Chapter 9: Differential Equations

Ans: Rewrite the equation in standard form:

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

Checking for homogeneity:

$F\left( x,y \right)=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

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

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

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

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

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

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

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

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

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

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

Separate the differentials:

$\frac{1+v}{1-v}dv=\frac{dx}{x}$ 

Integrate both side:

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

$\Rightarrow \int \frac{2-\left( 1-v \right)}{1-v}dv=\log x-\log k$ 

$\Rightarrow \int \frac{2}{1-v}dv-\int \frac{1-v}{1-v}dv=\log \frac{x}{k}$ 

$\Rightarrow -2\log \left( 1-v \right)-\int dv=\log \frac{x}{k}$ 

$\Rightarrow -2\log \left( 1-v \right)-v=\log \frac{x}{k}$ 

$\Rightarrow v=-\log \frac{x}{k}-2\log \left( 1-v \right)$ 

$\Rightarrow v=\log \left( \frac{k}{x{{\left( 1-v \right)}^{2}}} \right)$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

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

$\Rightarrow \frac{y}{x}=\log \left( \frac{kx}{{{\left( x-y \right)}^{2}}} \right)$ 

$\Rightarrow {{e}^{\frac{y}{x}}}=\frac{kx}{{{\left( x-y \right)}^{2}}}$

$\Rightarrow {{\left( x-y \right)}^{2}}=kx{{e}^{-\frac{y}{x}}}$  

 The solution of the given differential equation ${{\left( \text{x-y} \right)}^{\text{2}}}\text{=kx}{{\text{e}}^{\text{-}\frac{\text{y}}{\text{x}}}}$ .

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{x+y}{x}$ 

Checking for homogeneity:

.$F\left( x,y \right)=\frac{x+y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda x+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( x+y \right)}{\lambda \left( x \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{x+y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{x+\left( vx \right)}{x}$ 

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

$\Rightarrow v+x\frac{dv}{dx}=1+v$ 

$\Rightarrow x\frac{dv}{dx}=1$ 

Separate the differentials:

$dv=\frac{dx}{x}$ 

Integrate both side:

$\int dv=\int \frac{dx}{x}$ 

$\Rightarrow \int dv=\log x+\log k$ 

$\Rightarrow v=\log kx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{y}{x}=\log kx$ 

$\Rightarrow y=x\log kx$  

 The solution of the given differential equation $\text{y=x log kx}$ 

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{x+y}{x-y}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{x+y}{x-y}$ 

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

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

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

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{x+\left( vx \right)}{x-vx}$ 

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

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

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

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

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

Separate the differentials:

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

Integrate both side:

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

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

$\Rightarrow \int \frac{1}{1+{{v}^{2}}}dv-\int \frac{v}{1+{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v-\frac{1}{2}\int \frac{2v}{1+{{v}^{2}}}dv=\log x+C$ 

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

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

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

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

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

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

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

The solution of the given differential equation $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{y}}{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{log}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{+C}$.

Ans: Rewrite the equation in standard form:

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

Checking for homogeneity:

$F\left( x,y \right)=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

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

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

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

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

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

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

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

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

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

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

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

Separate the differentials:

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

Integrate both side:

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

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

$\Rightarrow \log \left( 1+{{v}^{2}} \right)+\log x=C$ 

$\Rightarrow \log \left[ x\left( 1+{{v}^{2}} \right) \right]=C$

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

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

$\Rightarrow \left( \frac{{{x}^{2}}+{{y}^{2}}}{x} \right)=k\,\,k={{e}^{C}}$  

$\Rightarrow {{x}^{2}}+{{y}^{2}}=kx$ 

 The solution of the given differential equation ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=kx}$.

Ans: Rewrite the equation in standard form:

\[\frac{dy}{dx}=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}\] 

Checking for homogeneity:

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

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

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

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

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

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

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

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

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

Separate the differentials:

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

Integrate both side:

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

$\Rightarrow \frac{1}{2}\int \frac{1}{\frac{1}{2}-{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow \frac{1}{2}\left( \frac{1}{2\times \frac{1}{\sqrt{2}}} \right)\log \frac{\left| \frac{1}{\sqrt{2}}+v \right|}{\left| \frac{1}{\sqrt{2}}-v \right|}=\log x+C$

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| 1+\sqrt{2}v \right|}{\left| 1-\sqrt{2}v \right|}=\log x+C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| 1+\sqrt{2}\left( \frac{y}{x} \right) \right|}{\left| 1-\sqrt{2}\left( \frac{y}{x} \right) \right|}=\log x+C$ 

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

 The solution of the given differential equation $\frac{\text{1}}{\text{2}\sqrt{\text{2}}}\text{log}\frac{\left| \text{x+}\sqrt{\text{2}}\text{y} \right|}{\left| \text{x-}\sqrt{\text{2}}\text{y} \right|}\text{=logx+C}$.

Ans: Rewrite the equation in standard form:

$xdy=\sqrt{{{x}^{2}}+{{y}^{2}}}dx+ydx$ 

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

Checking for homogeneity:

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

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

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

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

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

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{\sqrt{{{x}^{2}}+{{\left( vx \right)}^{2}}}+vx}{x}$ 

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

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

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

Separate the differentials:

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

Integrate both side:

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

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

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

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

$\Rightarrow \log \left| \frac{y+\sqrt{{{y}^{2}}+{{x}^{2}}}}{x} \right|=\log Cx$ 

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

Ans: Rewrite the equation in standard form:

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

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

Checking for homogeneity:

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

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\left\{ \lambda x\cos \left( \frac{\lambda y}{\lambda x} \right)+y\sin \left( \frac{\lambda y}{\lambda x} \right) \right\}\lambda y}{\left\{ \lambda y\sin \left( \frac{\lambda y}{\lambda x} \right)-\lambda x\cos \left( \frac{\lambda y}{\lambda x} \right) \right\}\lambda x}$ 

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

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{\left\{ x\cos \left( \frac{xv}{x} \right)+vx\sin \left( \frac{vx}{x} \right) \right\}vx}{\left\{ vx\sin \left( \frac{vx}{x} \right)-x\cos \left( \frac{vx}{x} \right) \right\}x}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left\{ \cos v+v\sin v \right\}v}{{{x}^{2}}\left\{ v\sin v-\cos v \right\}}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v}{v\sin v-\cos v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v}{v\sin v-\cos v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v-{{v}^{2}}\sin v+v\cos v}{v\sin v-\cos v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}$ 

Separate the differentials:

$\frac{v\sin v-\cos v}{v\cos v}dv=2\frac{dx}{x}$ 

Integrate both side:

$\int \frac{v\sin v-\cos v}{v\cos v}dv=2\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{v\sin v}{v\cos v}dv-\int \frac{\cos v}{v\cos v}dv=2\int \frac{dx}{x}$

$\Rightarrow \int \tan vdv-\int \frac{1}{v}dv=2\log \left| x \right|+\log C$ 

$\Rightarrow \log \left| \sec v \right|-\log \left| v \right|=\log C{{\left| x \right|}^{2}}$

$\Rightarrow \log \frac{\left| \sec v \right|}{\left| v \right|}=\log C{{\left| x \right|}^{2}}$ 

$\Rightarrow \sec v=Cv{{x}^{2}}$   

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

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

$\Rightarrow \cos \frac{y}{x}=\frac{k}{xy}\,\,k=\frac{1}{C}$ 

The solution of the given differential equation $\text{cos}\frac{\text{y}}{\text{x}}\text{=}\frac{\text{k}}{\text{xy}}$.

Ans: Rewrite the equation in standard form:

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

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

Checking for homogeneity:

$F\left( x,y \right)=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y-\lambda x\sin \left( \frac{\lambda y}{\lambda x} \right)}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( y-x\sin \left( \frac{y}{x} \right) \right)}{\lambda x}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{vx-x\sin \left( \frac{vx}{x} \right)}{x}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( v-\sin v \right)}{x}$ 

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

$\Rightarrow x\frac{dv}{dx}=-\sin v$ 

Separate the differentials:

$\frac{1}{\sin v}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \text{cosec}\,vdv=-\int \frac{dx}{x}$ 

$\Rightarrow \log \left| \text{cosec}\,v-\cot v \right|=-\log x+\log C$

$\Rightarrow \log \left| \text{cosec}\,v-\cot v \right|=\log \frac{C}{x}$ 

$\Rightarrow \text{cosec}\,v-\cot v=\frac{C}{x}$

$\Rightarrow \frac{1}{\sin v}-\frac{\cos v}{\sin v}=\frac{C}{x}$ 

$\Rightarrow 1-\cos v=\frac{C}{x}\sin v$   

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

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

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

The solution of the given differential equation $\text{x}\left( \text{1-cos}\frac{\text{y}}{\text{x}} \right)\text{=Csin}\left( \frac{\text{y}}{\text{x}} \right)$.

Ans: Rewrite the equation in standard form:

$ydx=2xdy-x\log \left( \frac{y}{x} \right)dy$ 

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

Checking for homogeneity:

$F\left( x,y \right)=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{2\lambda x-\lambda x\log \left( \frac{\lambda y}{\lambda x} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{\lambda \left( 2x-x\log \left( \frac{y}{x} \right) \right)}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{vx}{2x-x\log \left( \frac{vx}{x} \right)}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{xv}{x\left( 2-\log \left( v \right) \right)}$ 

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

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

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

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

Separate the differentials:

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

Integrate both side:

$\int \frac{2-\log v}{v\log v-v}dv=\int \frac{dx}{x}$ 

$\int \frac{1+1-\log v}{v\left( \log v-v \right)}dv=\log x+\log C$

$\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv+\int \frac{1-\log v}{v\left( \log v-1 \right)}dv=\log x+\log C$ 

$\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv-\int \frac{1}{v}dv=\log Cx$ ……(1)

Solving :

$\int \frac{1}{v\left( \log v-1 \right)}dv$

Substitute $\text{log v-1=t}$:

$\log v-1=t$   

$\Rightarrow \frac{1}{v}dv=dt$   

Thus the integral will be:

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\int \frac{dt}{t}\] 

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\log t\] 

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\log \left( \log v-1 \right)\] 

Using above result for solving (1):

$\Rightarrow \log \left( \log v-1 \right)-\log v=\log Cx$ 

$\Rightarrow \log \frac{\log v-1}{v}=\log Cx$ 

$\Rightarrow \frac{\log v-1}{v}=Cx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{\log \frac{y}{x}-1}{\frac{y}{x}}=Cx$ 

$\Rightarrow \log \frac{y}{x}-1=Cx\left( \frac{y}{x} \right)$ 

$\Rightarrow \log \frac{y}{x}-1=Cy$ 

 The solution of the given differential equation $\text{log}\frac{\text{y}}{\text{x}}\text{-1=Cy}$.

Ans: Rewrite the equation in standard form:

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

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

Checking for homogeneity:

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

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{e}^{\frac{\lambda x}{\lambda y}}}\left( 1-\frac{\lambda x}{\lambda y} \right)}{\left( 1+{{e}^{\frac{\lambda x}{\lambda y}}} \right)}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{x=vy}$:

\[\frac{d\left( vy \right)}{dy}=-\frac{{{e}^{\frac{vy}{y}}}\left( 1-\frac{vy}{y} \right)}{1+{{e}^{\frac{vy}{y}}}}\] 

$\Rightarrow v\frac{dy}{dy}+y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}$ 

$\Rightarrow v+y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}$ 

$\Rightarrow y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}-v$

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

$\Rightarrow y\frac{dv}{dy}=\frac{-\left( {{e}^{v}}+v \right)}{1+{{e}^{v}}}$  

Separate the differentials:

$\frac{1+{{e}^{v}}}{{{e}^{v}}+v}dv=-\frac{dy}{y}$ 

Integrate both side:

$\int \frac{1+{{e}^{v}}}{{{e}^{v}}+v}dv=-\int \frac{dy}{y}$ 

\[\int \frac{{{e}^{v}}+1}{{{e}^{v}}+v}dv=-\log y+\log C\] ……(1)

Solving the LHS integral. Substitute ${{\text{e}}^{\text{v}}}\text{+v=t}$:

\[{{e}^{v}}+v=t\] 

$\Rightarrow \left( {{e}^{v}}+1 \right)dv=dt$ 

Solving the expression (1):

\[\Rightarrow \int \frac{1}{t}dt=\log \frac{C}{y}\] 

\[\Rightarrow \log \left( t \right)=\log \frac{C}{y}\]

\[\Rightarrow \log \left( {{e}^{v}}+v \right)=\log \frac{C}{y}\] 

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

Back substitute $\text{v=}\frac{\text{x}}{\text{y}}$:

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

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

The solution of the given differential equation \[\text{y}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{+x=C}\].