Chapter 9: Differential Equations

Ans: The given differential equation is:

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

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

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

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

$p=\frac{1}{y}$ 

$Q=y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{\int \frac{1}{y}dy}}$ 

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

$\Rightarrow I.F=y$ 

General solution is of the form:

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

$\Rightarrow x\left( y \right)=\int \left( y \right)\left( y \right)dy+C$ 

$\Rightarrow xy=\int {{y}^{2}}dy+C$ 

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

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

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

Ans: The given differential equation is:

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

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

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

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

$p=-\frac{1}{y}$ 

$Q=3y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{-\int \frac{1}{y}dy}}$ 

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

$\Rightarrow I.F={{e}^{\log {{y}^{-1}}}}$ 

$\Rightarrow I.F=\frac{1}{y}$ 

General solution is of the form:

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

$\Rightarrow x\left( \frac{1}{y} \right)=\int \left( 3y \right)\left( \frac{1}{y} \right)dy+C$ 

$\Rightarrow \frac{x}{y}=3\int dy+C$ 

$\Rightarrow \frac{x}{y}=3y+C$ 

$\Rightarrow x=3{{y}^{2}}+Cy$ 

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

Ans: The given differential equation is:

$\frac{dy}{dx}+2y\tan x=\sin x$ 

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

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

$p=2\tan x$ 

$Q=\sin x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

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

$\Rightarrow I.F={{e}^{2\log \left| \sec x \right|}}$

\[\Rightarrow I.F={{e}^{\log {{\left( \sec x \right)}^{2}}}}\] 

\[\Rightarrow I.F={{\sec }^{2}}x\] 

General solution is of the form:

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

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

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

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

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

Given $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$:

$0=\cos \left( \frac{\pi }{3} \right)+C{{\cos }^{2}}\left( \frac{\pi }{3} \right)$ 

$\Rightarrow 0=\frac{1}{2}+C{{\left( \frac{1}{2} \right)}^{2}}$ 

$\Rightarrow C=-2$ 

Therefore the particular solution will be:

$\Rightarrow y=\cos x-2{{\cos }^{2}}x$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y=cosx-2co}{{\text{s}}^{\text{2}}}\text{x}$.

Ans: The given differential equation is:

$\left( 1+{{x}^{2}} \right)\frac{dy}{dx}+2xy=\frac{1}{1+{{x}^{2}}}$ 

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

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

$p=\frac{2x}{1+{{x}^{2}}}$ 

$Q=\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

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

$\Rightarrow I.F={{e}^{2\log \left| \sec x \right|}}$

$\Rightarrow I.F={{e}^{\log \left( 1+{{x}^{2}} \right)}}$ 

\[\Rightarrow I.F=1+{{x}^{2}}\] 

General solution is of the form:

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

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

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

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

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

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

$\Rightarrow C+\frac{\pi }{4}=0$ 

$\Rightarrow C=-\frac{\pi }{4}$ 

Therefore the particular solution will be:

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

The particular solution for given differential equation satisfying the given conditions is $\text{y}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\text{x-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}-3y\cot x=\sin 2x$ 

\[\frac{dy}{dx}+\left( -3\cot x \right)y=\sin 2x\] 

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

$p=-3\cot x$ 

$Q=\sin 2x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-3\int \cot xdx}}$ 

$\Rightarrow I.F={{e}^{-3\log \left| \sin x \right|}}$

$\Rightarrow I.F={{e}^{\log \left( \frac{1}{{{\sin }^{3}}x} \right)}}$ 

\[\Rightarrow I.F=\frac{1}{{{\sin }^{3}}x}\] 

General solution is of the form:

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

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

$\Rightarrow y\left( \frac{1}{{{\sin }^{3}}x} \right)=2\int \left( \sin x\cos x \right)\left( \frac{1}{{{\sin }^{3}}x} \right)dx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=2\int \left( \frac{\cos x}{{{\sin }^{2}}x} \right)dx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=2\int \cot x\text{cosec}xdx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=-2\text{cosec}\,x+C$ 

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

Given $\text{y=2}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

$2=-2{{\sin }^{2}}\left( \frac{\pi }{2} \right)+C{{\sin }^{3}}\left( \frac{\pi }{2} \right)$ 

$\Rightarrow C-2=2$ 

$\Rightarrow C=4$ 

Therefore the particular solution will be:

$\Rightarrow y=-2{{\sin }^{2}}x+4{{\sin }^{3}}x$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y=-2si}{{\text{n}}^{\text{2}}}\text{x+4si}{{\text{n}}^{\text{3}}}\text{x}$.

Ans: According to  the slope of tangent $\frac{\text{dy}}{\text{dx}}$ is equal to sum of the coordinates:

$\frac{dy}{dx}=x+y$ 

The given differential equation is:

$\frac{dy}{dx}=x+y$ 

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

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

$p=-1$ 

$Q=x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-1\int dx}}$ 

$\Rightarrow I.F={{e}^{-x}}$ 

General solution is of the form:

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

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

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

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

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

$\Rightarrow y=-x-1+C{{e}^{x}}$ 

$\Rightarrow y+x+1=C{{e}^{x}}$ 

Given $\text{y=0}$ when $\text{x=0}$ as it passes through origin:

$0+0+1=C{{e}^{0}}$ 

$\Rightarrow C=1$ 

Therefore the equation of the required curve is $\text{y+x+1=}{{\text{e}}^{\text{x}}}$.

Ans: Let the slope of tangent be $\frac{\text{dy}}{\text{dx}}$.

According to :

$x+y=\frac{dy}{dx}+5$ 

The given differential equation is:

$x+y=\frac{dy}{dx}+5$ 

\[\frac{dy}{dx}+\left( -1 \right)y=x-5\] 

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

$p=-1$ 

$Q=x-5$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-1\int dx}}$ 

$\Rightarrow I.F={{e}^{-x}}$ 

General solution is of the form:

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

$\Rightarrow y\left( {{e}^{-x}} \right)=\int \left( x-5 \right)\left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int x\left( {{e}^{-x}} \right)dx-5\int {{e}^{-x}}dx+C$ 

$\Rightarrow y{{e}^{-x}}=x\int {{e}^{-x}}dx-\int \left( \left( x \right)'\int {{e}^{-x}}dx \right)dx-5\int {{e}^{-x}}dxC$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+\int \left( {{e}^{-x}} \right)dx+5{{e}^{-x}}+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}-{{e}^{-x}}+5{{e}^{-x}}+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+4{{e}^{-x}}+C$ 

$\Rightarrow y=-x+4+C{{e}^{x}}$ 

$\Rightarrow y+x-4=C{{e}^{x}}$ 

Given as it passes through $\left( \text{0,2} \right)$ :

$2+0-4=C{{e}^{0}}$ 

$\Rightarrow C=-2$ 

Therefore the equation of the required curve is:

 $y+x-4=-2{{e}^{x}}$.

Ans: Given differential equation is:

$x\frac{dy}{dx}-y=2{{x}^{2}}$ 

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

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

$p=-\frac{1}{x}$ 

$I.F={{e}^{\int pdx}}$ 

\[\Rightarrow I.F={{e}^{-\int \frac{1}{x}dx}}\]

$\Rightarrow I.F={{e}^{-\log \left| x \right|}}$  

$\Rightarrow I.F={{e}^{\log {{x}^{-1}}}}$

$\Rightarrow I.F=\frac{1}{x}$

Therefore integrating factor is $\frac{\text{1}}{\text{x}}$. Thus the correct option is (C).

Ans: Given differential equation is:

$\left( 1-{{y}^{2}} \right)\frac{dx}{dy}+yx=ay$ 

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

Thus it is a linear differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$:

$p=\frac{y}{1-{{y}^{2}}}$ 

\[I.F={{e}^{\int pdy}}\] 

\[\Rightarrow I.F={{e}^{\int \frac{y}{1-{{y}^{2}}}dy}}\]

\[\Rightarrow I.F={{e}^{\frac{1}{-2}\int \frac{-2y}{1-{{y}^{2}}}dy}}\]  

\[\Rightarrow I.F={{e}^{\frac{1}{-2}\log \left( 1-{{y}^{2}} \right)}}\]

\[\Rightarrow I.F={{e}^{\log {{\left( 1-{{y}^{2}} \right)}^{-\frac{1}{2}}}}}\]

\[\Rightarrow I.F=\frac{1}{\sqrt{1-{{y}^{2}}}}\] 

Therefore integrating factor is $\frac{\text{1}}{\sqrt{\text{1-}{{\text{y}}^{\text{2}}}}}$. Thus the correct option is (D).