Chapter 9: Differential Equations

Ans: Given differential equation is:

$\left( x+y \right)dy+\left( x-y \right)dx=0$

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

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 $\text{y=vx}$:

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

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

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

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

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

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

Separate the differentials:

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

Integrate both side:

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

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

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

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

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

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

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

 Now $\text{y=1}$ and $\text{x=1}$:

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{1}^{2}}+{{1}^{2}}}{1} \right]+{{\tan }^{-1}}\frac{1}{1}=k$  

$k=\frac{1}{2}\log 2+\frac{\pi }{4}$ 

The required particular solution:

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{x}^{2}}+{{y}^{2}}}{x} \right]+{{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log 2+\frac{\pi }{4}$ 

Ans: Given differential equation is:

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

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

Checking for homogeneity:

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

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

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

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

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

Thus it is an homogenous equation.

Let $\text{y=vx}$:

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

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=-\frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}}\] 

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

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

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

Separate the differentials:

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

Integrate both side:

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

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

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

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

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

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

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

\[\Rightarrow \frac{v}{v+2}=\frac{{{C}^{2}}}{{{x}^{2}}}\] 

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

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

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

 Now $\text{y=1}$ and $\text{x=1}$:

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

\[\Rightarrow {{C}^{2}}=\frac{1}{3}\] 

The required particular solution:

\[\Rightarrow \frac{{{x}^{2}}y}{y+2x}=\frac{1}{3}\] 

\[\Rightarrow y+2x=3{{x}^{2}}y\] 

Ans: Given differential equation is:

$\left[ x{{\sin }^{2}}\left( \frac{x}{y} \right)-y \right]dx+xdy=0$

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

Checking for homogeneity:

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

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

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

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

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

Thus it is an homogenous equation.

Let $\text{y=vx}$:

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

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{-x{{\sin }^{2}}\left( v \right)+vx}{x}\] 

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

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

Separate the differentials:

$\text{cose}{{\text{c}}^{2}}vdv=-\frac{dx}{x}$ 

Integrate both side:

$\int \text{cose}{{\text{c}}^{2}}vdv=-\int \frac{dx}{x}$ 

$\Rightarrow -\cot v=-\log x-\log C$ 

$\Rightarrow \cot v=\log Cx$ 

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

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

 Now $\text{y=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ and $\text{x=1}$:

$\cot \frac{\frac{\pi }{4}}{1}=\log C\left( 1 \right)$  

\[\Rightarrow \log C=\cot \frac{\pi }{4}\] 

\[\Rightarrow \log C=1\] 

\[\Rightarrow C=e\] 

The required particular solution:

\[\Rightarrow \cot \frac{y}{x}=\log \left| ex \right|\] 

Ans: Given differential equation is:

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

\[\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)\] 

Checking for homogeneity:

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

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

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

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

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=v-\text{cosec}\left( v \right)$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=v-\text{cosec}\left( v \right)\] 

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

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

Separate the differentials:

$\sin vdv=-\frac{dx}{x}$ 

Integrate both side:

$\int \sin vdv=-\int \frac{dx}{x}$ 

\[\Rightarrow -\cos v=-\log \left| x \right|-\log C\] 

\[\cos v=\log \left| Cx \right|\] 

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

$\Rightarrow \cos \frac{y}{x}=\log \left| Cx \right|$ 

 Now $\text{y=0}$ and $\text{x=1}$:

$\Rightarrow \cos \frac{0}{1}=\log \left| C1 \right|$  

\[\Rightarrow \log C=1\] 

\[\Rightarrow C=e\] 

The required particular solution:

\[\Rightarrow \cos \frac{y}{x}=\log \left| ex \right|\].

Ans: Given differential equation is:

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

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

Checking for homogeneity:

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

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

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

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

Thus it is an homogenous equation.

Let $\text{y=vx}$:

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

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{2v+{{v}^{2}}}{2}\] 

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

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

Separate the differentials:

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

Integrate both side:

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

\[\Rightarrow \frac{{{v}^{-2+1}}}{-2+1}=\log \left| x \right|+C\] 

\[-\frac{2}{v}=\log \left| x \right|+C\] 

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

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

 Now $\text{y=2}$ and $\text{x=1}$:

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

$\Rightarrow C=-1$ 

The required particular solution:

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

$\Rightarrow y=\frac{2x}{1-\log \left| x \right|}\,\,\left( x\ne 0,e \right)$ 

Ans: The required substitution will be: 

$\frac{x}{y}=v$ 

$\Rightarrow x=vy$ 

The correct answer is (C).

Ans: For option (A):

$F\left( x,y \right)=\frac{3y+2x+4}{4x+6y+5}$

$F\left( \lambda x,\lambda y \right)=\frac{3\lambda y+2\lambda x+4}{4\lambda x+6\lambda y+5}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  

For option (B):

$F\left( x,y \right)=\frac{xy}{{{x}^{3}}+{{y}^{3}}}$

$F\left( \lambda x,\lambda y \right)=\frac{\left( \lambda x \right)\left( \lambda y \right)}{{{\left( \lambda x \right)}^{3}}+{{\left( \lambda y \right)}^{3}}}$ 

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

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  

For option (C):

$F\left( x,y \right)=-\frac{{{x}^{3}}+2{{y}^{2}}}{2xy}$

$F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{3}}{{x}^{3}}+2{{\lambda }^{2}}{{y}^{2}}}{2\left( \lambda x \right)\left( \lambda y \right)}$ 

$F\left( \lambda x,\lambda y \right)=-\frac{\lambda {{x}^{3}}+2{{y}^{2}}}{2xy}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  

For option (D):

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

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

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

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

Thus the correct answer is option (D).