{x+y}=e^{x} \cdot e^{y} \\ & \Rightarrow \frac{d y}{e^{y}}=e^{x} d x \\ & \Rightarrow e^{-y} d y=e^{x} d x \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & \int e^{-y} d y=\int e^{x} d x \\ & \Rightarrow-e^{-y}=e^{x}+k \\ & \Rightarrow e^{x}+e^{-y}=-k \\ & \Rightarrow e^{x}+e^{-y}=c \end{aligned} $

Hence, the correct answer is A.

Solution

The given differential equation i.e., $(x^{2}+x y) d y=(x^{2}+y^{2}) d x$ can be written as:

$\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}$

Let $F(x, y)=\frac{x^{2}+y^{2}}{x^{2}+x y}$.

Now, $F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}=\frac{x^{2}+y^{2}}{x^{2}+x y}=\lambda^{0} \cdot F(x, y)$

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{x^{2}+(v x)^{2}}{x^{2}+x(v x)} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{1+v^{2}}{1+v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}-v=\frac{(1+v^{2})-v(1+v)}{1+v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1-v}{1+v} \\ & \Rightarrow(\frac{1+v}{1-v})=d v=\frac{d x}{x} \\ & \Rightarrow(\frac{2-1+v}{1-v}) d v=\frac{d x}{x} \\ & \Rightarrow(\frac{2}{1-v}-1) d v=\frac{d x}{x} \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & -2 \log (1-v)-v=\log x-\log k \\ & \Rightarrow v=-2 \log (1-v)-\log x+\log k \\ & \Rightarrow v=\log [\frac{k}{x(1-v)^{2}}] \\ & \Rightarrow \frac{y}{x}=\log [\frac{k}{x(1-\frac{y}{x})^{2}}] \\ & \Rightarrow \frac{y}{x}=\log [\frac{k x}{(x-y)^{2}}] \\ & \Rightarrow \frac{k x}{(x-y)^{2}}=e^{\frac{y}{x}} \\ & \Rightarrow(x-y)^{2}=k x e^{-\frac{y}{x}} \end{aligned} $

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$y^{\prime}=\frac{x+y}{x}$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x}$

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

Now, $F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}=\lambda^{0} F(x, y)$

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x}$

$\Rightarrow v+x \frac{d v}{d x}=1+v$

$x \frac{d v}{d x}=1$

$\Rightarrow d v=\frac{d x}{x}$

Integrating both sides, we get:

$v=\log x+C$

$\Rightarrow \frac{y}{x}=\log x+C$

$\Rightarrow y=x \log x+C x$

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$(x-y) d y-(x+y) d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x-y}$

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

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}=\lambda^{0} \cdot F(x, y)$

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v}$

$x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1-v}$

$\Rightarrow \frac{1-v}{(1+v^{2})} d v=\frac{d x}{x}$

$\Rightarrow(\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}}) d v=\frac{d x}{x}$

Integrating both sides, we get:

$ \begin{aligned} & \tan ^{-1} v-\frac{1}{2} \log (1+v^{2})=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2} \log [1+(\frac{y}{x})^{2}]=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2} \log (\frac{x^{2}+y^{2}}{x^{2}})=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2}[\log (x^{2}+y^{2})-\log x^{2}]=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})=\frac{1}{2} \log (x^{2}+y^{2})+C \end{aligned} $

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$(x^{2}-y^{2}) d x+2 x y d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-(x^{2}-y^{2})}{2 x y}$

Let $F(x, y)=\frac{-(x^{2}-y^{2})}{2 x y}$.

$\therefore F(\lambda x, \lambda y)=[\frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)}]=\frac{-(x^{2}-y^{2})}{2 x y}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$ $\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=-[\frac{x^{2}-(v x)^{2}}{2 x \cdot(v x)}]$

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

$\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v=\frac{v^{2}-1-2 v^{2}}{2 v}$

$\Rightarrow x \frac{d v}{d x}=-\frac{(1+v^{2})}{2 v}$

$\Rightarrow \frac{2 v}{1+v^{2}} d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log (1+v^{2})=-\log x+\log C=\log \frac{C}{x}$

$\Rightarrow 1+v^{2}=\frac{C}{x}$

$\Rightarrow[1+\frac{y^{2}}{x^{2}}]=\frac{C}{x}$

$\Rightarrow x^{2}+y^{2}=C x$

This is the required solution of the given differential equation.

Solution

The given differential equation is:

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

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

$\therefore F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda x)(\lambda y)}{(\lambda x)^{2}}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x^{2}-2(v x)^{2}+x \cdot(v x)}{x^{2}}$

$\Rightarrow v+x \frac{d v}{d x}=1-2 v^{2}+v$

$\Rightarrow x \frac{d v}{d x}=1-2 v^{2}$

$\Rightarrow \frac{d v}{1-2 v^{2}}=\frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot \frac{d v}{\frac{1}{2}-v^{2}}=\frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot[\frac{d v}{(\frac{1}{\sqrt{2}})^{2}-v^{2}}]=\frac{d x}{x}$

Integrating both sides, we get:

$ \begin{aligned} & \frac{1}{2} \cdot \frac{1}{2 \times \frac{1}{\sqrt{2}}} \log |\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}|=\log |x|+C \\ & \Rightarrow \frac{1}{2 \sqrt{2}} \log |\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}|=\log |x|+C \\ & \Rightarrow \frac{1}{2 \sqrt{2}} \log |\frac{x+\sqrt{2} y}{x-\sqrt{2} y}|=\log |x|+C \end{aligned} $

This is the required solution for the given differential equation.

Solution

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

$\Rightarrow x d y=[y+\sqrt{x^{2}+y^{2}}] d x$

$\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$

Let $F(x, y)=\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+(v x)^{2}}}{x}$

$\Rightarrow v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}}$

$\Rightarrow \frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}$

Integrating both sides, we get:

$\log |v+\sqrt{1+v^{2}}|=\log |x|+\log C$

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

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

$\Rightarrow y+\sqrt{x^{2}+y^{2}}=C x^{2}$

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$$x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right) y d x=y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right) x d y$$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d y}{d x}=v+x=\frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{(x \cos v+v x \sin v) \cdot v x}{(v x \sin v-x \cos v) \cdot x} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v}-v \\ & \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v-v^{2} \sin v+v \cos v}{v \sin v-\cos v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\ & \Rightarrow[\frac{v \sin v-\cos v}{v \cos v}] d v=\frac{2 d x}{x} \\ & \Rightarrow(\tan v-\frac{1}{v}) d v=\frac{2 d x}{x} \end{aligned} $

Integrating both sides, we get:

$\log (\sec v)-\log v=2 \log x+\log C$

$\Rightarrow \log (\frac{\sec v}{v})=\log (C x^{2})$

$\Rightarrow(\frac{\sec v}{v})=C x^{2}$

$\Rightarrow \sec v=C x^{2} v$

$\Rightarrow \sec (\frac{y}{x})=C \cdot x^{2} \cdot \frac{y}{x}$

$\Rightarrow \sec (\frac{y}{x})=C x y$

$\Rightarrow \cos (\frac{y}{x})=\frac{1}{C x y}=\frac{1}{C} \cdot \frac{1}{x y}$

$\Rightarrow x y \cos (\frac{y}{x})=k \quad(k=\frac{1}{C})$

This is the required solution of the given differential equation.

Solution

$x \frac{d y}{d x}-y+x \sin (\frac{y}{x})=0$

$\Rightarrow x \frac{d y}{d x}=y-x \sin (\frac{y}{x})$

$\Rightarrow \frac{d y}{d x}=\frac{y-x \sin (\frac{y}{x})}{x}$

Let $F(x, y)=\frac{y-x \sin (\frac{y}{x})}{x}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y-\lambda x \sin (\frac{\lambda y}{\lambda x})}{\lambda x}=\frac{y-x \sin (\frac{y}{x})}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x-x \sin v}{x}$

$\Rightarrow v+x \frac{d v}{d x}=v-\sin v$

$\Rightarrow-\frac{d v}{\sin v}=\frac{d x}{x}$

$\Rightarrow cosec v d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log |cosec v-\cot v|=-\log x+\log C=\log \frac{C}{x}$

$\Rightarrow cosec(\frac{y}{x})-\cot (\frac{y}{x})=\frac{C}{x}$

$\Rightarrow \frac{1}{\sin (\frac{y}{x})}-\frac{\cos (\frac{y}{x})}{\sin (\frac{y}{x})}=\frac{C}{x}$

$\Rightarrow x[1-\cos (\frac{y}{x})]=C \sin (\frac{y}{x})$

This is the required solution of the given differential equation.

Solution

$y d x+x \log (\frac{y}{x}) d y-2 x d y=0$

$\Rightarrow y d x=[2 x-x \log (\frac{y}{x})] d y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{2 x-x \log (\frac{y}{x})}$

Let $F(x, y)=\frac{y}{2 x-x \log (\frac{y}{x})}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{2(\lambda x)-(\lambda x) \log (\frac{\lambda y}{\lambda x})}=\frac{y}{2 x-\log (\frac{y}{x})}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$ $\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x}{2 x-x \log v}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v$

$\Rightarrow x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v}$

$\Rightarrow \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x}$

$\Rightarrow[\frac{1+(1-\log v)}{v(\log v-1)}] d v=\frac{d x}{x}$

$\Rightarrow[\frac{1}{v(\log v-1)}-\frac{1}{v}] d v=\frac{d x}{x}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\int \frac{1}{x} d x \\ & \Rightarrow \int \frac{d v}{v(\log v-1)}-\log v=\log x+\log C \tag{2} \end{align*} $$

$\Rightarrow$ Let $\log v-1=t$

$\Rightarrow \frac{d}{d v}(\log v-1)=\frac{d t}{d v}$

$\Rightarrow \frac{1}{v}=\frac{d t}{d v}$

$\Rightarrow \frac{d v}{v}=d t$

Therefore, equation (1) becomes:

$\Rightarrow \int \frac{d t}{t}-\log v=\log x+\log C$

$\Rightarrow \log t-\log (\frac{y}{x})=\log (C x)$

$\Rightarrow \log [\log (\frac{y}{x})-1]-\log (\frac{y}{x})=\log (C x)$

$\Rightarrow \log [\frac{\log (\frac{y}{x})-1}{\frac{y}{x}}]=\log (C x)$

$\Rightarrow \frac{x}{y}[\log (\frac{y}{x})-1]=C x$

$\Rightarrow \log (\frac{y}{x})-1=C y$

This is the required solution of the given differential equation.

Solution

$(1+e^{\frac{x}{y}}) d x+e^{\frac{x}{y}}(1-\frac{x}{y}) d y=0$

$\Rightarrow(1+e^{\frac{x}{y}}) d x=-e^{\frac{x}{y}}(1-\frac{x}{y}) d y$ $\Rightarrow \frac{d x}{d y}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$

Let $F(x, y)=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$.

$\therefore F(\lambda x, \lambda y)=\frac{-e^{\frac{\lambda x}{\lambda y}}(1-\frac{\lambda x}{\lambda y})}{1+e^{\frac{\lambda x}{\lambda y}}}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$x=v y$

$\Rightarrow \frac{d}{d y}(x)=\frac{d}{d y}(v y)$

$\Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}$

Substituting the values of $x$ and $\frac{d x}{d y}$ in equation (1), we get:

$v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}}{1+e^{v}}-v$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}-v-v e^{v}}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=-[\frac{v+e^{v}}{1+e^{v}}]$

$\Rightarrow[\frac{1+e^{v}}{v+e^{v}}] d v=-\frac{d y}{y}$

Integrating both sides, we get: $\Rightarrow \log (v+e^{v})=-\log y+\log C=\log (\frac{C}{y})$

$\Rightarrow[\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}$

$\Rightarrow x+y e^{\frac{x}{y}}=C$

This is the required solution of the given differential equation.

Solution

$(x+y) d y+(x-y) d x=0$

$\Rightarrow(x+y) d y=-(x-y) d x$

$\Rightarrow \frac{d y}{d x}=\frac{-(x-y)}{x+y}$

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

$\therefore F(\lambda x, \lambda y)=\frac{-(\lambda x-\lambda y)}{\lambda x-\lambda y}=\frac{-(x-y)}{x+y}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get: $v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v-1}{v+1}$

$\Rightarrow x \frac{d v}{d x}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1}$

$\Rightarrow x \frac{d v}{d x}=\frac{v-1-v^{2}-v}{v+1}=\frac{-(1+v^{2})}{v+1}$

$\Rightarrow \frac{(v+1)}{1+v^{2}} d v=-\frac{d x}{x}$

$\Rightarrow[\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}] d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\frac{1}{2} \log (1+v^{2})+\tan ^{-1} v=-\log x+k$

$\Rightarrow \log (1+v^{2})+2 \tan ^{-1} v=-2 \log x+2 k$

$\Rightarrow \log [(1+v^{2}) \cdot x^{2}]+2 \tan ^{-1} v=2 k$

$\Rightarrow \log [(1+\frac{y^{2}}{x^{2}}) \cdot x^{2}]+2 \tan ^{-1} \frac{y}{x}=2 k$

$\Rightarrow \log (x^{2}+y^{2})+2 \tan ^{-1} \frac{y}{x}=2 k$

Now, $y=1$ at $x=1$.

$\Rightarrow \log 2+2 \tan ^{-1} 1=2 k$

$\Rightarrow \log 2+2 \times \frac{\pi}{4}=2 k$

$\Rightarrow \frac{\pi}{2}+\log 2=2 k$

Substituting the value of $2 k$ in equation (2), we get:

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

This is the required solution of the given differential equation.

Solution

$x^{2} d y+(x y+y^{2}) d x=0$

$\Rightarrow x^{2} d y=-(x y+y^{2}) d x$

$\Rightarrow \frac{d y}{d x}=\frac{-(x y+y^{2})}{x^{2}}$

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

$\therefore F(\lambda x, \lambda y)=\frac{[\lambda x \cdot \lambda y+(\lambda y)^{2}]}{(\lambda x)^{2}}=\frac{-(x y+y^{2})}{x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{-[x \cdot v x+(v x)^{2}]}{x^{2}}=-v-v^{2} \\ & \Rightarrow x \frac{d v}{d x}=-v^{2}-2 v=-v(v+2) \\ & \Rightarrow \frac{d v}{v(v+2)}=-\frac{d x}{x} \\ & \Rightarrow \frac{1}{2}[\frac{(v+2)-v}{v(v+2)}] d v=-\frac{d x}{x} \\ & \Rightarrow \frac{1}{2}[\frac{1}{v}-\frac{1}{v+2}] d v=-\frac{d x}{x} \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \frac{1}{2}[\log v-\log (v+2)]=-\log x+\log C \\ & \Rightarrow \frac{1}{2} \log (\frac{v}{v+2})=\log \frac{C}{x} \\ & \Rightarrow \frac{v}{v+2}=(\frac{C}{x})^{2} \\ & \Rightarrow \frac{\frac{y}{y}}{\frac{y}{x}+2}=(\frac{C}{x})^{2} \\ & \Rightarrow \frac{y}{y+2 x}=\frac{C^{2}}{x^{2}} \\ & \Rightarrow \frac{x^{2} y}{y+2 x}=C^{2} \tag{2} \end{align*} $$

Now, $y=1$ at $x=1$.

$\Rightarrow \frac{1}{1+2}=C^{2}$

$\Rightarrow C^{2}=\frac{1}{3}$

Substituting $C^{2}=\frac{1}{3}$ in equation (2), we get:

$ \begin{aligned} & \frac{x^{2} y}{y+2 x}=\frac{1}{3} \\ & \Rightarrow y+2 x=3 x^{2} y \end{aligned} $

This is the required solution of the given differential equation.

Solution

$[x \sin ^{2}(\frac{y}{x})-y] d x+x d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}$

Let $F(x, y)=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}$.

$\therefore F(\lambda x, \lambda y)=\frac{-[\lambda x \cdot \sin ^{2}(\frac{\lambda x}{\lambda y})-\lambda y]}{\lambda x}=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x=\frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{-[x \sin ^{2} v-v x]}{x}$

$\Rightarrow v+x \frac{d v}{d x}=-[\sin ^{2} v-v]=v-\sin ^{2} v$

$\Rightarrow x \frac{d v}{d x}=-\sin ^{2} v$

$\Rightarrow \frac{d v}{\sin ^{2} v}=-\frac{d x}{d x}$

$\Rightarrow cosec^{2} v d v=-\frac{d x}{x}$

Integrating both sides, we get: $-\cot v=-\log |x|-C$

$\Rightarrow \cot v=\log |x|+C$

$\Rightarrow \cot (\frac{y}{x})=\log |x|+\log C$

$\Rightarrow \cot (\frac{y}{x})=\log |Cx|$

Now, $y=\frac{\pi}{4}$ at $x=1$

$\Rightarrow \cot (\frac{\pi}{4})=\log |C|$

$\Rightarrow 1=\log C$

$\Rightarrow C=e^{1}=e$

Substituting $C=e$ in equation (2), we get:

$\cot (\frac{y}{x})=\log |e x|$

This is the required solution of the given differential equation.

Solution

$\frac{d y}{d x}-\frac{y}{x}+cosec(\frac{y}{x})=0$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x}-cosec(\frac{y}{x})$

Let $F(x, y)=\frac{y}{x}-cosec(\frac{y}{x})$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}-cosec(\frac{\lambda y}{\lambda x})$

$\Rightarrow F(\lambda x, \lambda y)=\frac{y}{x}-cosec(\frac{y}{x})=F(x, y)=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=v-cosec v$

$\Rightarrow-\frac{d v}{cosec v}=-\frac{d x}{x}$

$\Rightarrow-\sin v d v=\frac{d x}{x}$

Integrating both sides, we get:

$\cos v=\log x+\log C=\log |C x|$

$\Rightarrow \cos (\frac{y}{x})=\log |C x|$

This is the required solution of the given differential equation.

Now, $y=0$ at $x=1$.

$\Rightarrow \cos (0)=\log C$

$\Rightarrow 1=\log C$

$\Rightarrow C=e^{1}=e$

Substituting $C=e$ in equation (2), we get:

$\cos (\frac{y}{x})=\log |(e x)|$

This is the required solution of the given differential equation.

Solution

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

$\Rightarrow 2 x^{2} \frac{d y}{d x}=2 x y+y^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$

Let $F(x, y)=\frac{2 x y+y^{2}}{2 x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{2(\lambda x)(\lambda y)+(\lambda y)^{2}}{2(\lambda x)^{2}}=\frac{2 x y+y^{2}}{2 x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{2 x(v x)+(v x)^{2}}{2 x^{2}}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{2 v+v^{2}}{2}$

$\Rightarrow v+x \frac{d v}{d x}=v+\frac{v^{2}}{2}$

$\Rightarrow \frac{2}{v^{2}} d v=\frac{d x}{x}$

Integrating both sides, we get:

$$ \begin{align*} & 2 \cdot \frac{v^{-2+1}}{-2+1}=\log |x|+C \\ & \Rightarrow-\frac{2}{v}=\log |x|+C \\ & \Rightarrow-\frac{2}{\frac{y}{x}}=\log |x|+C \\ & \Rightarrow-\frac{2 x}{y}=\log |x|+C \\ & \text{ Now, } y=2 \text{ at } x=1 . \tag{2}\\ & \Rightarrow-1=\log (1)+C \\ & \Rightarrow C=-1 \end{align*} $$

Substituting $C=-1$ in equation (2), we get:

$ \begin{aligned} & -\frac{2 x}{y}=\log |x|-1 \\ & \Rightarrow \frac{2 x}{y}=1-\log |x| \\ & \Rightarrow y=\frac{2 x}{1-\log |x|},(x \neq 0, x \neq e) \end{aligned} $

This is the required solution of the given differential equation.

Solution

For solving the homogeneous equation of the form $\frac{d x}{d y}=h(\frac{x}{y})$, we need to make the substitution as $x=v y$.

Hence, the correct answer is $C$.

Solution

Function $F(x, y)$ is said to be the homogenous function of degree $n$, if $F(\lambda x, \lambda y)=\lambda^{n} F(x, y)$ for any non-zero constant $(\lambda)$.

Consider the equation given in alternativeD:

$y^{2} d x+(x^{2}-x y-y^{2}) d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}=\frac{y^{2}}{y^{2}+x y-x^{2}}$

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

$\Rightarrow F(\lambda x, \lambda y)=\frac{(\lambda y)^{2}}{(\lambda y)^{2}+(\lambda x)(\lambda y)-(\lambda x)^{2}}$

$=\frac{\lambda^{2} y^{2}}{\lambda^{2}(y^{2}+x y-x^{2})}$

$=\lambda^{0}(\frac{y^{2}}{y^{2}+x y-x^{2}})$

$=\lambda^{0} \cdot F(x, y)$

Hence, the differential equation given in alternative $\mathbf{D}$ is a homogenous equation.

Solution

The given differential equation is $\frac{d y}{d x}+2 y=\sin x$.

This is in the form of $\frac{d y}{d x}+p y=Q($ where $p=2$ and $Q=\sin x)$.

Now, I.F $=e^{\int p d x}=e^{\int 2 d x}=e^{2 x}$.

The solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{2 x}=\int \sin x \cdot e^{2 x} d x+C \tag{1} \end{align*} $$

Let $I=\int \sin x \cdot e^{2 x}$.

$\Rightarrow I=\sin x \cdot \int e^{2 x} d x-\int(\frac{d}{d x}(\sin x) \cdot \int e^{2 x} d x) d x$

$\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int(\cos x \cdot \frac{e^{2 x}}{2}) d x$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \int e^{2 x}-\int(\frac{d}{d x}(\cos x) \cdot \int e^{2 x} d x) d x]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \frac{e^{2 x}}{2}-\int[(-\sin x) \cdot \frac{e^{2 x}}{2}] d x]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} \int(\sin x \cdot e^{2 x}) d x$

$\Rightarrow I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I$

$\Rightarrow \frac{5}{4} I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)$

$\Rightarrow I=\frac{e^{2 x}}{5}(2 \sin x-\cos x)$

Therefore, equation (1) becomes:

$y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+C$

$\Rightarrow y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}$

This is the required general solution of the given differential equation.

Solution

$ \frac{d y}{d x}+p y=Q(\text{ where } p=3 \text{ and } Q=e^{-2 x}) \text{. } $

The given differential equation is

Now, I.F $=e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$.

The solution of the given differential equation is given by the relation,

$ \begin{aligned} & y(I . F .)=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{3 x}=\int(e^{-2 x} \times e^{3 x})+C \\ & \Rightarrow y e^{3 x}=\int e^{x} d x+C \\ & \Rightarrow y e^{3 x}=e^{x}+C \\ & \Rightarrow y=e^{-2 x}+C e^{-3 x} \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x}$ and $.Q=x^{2})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x$.

The solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(x)=\int(x^{2} \cdot x) d x+C$

$\Rightarrow x y=\int x^{3} d x+C$

$\Rightarrow x y=\frac{x^{4}}{4}+C$

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$\frac{d y}{d x}+p y=Q($ where $p=\sec x$ and $Q=\tan x)$

Now, I.F $=e^{\int p d x}=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$.

The general solution of the given differential equation is given by the relation,

$ y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C $

$\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+C$

$\Rightarrow y(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^{2} x d x+C$

$\Rightarrow y(\sec x+\tan x)=\sec x+\int(\sec ^{2} x-1) d x+C$

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

Solution

We have,

$\begin{aligned} & \cos ^2 x \frac{d y}{d x}+y=\tan x \ & \frac{d y}{d x}+\sec ^2 x(y)=\sec ^2 x \tan x \end{aligned}$

We know that the general equation

$\frac{d y}{d x}+P y=Q$

Here,

$P=\sec ^2 x, Q=\sec ^2 x \tan x$

Since, integrating factor I. $F=e^{\int P d x}$ I. $F=e^{\int \sec ^2 x d x}$ I. $F=e^{\tan x}$

We know that the general solution,

$\mathrm{y} \times \mathrm{I} . \mathrm{F}=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C}$

Therefore,

$y \times e^{\tan x}=\int\left(\sec ^2 x \tan x\right) e^{\tan x} d x+C$

Let $\mathrm{t}=\tan \mathrm{x}$

$\mathrm{dt}=\sec ^2 \mathrm{x} d \mathrm{x}$

Therefore,

$\begin{aligned} & y \times e^t=\int t e^t d t+C \\ & y \times e^t=t e^t-\int 1 e^t d t+C \\ & y \times e^t=t e^t-e^t+C \end{aligned}$

On putting the value of $t$, we get

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

Hence, this is the answer.

Solution

The given differential equation is:

$x \frac{d y}{d x}+2 y=x^{2} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$

This equation is in the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2}{x}$ and $.Q=x \log x)$

Now, I.F $=e^{\int p d x}=e^{\int_x^{2} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}$.

The general solution of the given differential equation is given by the relation, $y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot x^{2}=\int(x \log x \cdot x^{2}) d x+C$

$\Rightarrow x^{2} y=\int(x^{3} \log x) d x+C$

$\Rightarrow x^{2} y=\log x \cdot \int x^{3} d x-\int[\frac{d}{d x}(\log x) \cdot \int x^{3} d x] d x+C$

$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int(\frac{1}{x} \cdot \frac{x^{4}}{4}) d x+C$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+C$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+C$

$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+C$

$\Rightarrow y=\frac{1}{16} x^{2}(4 \log x-1)+C x^{-2}$

Solution

The given differential equation is:

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}$

This equation is the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x \log x}$ and $Q=\frac{2}{x^{2}}$ )

Now, I.F $=e^{\int \rho d x}=e^{\int \frac{1}{x \log } d x}=e^{\log (\log x)}=\log x$.

The general solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(I . F .)=\int(Q \times I \text{.F. }) d x+C \\ & \Rightarrow y \log x=\int(\frac{2}{x^{2}} \log x) d x+C \tag{1} \end{align*} $$

Now, $\int(\frac{2}{x^{2}} \log x) d x=2 \int(\log x \cdot \frac{1}{x^{2}}) d x$.

$ \begin{aligned} & =2[\log x \cdot \int \frac{1}{x^{2}} d x-\int{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^{2}} d x} d x] \\ & =2[\log x(-\frac{1}{x})-\int(\frac{1}{x} \cdot(-\frac{1}{x})) d x] \\ & =2[-\frac{\log x}{x}+\int \frac{1}{x^{2}} d x] \\ & =2[-\frac{\log x}{x}-\frac{1}{x}] \\ & =-\frac{2}{x}(1+\log x) \end{aligned} $

Substituting the value of $\int(\frac{2}{x^{2}} \log x) d x$ in equation (1), we get:

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

This is the required general solution of the given differential equation.

Solution

$(1+x^{2}) d y+2 x y d x=\cot x d x$

$\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}$

This equation is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2 x}{1+x^{2}}$ and $.Q=\frac{\cot x}{1+x^{2}})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log (1+x^{2})}=1+x^{2}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(1+x^{2})=\int[\frac{\cot x}{1+x^{2}} \times(1+x^{2})] d x+C$

$\Rightarrow y(1+x^{2})=\int \cot x d x+C$

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

Solution

$x \frac{d y}{d x}+y-x+x y \cot x=0$

$\Rightarrow x \frac{d y}{d x}+y(1+x \cot x)=x$

$\Rightarrow \frac{d y}{d x}+(\frac{1}{x}+\cot x) y=1$

This equation is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x}+\cot x$ and $.Q=1)$

Now, I.F $=e^{\int \rho d x}=e^{\int(\frac{1}{x}+\cot x) d x}=e^{\log x+\log (\sin x)}=e^{\log (x \sin x)}=x \sin x$.

The general solution of the given differential equation is given by the relation,

$ \begin{aligned} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y(x \sin x)=\int(1 \times x \sin x) d x+C \\ & \Rightarrow y(x \sin x)=\int(x \sin x) d x+C \\ & \Rightarrow y(x \sin x)=x \int \sin x d x-\int[\frac{d}{d x}(x) \cdot \int \sin x d x]+C \\ & \Rightarrow y(x \sin x)=x(-\cos x)-\int 1 \cdot(-\cos x) d x+C \\ & \Rightarrow y(x \sin x)=-x \cos x+\sin x+C \\ & \Rightarrow y=\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x} \\ & \Rightarrow y=-\cot \cdot x+\frac{1}{x}+\frac{C}{x \sin x} \end{aligned} $

Solution

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}$

$\Rightarrow \frac{d x}{d y}=x+y$

$\Rightarrow \frac{d x}{d y}-x=y$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p x=Q($ where $p=-1$ and $Q=y$ )

Now, I.F $=e^{\int p d y}=e^{\int-d y}=e^{-y}$.

The general solution of the given differential equation is given by the relation, $x($ I.F. $)=\int(Q \times I . F) d y+.C$

$\Rightarrow x e^{-y}=\int(y \cdot e^{-y}) d y+C$

$\Rightarrow x e^{-y}=y \cdot \int e^{-y} d y-\int[\frac{d}{d y}(y) \int e^{-y} d y] d y+C$

$\Rightarrow x e^{-y}=y(-e^{-y})-\int(-e^{-y}) d y+C$

$\Rightarrow x e^{-y}=-y e^{-y}+\int e^{-y} d y+C$

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

$\Rightarrow x=-y-1+Ce^{y}$

$\Rightarrow x+y+1=Ce^{y}$

Solution

$y d x+(x-y^{2}) d y=0$

$\Rightarrow y d x=(y^{2}-x) d y$

$\Rightarrow \frac{d x}{d y}=\frac{y^{2}-x}{y}=y-\frac{x}{y}$

$\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p x=Q(.$ where $p=\frac{1}{y}$ and $.Q=y)$

Now, I.F $=e^{\int \rho d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y$.

The general solution of the given differential equation is given by the relation, $x(I . F)=.\int(Q \times I . F) d y+.C$

$\Rightarrow x y=\int(y \cdot y) d y+C$

$\Rightarrow x y=\int y^{2} d y+C$

$\Rightarrow x y=\frac{y^{3}}{3}+C$

$\Rightarrow x=\frac{y^{2}}{3}+\frac{C}{y}$

Solution

$(x+3 y^{2}) \frac{d y}{d x}=y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^{2}}$

$\Rightarrow \frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y$

$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p x=Q(.$ where $p=-\frac{1}{y}$ and $.Q=3 y)$

Now, I.F $=e^{\int p d y}=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log (\frac{1}{y})}=\frac{1}{y}$.

The general solution of the given differential equation is given by the relation, $x(I . F)=.\int(Q \times I . F) d y+.C$

$\Rightarrow x \times \frac{1}{y}=\int(3 y \times \frac{1}{y}) d y+C$

$\Rightarrow \frac{x}{y}=3 y+C$

$\Rightarrow x=3 y^{2}+C y$

Solution

The given differential equation is $\frac{d y}{d x}+2 y \tan x=\sin x$.

This is a linear equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=2 \tan x$ and $Q=\sin x)$

Now, I.F $=e^{\int p d x}=e^{\int 2 tan x d x}=e^{2 \log |\sec x|}=e^{\log (\sec ^{2} x)}=\sec ^{2} x$.

The general solution of the given differential equation is given by the relation,

$y(I . F)=.\int(Q \times I . F) d x+.C$

$\Rightarrow y(\sec ^{2} x)=\int(\sin x \cdot \sec ^{2} x) d x+C$

$\Rightarrow y \sec ^{2} x=\int(\sec x \cdot \tan x) d x+C$

$\Rightarrow y \sec ^{2} x=\sec x+C$

Now, $y=0$ at $x=\frac{\pi}{3}$.

Therefore,

$0 \times \sec ^{2} \frac{\pi}{3}=\sec \frac{\pi}{3}+C$

$\Rightarrow 0=2+C$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation (1), we get: $y \sec ^{2} x=\sec x-2$

$\Rightarrow y=\cos x-2 \cos ^{2} x$

Hence, the required solution of the given differential equation is $y=\cos x-2 \cos ^{2} x$.

Solution

$(1+x^{2}) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}}$

$\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{(1+x^{2})^{2}}$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2 x}{1+x^{2}}$ and $.Q=\frac{1}{(1+x^{2})^{2}})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x d x}{1+x^{2}}}=e^{\log (1+x^{2})}=1+x^{2}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(1+x^{2})=\int[\frac{1}{(1+x^{2})^{2}} \cdot(1+x^{2})] d x+C$

$\Rightarrow y(1+x^{2})=\int \frac{1}{1+x^{2}} d x+C$

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

Now, $y=0$ at $x=1$.

Therefore,

$0=\tan ^{-1} 1+C$

$\Rightarrow C=-\frac{\pi}{4}$

Substituting $C=-\frac{\pi}{4}$ in equation (1), we get:

$y(1+x^{2})=\tan ^{-1} x-\frac{\pi}{4}$

This is the required general solution of the given differential equation.

Solution

The given differential equation is $\frac{d y}{d x}-3 y \cot x=\sin 2 x$.

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-3 \cot x$ and $Q=\sin 2 x)$

Now, I.F $=e^{\int p d x}=e^{-3 \int \cot x d x}=e^{-3 \log |\sin x|}=e^{\log |\frac{1}{\sin ^{3} x}|}=\frac{1}{\sin ^{3} x}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot \frac{1}{\sin ^{3} x}=\int[\sin 2 x \cdot \frac{1}{\sin ^{3} x}] d x+C$

$\Rightarrow y cosec^{3} x=2 \int(\cot x cosec x) d x+C$

$\Rightarrow y cosec^{3} x=2 cosec x+C$

$\Rightarrow y=-\frac{2}{cosec^{2} x}+\frac{3}{cosec^{3} x}$

$\Rightarrow y=-2 \sin ^{2} x+C \sin ^{3} x$

Now, $y=2$ at $x=\frac{\pi}{2}$.

Therefore, we get:

$2=-2+C$

$\Rightarrow C=4$

Substituting $C=4$ in equation (1), we get:

$y=-2 \sin ^{2} x+4 \sin ^{3} x$

$\Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x$

This is the required particular solution of the given differential equation.

Solution

Let $F(x, y)$ be the curve passing through the origin.

At point $(x, y)$, the slope of the curve will be $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}=x+y$

$\Rightarrow \frac{d y}{d x}-y=x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x)$

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{-x}=\int x e^{-x} d x+C \tag{1} \end{align*} $$

Now, $\int x e^{-x} d x=x \int e^{-x} d x-\int[\frac{d}{d x}(x) \cdot \int e^{-x} d x] d x$.

$ =-x e^{-x}-\int-e^{-x} d x $

$ \begin{aligned} & =-x e^{-x}+(-e^{-x}) \\ & =-e^{-x}(x+1) \end{aligned} $

Substituting in equation (1), we get:

$$ \begin{align*} & y e^{-x}=-e^{-x}(x+1)+C \\ & \Rightarrow y=-(x+1)+C e^{x} \\ & \Rightarrow x+y+1=C e^{x} \tag{2} \end{align*} $$

The curve passes through the origin.

Therefore, equation (2) becomes:

$1=C$

Substituting $C=1$ in equation (2), we get:

$\Rightarrow \quad x+y+1=e^{x}$

Hence, the required equation of curve passing through the origin is $x+y+1=e^{x}$.

Solution

Let $F(x, y)$ be the curve and let $(x, y)$ be a point on the curve. The slope of the tangent

to the curve at $(x, y)$ is $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}+5=x+y$

$\Rightarrow \frac{d y}{d x}-y=x-5$

This is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x-5)$

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general equation of the curve is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot e^{-x}=\int(x-5) e^{-x} d x+C$

Now, $\int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int[\frac{d}{d x}(x-5) \cdot \int e^{-x} d x] d x$.

$ \begin{aligned} & =(x-5)(-e^{-x})-\int(-e^{-x}) d x \\ & =(5-x) e^{-x}+(-e^{-x}) \\ & =(4-x) e^{-x} \end{aligned} $

Therefore, equation (1) becomes:

$$ \begin{align*} & y e^{-x}=(4-x) e^{-x}+C \\ & \Rightarrow y=4-x+C e^{x} \\ & \Rightarrow x+y-4=C e^{x} \tag{2} \end{align*} $$

The curve passes through point $(0,2)$.

Therefore, equation ( 2 ) becomes:

$0+2-4=Ce^{0}$

$\Rightarrow-2=C$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation (2), we get:

$ \begin{aligned} & x+y-4=-2 e^{x} \\ & \Rightarrow y=4-x-2 e^{x} \end{aligned} $

This is the required equation of the curve.

Solution

The given differential equation is:

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

$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=-\frac{1}{x}$ and $.Q=2 x)$

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore$ I.F $=e^{\int \frac{1}{x} d x}=e^{-\log x}=e^{\log (x^{-1})}=x^{-1}=\frac{1}{x}$

Hence, the correct answer is $C$.

Solution

The given differential equation is:

$(1-y^{2}) \frac{d x}{d y}+y x=a y$

$\Rightarrow \frac{d y}{d x}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p y=Q(.$ where $p=\frac{y}{1-y^{2}}$ and $Q=\frac{a y}{1-y^{2}}$ )

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore I . F=e^{\int p d y}=e^{\int \frac{y}{1-y^{2}} d y}=e^{-\frac{1}{2} \log (1-y^{2})}=e^{\log [\frac{1}{\sqrt{1-y^{2}}}]}=\frac{1}{\sqrt{1-y^{2}}}$

Hence, the correct answer is D.

Solution

(i) The differential equation is given as:

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}+5 x(\frac{d y}{d x})^{2}-6 y=\log x \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}+5 x(\frac{d y}{d x})^{2}-6 y-\log x=0 \end{aligned} $

The highest order derivative present in the differential equation is $\frac{d^{2} y}{d x^{2}}$. Thus, its order is two. The highest power raised to $\frac{d^{2} y}{d x^{2}}$ is one. Hence, its degree is one.

(ii) The differential equation is given as:

$ \begin{aligned} & (\frac{d y}{d x})^{3}-4(\frac{d y}{d x})^{2}+7 y=\sin x \\ & \Rightarrow(\frac{d y}{d x})^{3}-4(\frac{d y}{d x})^{2}+7 y-\sin x=0 \end{aligned} $

The highest order derivative present in the differential equation is $\frac{d y}{d x}$. Thus, its order is one. The highest power raised to $\frac{d y}{d x}$ is three. Hence, its degree is three.

(iii) The differential equation is given as:

$\frac{d^{4} y}{d x^{4}}-\sin (\frac{d^{3} y}{d x^{3}})=0$

The highest order derivative present in the differential equation is $\frac{d^{4} y}{d x^{4}}$. Thus, its order is four.

However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

Solution

(i) $y=a e^{x}+b e^{-x}+x^{2}$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=a \frac{d}{d x}(e^{x})+b \frac{d}{d x}(e^{-x})+\frac{d}{d x}(x^{2})$ $\Rightarrow \frac{d y}{d x}=a e^{x}-b e^{-x}+2 x$

Again, differentiating both sides with respect to $x$, we get:

$\frac{d^{2} y}{d x^{2}}=a e^{x}+b e^{-x}+2$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S.

$x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2$

$=x(a e^{x}+b e^{-x}+2)+2(a e^{x}-b e^{-x}+2 x)-x(a e^{x}+b e^{-x}+x^{2})+x^{2}-2$

$=(a x e^{x}+b x e^{-x}+2 x)+(2 a e^{x}-2 b e^{-x}+4 x)-(a x e^{x}+b x e^{-x}+x^{3})+x^{2}-2$

$=2 a e^{x}-2 b e^{-x}+x^{2}+6 x-2$

$\neq 0$

$\Rightarrow$ L.H.S. $\neq$ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii) $y=e^{x}(a \cos x+b \sin x)=a e^{x} \cos x+b e^{x} \sin x$

Differentiating both sides with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=a \cdot \frac{d}{d x}(e^{x} \cos x)+b \cdot \frac{d}{d x}(e^{x} \sin x) \\ & \Rightarrow \frac{d y}{d x}=a(e^{x} \cos x-e^{x} \sin x)+b \cdot(e^{x} \sin x+e^{x} \cos x) \\ & \Rightarrow \frac{d y}{d x}=(a+b) e^{x} \cos x+(b-a) e^{x} \sin x \end{aligned} $

Again, differentiating both sides with respect to $x$, we get:

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}=(a+b) \cdot \frac{d}{d x}(e^{x} \cos x)+(b-a) \frac{d}{d x}(e^{x} \sin x) \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=(a+b) \cdot[e^{x} \cos x-e^{x} \sin x]+(b-a)[e^{x} \sin x+e^{x} \cos x] \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}[(a+b)(\cos x-\sin x)+(b-a)(\sin x+\cos x)] \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}[a \cos x-a \sin x+b \cos x-b \sin x+b \sin x+b \cos x-a \sin x-a \cos x] \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=[2 e^{x}(b \cos x-a \sin x)] \end{aligned} $

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get:

$ \begin{aligned} & \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y \\ & =2 e^{x}(b \cos x-a \sin x)-2 e^{x}[(a+b) \cos x+(b-a) \sin x]+2 e^{x}(a \cos x+b \sin x) \\ & =e^{x} \begin{bmatrix} (2 b \cos x-2 a \sin x)-(2 a \cos x+2 b \cos x) \\ -(2 b \sin x-2 a \sin x)+(2 a \cos x+2 b \sin x) \end{bmatrix} \\ & =e^{x}[(2 b-2 a-2 b+2 a) \cos x]+e^{x}[(-2 a-2 b+2 a+2 b) \sin x] \\ & =0 \end{aligned} $

Hence, the given function is a solution of the corresponding differential equation.

(iii) $y=x \sin 3 x$

Differentiating both sides with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(x \sin 3 x)=\sin 3 x+x \cdot \cos 3 x \cdot 3 \\ & \Rightarrow \frac{d y}{d x}=\sin 3 x+3 x \cos 3 x \end{aligned} $

Again, differentiating both sides with respect to $x$, we get: $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\sin 3 x)+3 \frac{d}{d x}(x \cos 3 x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=3 \cos 3 x+3[\cos 3 x+x(-\sin 3 x) \cdot 3]$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=6 \cos 3 x-9 x \sin 3 x$

Substituting the value of $\frac{d^{2} y}{d x^{2}}$ in the L.H.S. of the given differential equation, we get:

$\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x$

$=(6 \cdot \cos 3 x-9 x \sin 3 x)+9 x \sin 3 x-6 \cos 3 x$

$=0$

Hence, the given function is a solution of the corresponding differential equation.

(iv) $x^{2}=2 y^{2} \log y$

Differentiating both sides with respect to $x$, we get:

$2 x=2 \cdot \frac{d}{d x}=[y^{2} \log y]$

$\Rightarrow x=[2 y \cdot \log y \cdot \frac{d y}{d x}+y^{2} \cdot \frac{1}{y} \cdot \frac{d y}{d x}]$

$\Rightarrow x=\frac{d y}{d x}(2 y \log y+y)$

$\Rightarrow \frac{d y}{d x}=\frac{x}{y(1+2 \log y)}$

Substituting the value of $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get: $(x^{2}+y^{2}) \frac{d y}{d x}-x y$

$=(2 y^{2} \log y+y^{2}) \cdot \frac{x}{y(1+2 \log y)}-x y$

$=y^{2}(1+2 \log y) \cdot \frac{x}{y(1+2 \log y)}-x y$

$=x y-x y$

$=0$

Hence, the given function is a solution of the corresponding differential equation.

Solution

$(x^{3}-3 x y^{2}) d x=(y^{3}-3 x^{2} y) d y$

$\Rightarrow \frac{d y}{d x}=\frac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}$

This is a homogeneous equation. To simplify it, we need to make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d v}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x^{3}-3 x(v x)^{2}}{(v x)^{3}-3 x^{2}(v x)}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{1-3 v^{2}}{v^{3}-3 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1-3 v^{2}}{v^{3}-3 v}-v$

$\Rightarrow x \frac{d v}{d x}=\frac{1-3 v^{2}-v(v^{3}-3 v)}{v^{3}-3 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1-v^{4}}{v^{3}-3 v}$

$\Rightarrow(\frac{v^{3}-3 v}{1-v^{4}}) d v=\frac{d x}{x}$

Integrating both sides, we get:

$\int(\frac{v^{3}-3 v}{1-v^{4}}) d v=\log x+\log C^{\prime}$

Now, $\int(\frac{v^{3}-3 v}{1-v^{4}}) d v=\int \frac{v^{3} d v}{1-v^{4}}-3 \int \frac{v d v}{1-v^{4}}$

$\Rightarrow \int(\frac{v^{3}-3 v}{1-v^{4}}) d v=I_1-3 I_2$, where $I_1=\int \frac{v^{3} d v}{1-v^{4}}$ and $I_2=\int \frac{v d v}{1-v^{4}}$

Let $1-v^{4}=t$.

$\therefore \frac{d}{d v}(1-v^{4})=\frac{d t}{d v}$

$\Rightarrow-4 v^{3}=\frac{d t}{d v}$

$\Rightarrow v^{3} d v=-\frac{d t}{4}$

Now, $I_1=\int \frac{-d t}{4 t}=-\frac{1}{4} \log t=-\frac{1}{4} \log (1-v^{4})$

And, $I_2=\int \frac{v d v}{1-v^{4}}=\int \frac{v d v}{1-(v^{2})^{2}}$

Let $v^{2}=p$.

$\therefore \frac{d}{d v}(v^{2})=\frac{d p}{d v}$

$\Rightarrow 2 v=\frac{d p}{d v}$

$\Rightarrow v d v=\frac{d p}{2}$

$\Rightarrow I_2=\frac{1}{2} \int \frac{d p}{1-p^{2}}=\frac{1}{2 \times 2} \log |\frac{1+p}{1-p}|=\frac{1}{4} \log |\frac{1+v^{2}}{1-v^{2}}|$

Substituting the values of $I_1$ and $I_2$ in equation (3), we get:

$\int(\frac{v^{3}-3 v}{1-v^{4}}) d v=-\frac{1}{4} \log (1-v^{4})-\frac{3}{4} \log |\frac{1-v^{2}}{1+v^{2}}|$

Therefore, equation (2) becomes:

$ \begin{aligned} & \frac{1}{4} \log (1-v^{4})-\frac{3}{4} \log |\frac{1+v^{2}}{1-v^{2}}|=\log x+\log C^{\prime} \\ & \Rightarrow-\frac{1}{4} \log [(1-v^{4})(\frac{1+v^{2}}{1-v^{2}})^{3}]=\log C^{\prime} x \\ & \Rightarrow \frac{(1+v^{2})^{4}}{(1-v^{2})^{2}}=(C^{\prime} x)^{-4} \\ & \Rightarrow \frac{(1+\frac{y^{2}}{x^{2}})^{4}}{(1-\frac{y^{2}}{x^{2}})^{2}}=\frac{1}{C^{\prime 4} x^{4}} \\ & \Rightarrow \frac{(x^{2}+y^{2})^{4}}{x^{4}(x^{2}-y^{2})^{2}}=\frac{1}{C^{\prime 4} x^{4}} \\ & \Rightarrow(x^{2}-y^{2})^{2}=C^{\prime 4}(x^{2}+y^{2})^{4} \\ & \Rightarrow(x^{2}-y^{2})=C^{\prime 2}(x^{2}+y^{2})^{2} \\ & \Rightarrow x^{2}-y^{2}=C(x^{2}+y^{2})^{2}, \text{ where } C=C^{\prime 2} \end{aligned} $

Hence, the given result is proved.

Solution

$\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\frac{-d x}{\sqrt{1-x^{2}}}$

Integrating both sides, we get:

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

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

Solution

$ \begin{aligned} & \frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0 \\ & \Rightarrow \frac{d y}{d x}=-\frac{(y^{2}+y+1)}{x^{2}+x+1} \\ & \Rightarrow \frac{d y}{y^{2}+y+1}=\frac{-d x}{x^{2}+x+1} \\ & \Rightarrow \frac{d y}{y^{2}+y+1}+\frac{d x}{x^{2}+x+1}=0 \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & \int \frac{d y}{y^{2}+y+1}+\int \frac{d x}{x^{2}+x+1}=C \\ & \Rightarrow \int \frac{d y}{(y+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}+\int \frac{d x}{(x+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}=C \\ & \Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}[\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}]+\frac{2}{\sqrt{3}} \tan ^{-1}[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}]=C \\ & \Rightarrow \tan ^{-1}[\frac{2 y+1}{\sqrt{3}}]+\tan ^{-1}[\frac{2 x+1}{\sqrt{3}}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{(2 y+1)}{\sqrt{3}} \cdot \frac{(2 x+1)}{\sqrt{3}}}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{\frac{2 x+2 y+2}{\sqrt{3}}}{1-(\frac{4 x y+2 x+2 y+1}{3})}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \tan ^{-1}[\frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}]=\frac{\sqrt{3} C}{2} \\ & \Rightarrow \frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}=\tan (\frac{\sqrt{3} C}{2})=B, \text{ where } B=\tan (\frac{\sqrt{3} C}{2}) \\ & \Rightarrow x+y+1=\frac{2 B}{\sqrt{3}}(1-x y-2 x y) \\ & \Rightarrow x+y+1=A(1-x-y-2 x y) \text{, where } A=\frac{2 B}{\sqrt{3}} \end{aligned} $

Hence, the given result is proved.

Solution

The differential equation of the given curve is:

$\sin x \cos y d x+\cos x \sin y d y=0$

$\Rightarrow \frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0$

$\Rightarrow \tan x d x+\tan y d y=0$

Integrating both sides, we get:

$\log (\sec x)+\log (\sec y)=\log C$

$\log (\sec x \cdot \sec y)=\log C$

$\Rightarrow \sec x \cdot \sec y=C$

The curve passes through point $(0, \frac{\pi}{4})$.

$\therefore 1 \times \sqrt{2}=C$

$\Rightarrow C=\sqrt{2}$

On substituting in equation (1), we get:

$\sec x \cdot \sec y=\sqrt{2}$

$\Rightarrow \sec x \cdot \frac{1}{\cos y}=\sqrt{2}$

$\Rightarrow \cos y=\frac{\sec x}{\sqrt{2}}$

Hence, the required equation of the curve is $\cos y=\frac{\sec x}{\sqrt{2}}$.

Solution

$(1+e^{2 x}) d y+(1+y^{2}) e^{x} d x=0$

$\Rightarrow \frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}=0$

Integrating both sides, we get:

$\tan ^{-1} y+\int \frac{e^{x} d x}{1+e^{2 x}}=C$

Let $e^{x}=t \Rightarrow e^{2 x}=t^{2}$.

$\Rightarrow \frac{d}{d x}(e^{x})=\frac{d t}{d x}$

$\Rightarrow e^{x}=\frac{d t}{d x}$

$\Rightarrow e^{x} d x=d t$

Substituting these values in equation (1), we get:

$\tan ^{-1} y+\int \frac{d t}{1+t^{2}}=C$

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

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

Now, $y=1$ at $x=0$.

Therefore, equation (2) becomes:

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

$\Rightarrow \frac{\pi}{4}+\frac{\pi}{4}=C$

$\Rightarrow C=\frac{\pi}{2}$

Substituting $C=\frac{\pi}{2}$ in equation (2), we get:

$\tan ^{-1} y+\tan ^{-1}(e^{x})=\frac{\pi}{2}$

This is the required particular solution of the given differential equation.

Solution

$y e^{\frac{x}{y}} d x=(x e^{\frac{x}{y}}+y^{2}) d y$

$\Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y}=x e^{\frac{x}{y}}+y^{2}$

$\Rightarrow e^{\frac{x}{y}}[y \cdot \frac{d x}{d y}-x]=y^{2}$

$\Rightarrow e^{\frac{x}{y}} \frac{[y \cdot \frac{d x}{d y}-x]}{y^{2}}=1$

Let $e^{\frac{x}{y}}=z$.

Differentiating it with respect to $y$, we get:

$\frac{d}{d y}(e^{\frac{x}{y}})=\frac{d z}{d y}$

$\Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}(\frac{x}{y})=\frac{d z}{d y}$

$\Rightarrow e^{\frac{x}{y}} \cdot[\frac{y \cdot \frac{d x}{d y}-x}{y^{2}}]=\frac{d z}{d y}$

From equation (1) and equation (2), we get:

$\frac{d z}{d y}=1$

$\Rightarrow d z=d y$

Integrating both sides, we get: $z=y+C$

$\Rightarrow e^{\frac{x}{y}}=y+C$

Solution

$(x-y)(d x+d y)=d x-d y$

$\Rightarrow(x-y+1) d y=(1-x+y) d x$

$\Rightarrow \frac{d y}{d x}=\frac{1-x+y}{x-y+1}$

$\Rightarrow \frac{d y}{d x}=\frac{1-(x-y)}{1+(x-y)}$

Let $x-y=t$.

$\Rightarrow \frac{d}{d x}(x-y)=\frac{d t}{d x}$

$\Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow 1-\frac{d t}{d x}=\frac{d y}{d x}$

Substituting the values of $x-y$ and $\frac{d y}{d x}$ in equation (1), we get: $1-\frac{d t}{d x}=\frac{1-t}{1+t}$

$\Rightarrow \frac{d t}{d x}=1-(\frac{1-t}{1+t})$

$\Rightarrow \frac{d t}{d x}=\frac{(1+t)-(1-t)}{1+t}$

$\Rightarrow \frac{d t}{d x}=\frac{2 t}{1+t}$

$\Rightarrow(\frac{1+t}{t}) d t=2 d x$

$\Rightarrow(1+\frac{1}{t}) d t=2 d x$

Integrating both sides, we get:

$t+\log |t|=2 x+C$

$\Rightarrow(x-y)+\log |x-y|=2 x+C$

$\Rightarrow \log |x-y|=x+y+C$

Now, $y=-1$ at $x=0$.

Therefore, equation (3) becomes:

$\log 1=0-1+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (3) we get:

$\log |x-y|=x+y+1$

This is the required particular solution of the given differential equation.

Solution

$[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}] \frac{d x}{d y}=1$

$\Rightarrow \frac{d y}{d x}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$

$\Rightarrow \frac{d y}{d x}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$

This equation is a linear differential equation of the form

$\frac{d y}{d x}+P y=Q$, where $P=\frac{1}{\sqrt{x}}$ and $Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$.

Now, I.F $=e^{\int P d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}$

The general solution of the given differential equation is given by,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y e^{2 \sqrt{x}}=\int(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \times e^{2 \sqrt{x}}) d x+C$

$\Rightarrow y e^{2 \sqrt{x}}=\int \frac{1}{\sqrt{x}} d x+C$

$\Rightarrow y e^{2 \sqrt{x}}=2 \sqrt{x}+C$

Solution

The given differential equation is:

$\frac{d y}{d x}+y \cot x=4 x cosec x$

This equation is a linear differential equation of the form $\frac{d y}{d x}+p y=Q$, where $p=\cot x$ and $Q=4 x cosec x$.

Now, I.F $=e^{\int p p d x}=e^{\int \cot x d x}=e^{\log |\sin x|}=\sin x$

The general solution of the given differential equation is given by,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y \sin x=\int(4 x cosec x \cdot \sin x) d x+C$

$\Rightarrow y \sin x=4 \int x d x+C$

$\Rightarrow y \sin x=4 \cdot \frac{x^{2}}{2}+C$

$\Rightarrow y \sin x=2 x^{2}+C$

Now, $y=0$ at $x=\frac{\pi}{2}$.

Therefore, equation (1) becomes:

$0=2 \times \frac{\pi^{2}}{4}+C$

$\Rightarrow C=-\frac{\pi^{2}}{2}$

Substituting $C=-\frac{\pi^{2}}{2}$ in equation (1), we get:

$y \sin x=2 x^{2}-\frac{\pi^{2}}{2}$

This is the required particular solution of the given differential equation.

Solution

$ \begin{aligned} & (x+1) \frac{d y}{d x}=2 e^{-y}-1 \\ & \Rightarrow \frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1} \\ & \Rightarrow \frac{e^{y} d y}{2-e^{y}}=\frac{d x}{x+1} \end{aligned} $

Integrating both sides, we get:

$$ \begin{equation*} \int \frac{e^{y} d y}{2-e^{y}}=\log |x+1|+\log C \tag{1} \end{equation*} $$

Let $2-e^{y}=t$.

$\therefore \frac{d}{d y}(2-e^{y})=\frac{d t}{d y}$

$\Rightarrow-e^{y}=\frac{d t}{d y}$

$\Rightarrow e^{y} d t=-d t$

Substituting this value in equation (1), we get:

$$ \begin{align*} & \int \frac{-d t}{t}=\log |x+1|+\log C \\ & \Rightarrow-\log |t|=\log |C(x+1)| \\ & \Rightarrow-\log |2-e^{y}|=\log |C(x+1)| \\ & \Rightarrow \frac{1}{2-e^{y}}=C(x+1) \\ & \Rightarrow 2-e^{y}=\frac{1}{C(x+1)} \tag{2} \end{align*} $$

Now, at $x=0$ and $y=0$, equation (2) becomes:

$\Rightarrow 2-1=\frac{1}{C}$

$\Rightarrow C=1$

Substituting $C=1$ in equation (2), we get: $2-e^{y}=\frac{1}{x+1}$

$\Rightarrow e^{y}=2-\frac{1}{x+1}$

$\Rightarrow e^{y}=\frac{2 x+2-1}{x+1}$

$\Rightarrow e^{y}=\frac{2 x+1}{x+1}$

$\Rightarrow y=\log |\frac{2 x+1}{x+1}|,(x \neq-1)$

This is the required particular solution of the given differential equation.

Solution

The given differential equation is:

$ \begin{aligned} & \frac{y d x-x d y}{y}=0 \\ & \Rightarrow \frac{y d x-x d y}{x y}=0 \\ & \Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0 \end{aligned} $

Integrating both sides, we get:

$\log |x|-\log |y|=\log k$

$\Rightarrow \log |\frac{x}{y}|=\log k$

$\Rightarrow \frac{x}{y}=k$

$\Rightarrow y=\frac{1}{k} x$

$\Rightarrow y=C x$ where $C=\frac{1}{k}$

Hence, the correct answer is C.

Solution

The integrating factor of the given differential equation $\frac{d x}{d y}+P_1 x=Q_1$ is $e^{\int P_P d y}$.

The general solution of the differential equation is given by,

$ \begin{aligned} & x(\text{ I.F. })=\int(Q \times \text{ I.F. }) d y+C \\ & \Rightarrow x \cdot e^{\int P_1 d y}=\int(Q_1 e^{\int P_1 d y}) d y+C \end{aligned} $

Hence, the correct answer is C.

Solution

The given differential equation is:

$ \begin{aligned} & e^{x} d y+(y e^{x}+2 x) d x=0 \\ & \Rightarrow e^{x} \frac{d y}{d x}+y e^{x}+2 x=0 \\ & \Rightarrow \frac{d y}{d x}+y=-2 x e^{-x} \end{aligned} $

This is a linear differential equation of the form

$\frac{d y}{d x}+P y=Q$, where $P=1$ and $Q=-2 x e^{-x}$.

Now, I.F $=e^{\int P d t}=e^{\int d x}=e^{x}$

The general solution of the given differential equation is given by,

$ \begin{aligned} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{x}=\int(-2 x e^{-x} \cdot e^{x}) d x+C \\ & \Rightarrow y e^{x}=-\int 2 x d x+C \\ & \Rightarrow y e^{x}=-x^{2}+C \\ & \Rightarrow y e^{x}+x^{2}=C \end{aligned} $

Hence, the correct answer is $C$.