Chapter 9: Differential Equations

Ans: The given differential equation is:

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

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

$p=2$ 

$Q=\sin x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int 2dx}}$ 

$\Rightarrow I.F={{e}^{2x}}$ 

General solution is of the form:

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

$y{{e}^{2x}}=\int \sin x\left( {{e}^{2x}} \right)dx+C$ 

\[\Rightarrow y{{e}^{2x}}=I+C\,\,\,\left( I=\int \sin x\left( {{e}^{2x}} \right)dx \right)\] ……(1)

\[I=\int \sin x\left( {{e}^{2x}} \right)dx\] 

$\Rightarrow I=\left( \sin x \right)\int {{e}^{2x}}dx-\int \left( \left( \sin x \right)'\int {{e}^{2x}}dx \right)dx$ 

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

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \cos x\int {{e}^{2x}}dx-\int \left( \left( \cos x \right)'\left( \int {{e}^{2x}}dx \right) \right)dx \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \frac{{{e}^{2x}}}{2}\cos x+\frac{1}{2}\int {{e}^{2x}}\left( \sin x \right)dx \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \frac{{{e}^{2x}}}{2}\cos x+\frac{1}{2}I \right]$ 

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

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

\[\Rightarrow I=\frac{2{{e}^{2x}}}{5}\sin x-\frac{{{e}^{2x}}}{5}\cos x\] 

\[\Rightarrow I=\frac{{{e}^{2x}}}{5}\left[ 2\sin x-\cos x \right]\] 

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

\[\Rightarrow y{{e}^{2x}}=\frac{{{e}^{2x}}}{5}\left[ 2\sin x-\cos x \right]+C\,\] 

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

The general solution for given differential equation is \[\text{y=}\frac{\text{1}}{\text{5}}\left( \text{2sinx-cosx} \right)\text{+C}{{\text{e}}^{\text{-2x}}}\].

Ans: The given differential equation is:

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

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

$p=3$ 

$Q={{e}^{-2x}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int 3dx}}$ 

$\Rightarrow I.F={{e}^{3x}}$ 

General solution is of the form:

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

$\Rightarrow y{{e}^{3x}}=\int {{e}^{-2x}}\left( {{e}^{3x}} \right)dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{-2x+3x}}dx+C$ 

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

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

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

The general solution for given differential equation is $\text{y=}{{\text{e}}^{\text{-2x}}}\text{+C}{{\text{e}}^{\text{-3x}}}$.

Ans: The given differential equation is:

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

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

$p=\frac{1}{x}$ 

$Q={{x}^{2}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

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

$\Rightarrow I.F={{e}^{\log x}}$

$\Rightarrow I.F=x$  

General solution is of the form:

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

$\Rightarrow yx=\int {{x}^{2}}\left( x \right)dx+C$ 

$\Rightarrow xy=\int {{x}^{3}}dx+C$ 

$\Rightarrow xy=\frac{{{x}^{3+1}}}{3+1}+C$ 

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

The general solution for given differential equation is $\text{xy=}\frac{{{\text{x}}^{\text{4}}}}{\text{4}}\text{+C}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+\left( \sec x \right)y=\tan x$

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

$p=\sec x$ 

$Q=\tan x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \sec xdx}}$ 

$\Rightarrow I.F={{e}^{\log \left( \sec x+\tan x \right)}}$

$\Rightarrow I.F=\left( \sec x+\tan x \right)$  

General solution is of the form:

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

$\Rightarrow y\left( \sec x+\tan x \right)=\int \tan x\left( \sec x+\tan x \right)dx+C$ 

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

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

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\int {{\sec }^{2}}xdx-\int dx+C$ 

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

The general solution for given differential equation is $\text{y}\left( \text{secx+tanx} \right)\text{=secx+tanx-x+C}$.

Ans: The given differential equation is:

${{\cos }^{2}}x\frac{dy}{dx}+y=\tan x\,$

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

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

$p={{\sec }^{2}}x$ 

$Q={{\sec }^{2}}x\tan x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int {{\sec }^{2}}xdx}}$ 

$\Rightarrow I.F={{e}^{\tan x}}$ 

General solution is of the form:

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

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

$\Rightarrow y{{e}^{\tan x}}=I+C\,\,\left( I=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx \right)$ ……(1)

Solving the integral $\text{I}$:

$I=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx$ 

Substitute $\text{tanx=t}$ :
$\tan x=t$ 

$\Rightarrow {{\sec }^{2}}xdx=dt$ 

$\Rightarrow I=\int {{e}^{t}}tdt$ 

$\Rightarrow I=t\int {{e}^{t}}dt-\int \left( \left( t \right)'\int {{e}^{t}}dt \right)dt$ 

$\Rightarrow I=t{{e}^{t}}-\int \left( {{e}^{t}} \right)dt$ 

$\Rightarrow I=t{{e}^{t}}-{{e}^{t}}$ 

Back substitute $\text{t}$:

$I=\tan x{{e}^{\tan x}}-{{e}^{\tan x}}$ 

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

$\Rightarrow y{{e}^{\tan x}}=I+C\,\,$ 

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

The general solution for given differential equation is $\text{y}{{\text{e}}^{\text{tanx}}}\text{=tanx}{{\text{e}}^{\text{tanx}}}\text{-}{{\text{e}}^{\text{tanx}}}\text{+C}\,\,$.

Ans: The given differential equation is:

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

\[\frac{dy}{dx}+\frac{2}{x}y=x\log x\] 

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

$p=\frac{2}{x}$ 

$Q=x\log x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{2}{x}dx}}$ 

$\Rightarrow I.F={{e}^{2\log x}}$

$\Rightarrow I.F={{e}^{\log {{x}^{2}}}}$ 

$\Rightarrow I.F={{x}^{2}}$  

General solution is of the form:

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

$\Rightarrow y{{x}^{2}}=\int {{x}^{2}}\left( x\log x \right)dx+C$ 

\[\Rightarrow y{{x}^{2}}=\int {{x}^{3}}\log xdx+C\] 

\[\Rightarrow y{{x}^{2}}=I+C\,\,\left( I=\int {{x}^{3}}\log xdx \right)\] ……(1)

Solving the integral $\text{I}$:

\[I=\int {{x}^{3}}\log xdx\] 

\[\Rightarrow I=\log x\int {{x}^{3}}dx-\int \left( \left( \log x \right)'\int {{x}^{3}}dx \right)dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\int \left( \frac{1}{x}\left( \frac{{{x}^{4}}}{4} \right) \right)dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{1}{4}\int {{x}^{3}}dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{1}{4}\left( \frac{{{x}^{4}}}{4} \right)\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}\] 

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

\[\Rightarrow y{{x}^{2}}=I+C\] 

\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}+C\] 

\[\Rightarrow y=\frac{{{x}^{2}}}{16}\left( 4\log x-1 \right)+C{{x}^{-2}}\] 

The general solution for given differential equation is \[\text{y=}\frac{{{\text{x}}^{\text{2}}}}{\text{16}}\left( \text{4logx-1} \right)\text{+C}{{\text{x}}^{\text{-2}}}\].

Ans: The given differential equation is:

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

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

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

$p=\frac{1}{x\log x}$ 

$Q=\frac{2}{{{x}^{2}}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

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

Substitute $\text{logx=t}$:
$\log x=t$ 

$\Rightarrow \frac{1}{x}dx=dt$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{t}dt}}$

$\Rightarrow I.F={{e}^{\log t}}$ 

$\Rightarrow I.F=t$

$\Rightarrow I.F=\log x$   

General solution is of the form:

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

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

$\Rightarrow y\log x=I+C\,\,\left( I=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx \right)$ ……(1)

Solving the integral $\text{I}$:

$I=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx$ 

\[I=2\left[ \log x\int \frac{1}{{{x}^{2}}}dx-\int \left( \left( \log x \right)'\int \frac{1}{{{x}^{2}}}dx \right)dx \right]\] 

\[I=2\left[ \log x\left( \frac{-1}{x} \right)-\int \left( \frac{1}{x}\left( \frac{-1}{x} \right) \right)dx \right]\] 

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

\[I=2\left[ -\frac{\log x}{x}-\frac{1}{x} \right]\] 

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

$y\log x=I+C$ 

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

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

The general solution for given differential equation is \[\text{ylogx=-}\frac{\text{2}}{\text{x}}\left( \text{logx+1} \right)\text{+C}\].

Ans: The given differential equation is:

$\left( 1+{{x}^{2}} \right)dy+2xydx=\cot xdx\,\,\left( x\ne 0 \right)$

$\Rightarrow \frac{dy}{dx}+\frac{2xy}{1+{{x}^{2}}}=\frac{\cot x}{1+{{x}^{2}}}$ 

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

$p=\frac{2x}{1+{{x}^{2}}}$ 

$Q=\frac{\cot x}{1+{{x}^{2}}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{2x}{1+{{x}^{2}}}dx}}$ 

Substitute $\text{logx=t}$:
$1+{{x}^{2}}=t$ 

$\Rightarrow 2xdx=dt$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{t}dt}}$

$\Rightarrow I.F={{e}^{\log t}}$ 

$\Rightarrow I.F=t$

$\Rightarrow I.F=1+{{x}^{2}}$   

General solution is of the form:

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

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \frac{\cot x}{1+{{x}^{2}}}\left( 1+{{x}^{2}} \right)dx+C$ 

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

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

The general solution for given differential equation is $\text{y}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{=log}\left| \text{sinx} \right|\text{+C}$.

Ans: The given differential equation is:

$x\frac{dy}{dx}+y-x+xy\cot x=0\,$

$\Rightarrow \frac{dy}{dx}+\frac{y}{x}-1+y\cot x=0$ 

$\Rightarrow \frac{dy}{dx}+\left( \frac{1}{x}+\cot x \right)y=1$ 

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

$p=\left( \frac{1}{x}+\cot x \right)$ 

$Q=1$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \left( \frac{1}{x}+\cot x \right)dx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x}dx+\int \cot xdx}}$

$\Rightarrow I.F={{e}^{\log x+\log \left( \sin x \right)}}$ 

$\Rightarrow I.F={{e}^{\log \left( x\sin x \right)}}$

$\Rightarrow I.F=x\sin x$   

General solution is of the form:

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

$\Rightarrow y\left( x\sin x \right)=\int \left( 1 \right)\left( x\sin x \right)dx+C$ 

$\Rightarrow xy\sin x=I+C\,\,\left( I=\int x\sin xdx \right)$ ……(1) 

Solving the integral $\text{I}$:

$I=\int x\sin xdx$ 

\[\Rightarrow I=x\int \sin xdx-\int \left( \left( x \right)'\int \sin xdx \right)dx\] 

\[\Rightarrow I=x\left( -\cos x \right)+\int \left( \cos x \right)dx\] 

\[\Rightarrow I=-x\cos x+\sin x\] 

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

$xy\sin x=I+C\,$ 

\[\Rightarrow xy\sin x=-x\cos x+\sin x+C\,\] 

\[\Rightarrow y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}\,\] 

The general solution for given differential equation is \[\text{y=-cotx+}\frac{\text{1}}{\text{x}}\text{+}\frac{\text{C}}{\text{xsinx}}\,\].

Ans: The given differential equation is:

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

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

$\Rightarrow \frac{dx}{dy}=x+y$ 

$\Rightarrow \frac{dx}{dy}-x=y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=-1$ 

$Q=y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{\int -1dy}}$ 

$\Rightarrow I.F={{e}^{-y}}$

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

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

$\Rightarrow x{{e}^{-y}}=I+C\,\,\left( I=\int y{{e}^{-y}}dy \right)$ ……(1) 

Solving the integral $\text{I}$:

$I=\int y{{e}^{-y}}dy$ 

$\Rightarrow I=y\int {{e}^{-y}}dy-\int \left( \left( y \right)'\int {{e}^{-y}}dy \right)dy$ 

$\Rightarrow I=-y{{e}^{-y}}+\int \left( \left( 1 \right){{e}^{-y}} \right)dy$ 

$\Rightarrow I=-y{{e}^{-y}}+\int {{e}^{-y}}dy$

$\Rightarrow I=-y{{e}^{-y}}-{{e}^{-y}}$  

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

$x{{e}^{-y}}=I+C$ 

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

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

\[\Rightarrow x+y+1=C{{e}^{y}}\]  

The general solution for given differential equation is \[\text{x+y+1=C}{{\text{e}}^{\text{y}}}\].