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NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 9- Differential Equation

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NCERT Solutions for Class 12 Maths Chapter 9 Miscellaneous Exercise - Free PDF Download

Class 12 Maths NCERT Solutions for Chapter 9 Differential Equation includes solutions to all Miscellaneous Exercise problems. Differential Equation Class 12 NCERT Solutions Miscellaneous Exercises are based on the ideas presented in Maths Chapter 9. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the latest CBSE Class 12 Maths Syllabus in PDF format and practice it offline.

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Access NCERT Class 12 Maths Chapter 9 Differential Equation

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. 

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

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.


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

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. 

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

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


CBSE Class 12 Maths Chapter 9 Other Study Materials


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.


FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 9- Differential Equation

1. What kind of problems are covered in the miscellaneous exercise?

The miscellaneous exercise in Chapter 9 deals with various aspects of solving differential equations, including:

  • Identifying the order and degree of a differential equation.

  • Finding general and particular solutions of differential equations using different methods (separation of variables,homogeneous equations, integrating factors, etc.).

  • Applying differential equations to solve real-world problems in various fields (e.g., population growth, motion,electrical circuits).

2. How do the NCERT solutions approach these problems?

The NCERT solutions typically:

  • Briefly remind you of the relevant concepts from differential equations, including order, degree, general and particular solutions, and different solution methods.

  • Guide you through the process of classifying the differential equation based on its order and degree.

  • Demonstrate how to apply the appropriate solution method for the given differential equation. This might involve separation of variables for separable equations, using integrating factors for specific types of equations, or other relevant methods.

  • Show you how to find the general solution and then apply initial conditions (if provided) to obtain the particular solution.

3. Where can I find additional resources for practising miscellaneous exercise problems?

  • The NCERT textbook itself might provide solutions to some problems within the miscellaneous exercise section.

  • Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 9 Miscellaneous Exercise Differential Equations."

4. Differential Equation sample questions?

While I cannot provide specific solutions due to copyright, here are 2 sample questions to illustrate the types of problems you might encounter:

  1. Solve the differential equation dy/dx = x^2 + y (This question might involve using separation of variables)

  2. Find the general solution of the differential equation d^2y/dx^2 + 4y = 0 (This question requires identifying the order and degree and then solving the homogeneous equation)