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 $C$.