Chapter 9: Differential Equations

Ans: The given differential equation is:

 $\frac{{{d}^{2}}y}{d{{x}^{2}}}+5x{{\left( \frac{dy}{dx} \right)}^{2}}-6y-\log x=0$ 

The highest order derivative in the equation is of the term $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order of the equation is $\text{2}$ and its highest power is $\text{1}$. Therefore its degree is $\text{1}$.

Ans: The given differential equation is:

 ${{\left( \frac{dy}{dx} \right)}^{3}}-4{{\left( \frac{dy}{dx} \right)}^{2}}+7y-\sin x=0$ 

The highest order derivative in the equation is of the term ${{\left( \frac{\text{dy}}{\text{dx}} \right)}^{3}}$, thus the order of the equation is $1$ and its highest power is $3$. Therefore its degree is $3$.

Ans: The given differential equation is:

 $\frac{{{d}^{4}}y}{d{{x}^{4}}}-\sin \left( \frac{{{d}^{3}}y}{d{{x}^{3}}} \right)=0$ 

The highest order derivative in the equation is of the term $\frac{{{\text{d}}^{4}}\text{y}}{\text{d}{{\text{x}}^{4}}}$, thus the order of the equation is $4$.

As the differential equation is not polynomial in its derivative, therefore its degree is not defined.

Ans: The given function is:
$xy=a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}}$ 

Take derivative on both side:

$\Rightarrow y+x\frac{dy}{dx}=a{{e}^{x}}-b{{e}^{-x}}+2x$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}+x\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}+2$ 

$\Rightarrow x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}+2$ ……(1)

The given differential equation is:
$x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}-xy+{{x}^{2}}-2=0$ 

Solving LHS:

Substitute $\text{x}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+2}\frac{\text{dy}}{\text{dx}}$ from the result (1) and $\text{xy}$:

$\Rightarrow \left( x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx} \right)-xy+{{x}^{2}}-2$ 

$\Rightarrow \left( a{{e}^{x}}+b{{e}^{-x}}+2 \right)-\left( a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}} \right)+{{x}^{2}}-2$ 

$\Rightarrow 2-{{x}^{2}}+{{x}^{2}}-2$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

1. $\text{y=}{{\text{e}}^{\text{x}}}\left( \text{acosx+bsinx} \right)\,\,\text{:}\,\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-2}\frac{\text{dy}}{\text{dx}}\text{+2y=0}$ 

Ans: The given function is:
$y={{e}^{x}}\left( a\cos x+b\sin x \right)$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\left( a\cos x+b\sin x \right)+{{e}^{x}}\left( -a\sin x+b\cos x \right)$ 

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)$ 

Take derivative on both side:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)+{{e}^{x}}\left( -\left( a+b \right)\sin x+\left( b-a \right)\cos x \right)$

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( \left( a+b+b-a \right)\cos x+\left( b-a-a-b \right)\sin x \right)$ 

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( 2b\cos x-2a\sin x \right)$  

The given differential equation is:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0$ 

Solving LHS:

$\Rightarrow {{e}^{x}}\left( 2b\cos x-2a\sin x \right)-2{{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)+2y$ 

$\Rightarrow {{e}^{x}}\left( \left( 2b-2a-2b \right)\cos x+\left( -2a-2b+2a \right)\sin x \right)-2y$ 

$\Rightarrow {{e}^{x}}\left( -2a\cos x-2b\sin x \right)-2y$ 

$\Rightarrow -2{{e}^{x}}\left( a\cos x+b\sin x \right)-2y$

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

1. $\text{y=x}\,\text{sin3x}\,\,\text{:}\,\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+9y-6cos3x=0}$ 

Ans: The given function is:
$y=x\sin 3x$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}=\sin 3x+3x\cos 3x$ 

Take derivative on both side:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=3\cos 3x+3\left( \cos 3x+x\left( -3\sin 3x \right) \right)$

\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=3\cos 3x+3\cos 3x-9x\sin 3x\] 

\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=6\cos 3x-9x\sin 3x\]  

The given differential equation is:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x=0$ 

Solving LHS:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x$ 

$\Rightarrow \left( 6\cos 3x-9x\sin 3x \right)+9\left( x\sin 3x \right)-6\cos 3x$ 

$\Rightarrow 6\cos 3x-9x\sin 3x+9x\sin 3x-6\cos 3x$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

1. ${{\text{x}}^{\text{2}}}\text{=2}{{\text{y}}^{\text{2}}}\,\text{log y}\,\,\text{:}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{-xy=0}$ 

Ans: The given function is:
${{x}^{2}}=2{{y}^{2}}\log y$ 

Take derivative on both side:

\[\Rightarrow 2x=2\left( 2y\log y+{{y}^{2}}\left( \frac{1}{y} \right) \right)\frac{dy}{dx}\] 

\[\Rightarrow \frac{dy}{dx}=\frac{x}{\left( 2y\log y+y \right)}\]

Multiply numerator and denominator by $\text{y}$:

\[\Rightarrow \frac{dy}{dx}=\frac{xy}{\left( 2{{y}^{2}}\log y+{{y}^{2}} \right)}\] 

\[\Rightarrow \frac{dy}{dx}=\frac{xy}{\left( {{x}^{2}}+{{y}^{2}} \right)}\]  

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

Solving LHS:

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

$\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\left( \frac{xy}{{{x}^{2}}+{{y}^{2}}} \right)-xy$ 

$\Rightarrow xy-xy$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

Ans: Given differential equation:

$\left( {{x}^{3}}-3x{{y}^{2}} \right)dx=\left( {{y}^{3}}-3{{x}^{2}}y \right)dy$ 

$\Rightarrow \frac{dy}{dx}=\frac{{{x}^{3}}-3x{{y}^{2}}}{{{y}^{3}}-3{{x}^{2}}y}$ 

As it can be seen that this is an homogenous equation. Substitute $\text{y=vx}$:

$\Rightarrow \frac{d\left( vx \right)}{dx}=\frac{{{x}^{3}}-3x{{\left( vx \right)}^{2}}}{{{\left( vx \right)}^{3}}-3{{x}^{2}}\left( vx \right)}$ 

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

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

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

$\Rightarrow x\frac{dv}{dx}=\frac{1-3{{v}^{2}}-{{v}^{4}}+3{{v}^{2}}}{{{v}^{3}}-3v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-{{v}^{4}}}{{{v}^{3}}-3v}$ 

Separate the differentials:

$\frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\log x+\log C$ 

\[\Rightarrow I=\log x+\log C\,\,\left( I=\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv \right)\] ……(1)

Solving integral $\text{I}$:

\[\Rightarrow I=\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv\] 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{{{v}^{3}}-3v}{\left( 1-{{v}^{2}} \right)\left( 1+{{v}^{2}} \right)}\] 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{{{v}^{3}}-3v}{\left( 1-v \right)\left( 1+v \right)\left( 1+{{v}^{2}} \right)}\] 

Using partial fraction:

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{A}{1-v}+\frac{B}{1+v}+\frac{Cv+D}{1+{{v}^{2}}}\] 

Solving for $\text{A,B,C}\,\text{and}\,\text{D}$:

$A=-\frac{1}{2}$ 

$B=\frac{1}{2}$ 

$C=-2$ 

$D=0$ 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{-\frac{1}{2}}{1-v}+\frac{\frac{1}{2}}{1+v}+\frac{-2v+0}{1+{{v}^{2}}}\] 

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

$I=-\frac{1}{2}\left( -\log \left( 1-v \right) \right)+\frac{1}{2}\left( \log \left( 1+v \right) \right)-\log \left( 1+{{v}^{2}} \right)$ 

\[I=\frac{1}{2}\left( \log \left( 1-{{v}^{2}} \right) \right)-\frac{2}{2}\log \left( 1+{{v}^{2}} \right)\] 

\[I=\frac{1}{2}\left( \log \frac{\left( 1-{{v}^{2}} \right)}{{{\left( 1+{{v}^{2}} \right)}^{2}}} \right)\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( 1-\frac{{{y}^{2}}}{{{x}^{2}}} \right)}{{{\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right)}^{2}}} \right)\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{{{x}^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)\]

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\frac{1}{2}\log {{x}^{2}}\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\log x\] 

Back substitute $\text{I}$ in expression (1):

\[\Rightarrow I=\log x+\log C\,\] 

\[\Rightarrow \frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\log x=\log x+\log C\]

\[\Rightarrow \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}=2\log C\] 

\[\Rightarrow \frac{{{x}^{2}}-{{y}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}={{C}^{2}}\] 

\[\Rightarrow {{x}^{2}}-{{y}^{2}}=c{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}\,\,\left( c={{C}^{2}} \right)\] 

Thus for given differential equation, its general solution is \[{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}\].

Ans: The given differential equation is:

 $\frac{dy}{dx}+\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}=0$ 

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

$\Rightarrow \frac{dy}{\sqrt{1-{{y}^{2}}}}=-\frac{dx}{\sqrt{1-{{x}^{2}}}}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{\sqrt{1-{{y}^{2}}}}=-\int \frac{dx}{\sqrt{1-{{x}^{2}}}}$ 

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

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

Thus the general solution of given differential equation is $\text{si}{{\text{n}}^{\text{-1}}}\text{y+si}{{\text{n}}^{\text{-1}}}\text{x=C}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}=0$

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

$\Rightarrow \frac{dy}{dx}=-\frac{{{y}^{2}}+2\left( \frac{1}{2} \right)y+\frac{1}{4}-\frac{1}{4}+1}{{{x}^{2}}+2\left( \frac{1}{2} \right)x+\frac{1}{4}-\frac{1}{4}+1}$ 

$\Rightarrow \frac{dy}{dx}=-\frac{{{\left( y+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}{{{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}$   

$\Rightarrow \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}=-\frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}=-\int \frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \frac{1}{\left( \frac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left[ \frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]=-\frac{1}{\left( \frac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left[ \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]+C$ 

$\Rightarrow \frac{2}{\sqrt{3}}\left( {{\tan }^{-1}}\left[ \frac{2y+1}{\sqrt{3}} \right]+{{\tan }^{-1}}\left[ \frac{2x+1}{\sqrt{3}} \right] \right)=C$ 

$\Rightarrow {{\tan }^{-1}}\left[ \frac{2y+1}{\sqrt{3}} \right]+{{\tan }^{-1}}\left[ \frac{2x+1}{\sqrt{3}} \right]=\frac{\sqrt{3}}{2}C$ 

Thus the general solution for given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2y+1}}{\sqrt{\text{3}}} \right]\text{+ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2x+1}}{\sqrt{\text{3}}} \right]\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\text{C}$.

Ans: Given differential equation is:

$\sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy=0$

$\Rightarrow \sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy=0$

Divide both side by $\text{cos}\,\text{x}\,\text{cos}\,\text{y}$:

$\Rightarrow \frac{\sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy}{\cos x\cos y}=0$ 

$\Rightarrow \tan xdx+\tan ydy=0$ 

$\Rightarrow \tan ydy=-\tan xdx$ 

Integrate both side:

$\Rightarrow \int \tan ydy=-\int \tan xdx$ 

$\Rightarrow \log \left( \sec y \right)=-\log \left( \sec x \right)+C$ 

$\Rightarrow \log \left( \sec y \right)+\log \left( \sec x \right)=C$ 

$\Rightarrow \log \left( \sec x\sec y \right)=C$ 

$\Rightarrow \sec x\sec y=k\,\,\left( k={{e}^{C}} \right)$ 

As curve passes through $\left( \text{0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$:

$\sec 0\sec \left( \frac{\pi }{4} \right)=k$ 

$\Rightarrow k=\sqrt{2}$ 

$\Rightarrow \sec x\sec y=\sqrt{2}$ 

Thus the equation of required curve is $\text{sec}\,\text{x}\,\text{sec}\,\text{y=}\sqrt{\text{2}}$.

Ans: The given differential equation is:

$\left( 1+{{e}^{2x}} \right)dy+\left( 1+{{y}^{2}} \right){{e}^{x}}dx=0$ 

Divide both side $\left( \text{1+}{{\text{e}}^{\text{2x}}} \right)\left( \text{1+}{{\text{y}}^{\text{2}}} \right)$:

\[\frac{dy}{\left( 1+{{y}^{2}} \right)}+\frac{{{e}^{x}}}{\left( 1+{{e}^{2x}} \right)}dx=0\] 

\[\int \frac{dy}{\left( 1+{{y}^{2}} \right)}=-\int \frac{{{e}^{x}}}{\left( 1+{{e}^{2x}} \right)}dx\] 

\[{{\tan }^{-1}}y=-\int \frac{{{e}^{x}}}{\left( 1+{{\left( {{e}^{x}} \right)}^{2}} \right)}dx\]

Substitute $t={{e}^{x}}$:

$dt={{e}^{x}}dx$ 

\[\Rightarrow {{\tan }^{-1}}y=-\int \frac{1}{\left( 1+{{t}^{2}} \right)}dt\] 

\[\Rightarrow {{\tan }^{-1}}y=-{{\tan }^{-1}}t+C\] 

\[\Rightarrow {{\tan }^{-1}}y=-{{\tan }^{-1}}{{e}^{x}}+C\] 

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

As $\text{y=1}$ when $\text{x=0}$:

${{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( {{e}^{0}} \right)=C$ 

\[\Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\] 

\[\Rightarrow C=\frac{\pi }{2}\] 

$\Rightarrow {{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=\frac{\pi }{2}$ 

Thus the required particular solution is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y+ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{x}}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Ans: The given differential equation is:

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

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

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

$\Rightarrow {{e}^{\frac{x}{y}}}\frac{\left[ y\frac{dx}{dy}-x \right]}{{{y}^{2}}}=1$ 

Substitute $\text{z=}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}$:

$z={{e}^{\frac{x}{y}}}$ 

$\frac{d}{dy}z=\frac{d}{dy}{{e}^{\frac{x}{y}}}$ 

$\Rightarrow \frac{dz}{dy}=\frac{d}{dy}\left( {{e}^{\frac{x}{y}}} \right)$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\frac{d}{dy}\left( \frac{x}{y} \right)$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\left[ \left( \frac{1}{y} \right)\frac{dx}{dy}-\frac{x}{{{y}^{2}}} \right]$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\left[ \frac{y\frac{dx}{dy}-x}{{{y}^{2}}} \right]$ 

$\Rightarrow \frac{dz}{dy}=1$ 

\[\Rightarrow dz=dy\] 

\[\Rightarrow \int dz=\int dy\]

\[\Rightarrow z=y+C\] 

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

Thus the required general solution is \[{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{=y+C}\].

Ans: Given differential equation is:

$\left( x-y \right)\left( dx+dy \right)=dx-dy$ 

$\Rightarrow \left( x-y \right)dx-dx=\left( y-x \right)dy-dy$

$\Rightarrow \left( x-y+1 \right)dy=\left( 1-x+y \right)dx$ 

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

Put $\text{x-y=t}$:

$x-y=t$ 

$\Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}$ 

$\Rightarrow 1-\frac{dt}{dx}=\frac{dy}{dx}$ 

$\Rightarrow 1-\frac{dt}{dx}=\frac{1-t}{1+t}$ 

$\Rightarrow \frac{dt}{dx}=1-\frac{1-t}{1+t}$

$\Rightarrow \frac{dt}{dx}=\frac{1+t-1+t}{1+t}$ 

$\Rightarrow \frac{dt}{dx}=\frac{2t}{1+t}$ 

$\Rightarrow \frac{1+t}{t}dt=2dx$ 

Integrate both side:

$\Rightarrow \int \frac{1+t}{t}dt=2\int dx$

$\Rightarrow \int \frac{1}{t}dt+\int dt=2x+C$ 

$\Rightarrow \log \left| t \right|+t=2x+C$ 

$\Rightarrow \log \left| x-y \right|+x-y=2x+C$ 

$\Rightarrow \log \left| x-y \right|-y=x+C$ 

As $\text{y=-1}$ when $\text{x=0}$:

$\Rightarrow \log \left| 0-\left( -1 \right) \right|-\left( -1 \right)=0+C$ 

$\Rightarrow \log 1+1=C$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

$\log \left| x-y \right|-y=x+1$.

Ans: Given differential equation is:
$\left[ \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right]\frac{dx}{dy}=1$ 

$\Rightarrow \frac{dy}{dx}=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$ 

$\Rightarrow \frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}$ 

It is linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=\frac{1}{\sqrt{x}}$ 

$Q=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}$ 

Calculating integrating factor:

$I.F={{e}^{\int pdx}}$

$I.F={{e}^{\int \frac{1}{\sqrt{x}}dx}}$ 

$I.F={{e}^{2\sqrt{x}}}$  

The general solution is given by:

$y\times I.F=\int \left( Q\times I.F \right)dx+C$ 

$y\times \left( {{e}^{2\sqrt{x}}} \right)=\int \left( \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}\times {{e}^{2\sqrt{x}}} \right)dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=\int \left( \frac{{{e}^{-2\sqrt{x}+2\sqrt{x}}}}{\sqrt{x}} \right)dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=\int \frac{1}{\sqrt{x}}dx+C$ 

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

Thus the general solution for the given differential equation is

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