NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations - 2025-26

Miscellaneous Exercise

1. For each of the differential equations given below, indicate its order and degree (if defined).

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

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 $\text{1}$.

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

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 $3$.

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

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.

2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

1. $\text{xy=a}{{\text{e}}^{\text{x}}}\text{+b}{{\text{e}}^{\text{-x}}}\text{+}{{\text{x}}^{\text{2}}}\,\,\,\text{:x}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+2}\frac{\text{dy}}{\text{dx}}\text{-xy+}{{\text{x}}^{\text{2}}}\text{-2=0}$ 

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. 

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

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.

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

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. 

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

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.

3. Prove that ${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}$  is the general solution of differential equation $\left( {{\text{x}}^{\text{3}}}\text{-3x}{{\text{y}}^{\text{2}}} \right)\text{dx=}\left( {{\text{y}}^{\text{3}}}\text{-3}{{\text{x}}^{\text{2}}}\text{y} \right)\text{dy}$ , where $\text{c}$ is a parameter.

 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 \[{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}\].

4. Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\sqrt{\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1-}{{\text{x}}^{\text{2}}}}}\text{=0}$.

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 $\text{si}{{\text{n}}^{\text{-1}}}\text{y+si}{{\text{n}}^{\text{-1}}}\text{x=C}$.

5. Show that the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\frac{{{\text{y}}^{\text{2}}}\text{+y+1}}{{{\text{x}}^{\text{2}}}\text{+x+1}}\text{=0}$ is given by $\left( \text{x+y+1} \right)\text{=A}\left( \text{1-x-y-2xy} \right)$  where $\text{A}$ is a parameter.

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 $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2y+1}}{\sqrt{\text{3}}} \right]\text{+ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2x+1}}{\sqrt{\text{3}}} \right]\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\text{C}$.

6.Find the equation of the curve passing through the point $\left( 0,\frac{\pi }{4} \right)$ whose differential equation is $\text{sin}\,\text{x}\,\text{cos}\,\text{ydx+cos}\,\text{x}\,\text{sin}\,\text{ydy=0}$.

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 $\text{sec}\,\text{x}\,\text{sec}\,\text{y=}\sqrt{\text{2}}$.

7. Find the particular solution of the differential equation $\left( \text{1+}{{\text{e}}^{\text{2x}}} \right)\text{dy+}\left( \text{1+}{{\text{y}}^{\text{2}}} \right){{\text{e}}^{\text{x}}}\text{dx=0}$ given that $\text{y=1}$ when $\text{x=0}$.

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 $\text{ta}{{\text{n}}^{\text{-1}}}\text{y+ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{x}}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

8. Solve the differential equation $\text{y}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{dx=}\left( \text{x}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{dy}\,\,\left( \text{y}\ne \text{0} \right)$.

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

9. Find a particular solution of the differential equation $\left( \text{x-y} \right)\left( \text{dx+dy} \right)\text{=dx-dy}$ given that $\text{y=-1}$ when $\text{x=0}$. Hint (put $\text{x-y=t}$).

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:

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

10. Solve the differential equation $\left[ \frac{{{\text{e}}^{\text{-2}\sqrt{\text{x}}}}}{\sqrt{\text{x}}}\text{-}\frac{\text{y}}{\sqrt{\text{x}}} \right]\frac{\text{dx}}{\text{dy}}\text{=1}\,\left( \text{x}\ne \text{0} \right)$.

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

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

11. Find a particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+ycotx=4x}\,\text{cosec}\,\text{x}\left( \text{x}\ne \text{0} \right)$ given that $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Ans:
The given differential equation is:

 $\frac{dy}{dx}+y\cot x=4x\,\text{cosec}x$ 

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

$p=\cot x$ 

$Q=4x\text{cosec}\,x$ 

Calculating integrating factor:

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

$I.F={{e}^{\int \cot xdx}}$ 

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

$I.F=\sin x$ 

The general solution is given by:

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

$\Rightarrow y\times \sin x=\int \left( 4x\cos \text{ec}x \right)\sin xdx+C$ 

$\Rightarrow y\sin x=4\int xdx+C$ 

$\Rightarrow y\sin x=4\left( \frac{{{x}^{2}}}{2} \right)+C$ 

$\Rightarrow y\sin x=2{{x}^{2}}+C$ 

As $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

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

$C=-2\left( \frac{{{\pi }^{2}}}{4} \right)$ 

$C=-\frac{{{\pi }^{2}}}{2}$ 

Thus the required particular solution is:

$y\sin x=2{{x}^{2}}-\frac{{{\pi }^{2}}}{2}$

12. Find a particular solution of the differential equation $\left( \text{x+1} \right)\frac{\text{dy}}{\text{dx}}\text{=2}{{\text{e}}^{\text{-y}}}\text{-1}$ given that $\text{y=0}$ when $\text{x=0}$.

Ans:
The given differential equation is:

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

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

Integrate both side:

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

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

Evaluating LHS integral:

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\int \frac{{{e}^{y}}dy}{2-{{e}^{y}}}$ 

Put $\text{t=2-}{{\text{e}}^{\text{y}}}$:

$t=2-{{e}^{y}}$

$dt=-{{e}^{y}}dy$  

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\int \frac{dt}{t}$

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\log \left( t \right)$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{t}$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{2-{{e}^{y}}}$ 

Back substituting in expression (1):

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

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

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

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

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

$\Rightarrow 2-1=\frac{1}{C}$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

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

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

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

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

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

Thus for given conditions the particular solution is $\text{y=log}\left( \frac{\text{2x+1}}{\text{x+1}} \right)$ .

13. The general solution of the differential equation $\frac{\text{ydx-xdy}}{\text{y}}\text{=0}$.

A. $xy=C$ 

B. $x=C{{y}^{2}}$ 

C. $y=Cx$ 

D. $y=C{{x}^{2}}$ 

Ans: Given differential equation:

$\frac{ydx-xdy}{y}=0$

Divide both side by $\text{x}$ :

$\Rightarrow \frac{ydx-xdy}{xy}=0$

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

Integrate both side:

$\Rightarrow \int \frac{dx}{x}-\int \frac{dy}{y}=0$ 

$\Rightarrow \log \left| x \right|-\log \left| y \right|=\log k$ 

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

$\Rightarrow \frac{x}{y}=k$ 

$\Rightarrow y=Cx\,\,\,\left( C=\frac{1}{k} \right)$

Thus the correct option is ©

14. Find the general solution of a differential equation of the type $\frac{\text{dx}}{\text{dy}}\text{+}{{\text{P}}_{\text{1}}}\text{x=}{{\text{Q}}_{\text{1}}}$.

A. \[y{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C\]   

B. \[y{{e}^{\int {{P}_{1}}dx}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dx}} \right)dy+C\] 

C. \[x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C\]   

D. \[x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dx}} \right)dy+C\]   

Ans: The given differential equation is:

$\frac{dx}{dy}+{{P}_{1}}x={{Q}_{1}}$ 

It is a linear differential equation and its general solution is:

$x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C$ 

With integrating factor $I.F={{e}^{\int {{P}_{1}}dy}}$.

Thus the correct option is (C).

15. Find the general solution of the differential equation ${{\text{e}}^{\text{x}}}\text{dy+}\left( \text{y}{{\text{e}}^{\text{x}}}\text{+2x} \right)\text{dx=0}$.

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

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

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

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

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

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

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

The given differential equation is of the form:

$\frac{dy}{dx}+Py=Q$ 

$\Rightarrow P=1$ 

$\Rightarrow Q=-2x{{e}^{-x}}$ 

Calculating integrating factor:
$I.F={{e}^{\int Pdx}}$

$\Rightarrow I.F={{e}^{\int dx}}$ 

$\Rightarrow I.F={{e}^{x}}$  

It is a linear differential equation and its general solution is:

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

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

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

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

$\Rightarrow y{{e}^{x}}+{{x}^{2}}=C$ 

Thus the correct answer is option (C).

Conclusion

NCERT solutions for class 12 maths Differential Equation miscellaneous exercise is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

Class 12 Maths Chapter 9: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 12 Maths