NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations Exercise 9.3

Exercise 9.3

1. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$.

Use trigonometric half–angle identities to simplify:

$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{\sin }^{2}}\frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}$ 

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

$\Rightarrow \frac{dy}{dx}={{\sec }^{2}}\frac{x}{2}-1$ 

Separate the differentials and integrate:

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

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

$\Rightarrow y=2\tan \frac{x}{2}-x+C$ 

Thus the general solution of given differential equation is $\text{y = 2tan}\frac{\text{x}}{\text{2}}\text{ - x + C}$.

2. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$.

Simplify the expression:

$\frac{dy}{dx}=\sqrt{4-{{y}^{2}}}$ 

$\Rightarrow \frac{dy}{\sqrt{4-{{y}^{2}}}}=dx$ 

Use standard integration:

$\int \frac{dy}{\sqrt{4-{{y}^{2}}}}=\int dx$ 

$\Rightarrow {{\sin }^{-1}}\frac{y}{2}=x+C$ 

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

$\Rightarrow y=2\sin \left( x+C \right)$ 

Thus the general solution of given differential equation is $\text{y = 2 sin}\left( \text{x + C} \right)$.

3. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$.

Simplify the expression:

$\frac{dy}{dx}+y=1$ 

$\Rightarrow \frac{dy}{dx}=1-y$ 

$\Rightarrow \frac{dy}{1-y}=dx$ 

Use standard integration:

$\int \frac{dy}{1-y}=\int dx$ 

$\Rightarrow -\log \left( 1-y \right)=x+C$ 

$\Rightarrow \log \left( 1-y \right)=-\left( x+C \right)$ 

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

$y=1-A{{e}^{-x}}\,\left( A={{e}^{-C}} \right)$ 

Thus the general solution of given differential equation is $\text{y=1- A}{{\text{e}}^{\text{-x}}}\,$.

4. Find the general solution for $\text{se}{{\text{c}}^{\text{2}}}\text{x}\,\text{tany}\,\text{dx+se}{{\text{c}}^{\text{2}}}\text{y}\,\text{tanx}\,\text{dy=0}$.

Ans: The given differential equation is:

${{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy=0$.

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

$\frac{{{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy}{\tan x\tan y}=0$ 

\[\Rightarrow \frac{{{\sec }^{2}}x}{\tan x}dx+\frac{{{\sec }^{2}}y}{\tan y}dy=0\] 

Integrate both side:

$\int \frac{{{\sec }^{2}}x}{\tan x}dx+\int \frac{{{\sec }^{2}}y}{\tan y}dy=0$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{\sec }^{2}}x}{\tan x}dx$       ……(1)

Use a substitution method for integration. Substitute $\text{tan x = u}$:

For integral on RHS:

$\Rightarrow \tan x=u$ 

$\Rightarrow {{\sec }^{2}}xdx=du$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log u$ 

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

Thus evaluating result form (1):

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

$\Rightarrow \log \left( \tan y \right)=\log \left( \frac{C}{\tan x} \right)$ 

$\Rightarrow \tan x\tan y=C$ 

Thus the general solution of given differential equation is $\text{tan x tan y = C}$.

5. Find the general solution for $\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{dy-}\left( {{\text{e}}^{\text{x}}}\text{-}{{\text{e}}^{\text{-x}}} \right)\text{dx=0}$ .

Ans: The given differential equation is:

$\left( {{e}^{x}}+{{e}^{-x}} \right)dy-\left( {{e}^{x}}-{{e}^{-x}} \right)dx=0$.

Simplify the expression:

$dy=\left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ 

Integrate both side:

$\int dy=\int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ ……(1)

Use a substitution method for integration. Substitute \[{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}}\text{=t}\]:

For integral on RHS:

$\Rightarrow {{e}^{x}}+{{e}^{-x}}=t$ 

$\Rightarrow \left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\int \frac{dt}{t}$ s

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\ln t+C$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus evaluating result form (1):

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

Thus the general solution of given differential equation is $\text{y=log}\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{+C}$.

6. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\left( \text{1+}{{\text{y}}^{\text{2}}} \right)$.

Ans: The given differential equation is:

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

Simplify the expression:

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

Integrate both side:

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

Use standard integration:

${{\tan }^{-1}}y=\int dx+\int {{x}^{2}}dx$ 

$\Rightarrow {{\tan }^{-1}}y=x+\frac{{{x}^{3}}}{3}+C$ 

Thus the general solution of given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y=x+}\frac{{{\text{x}}^{\text{3}}}}{\text{3}}\text{+C}$.

7. Find the general solution for $\text{ylog y dx-xdy=0}$.

Ans: The given differential equation is:

$y\log ydx-xdy=0$.

Simplify the expression:

$y\log ydx=xdy$ 

$\Rightarrow \frac{dx}{x}=\frac{dy}{y\log y}$ 

Integrate both side:

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$ ……(1)

Use substitution method for integration on LHS. Substitute \[\text{log y=t}\]:

$\log y=t$ 

$\Rightarrow \frac{1}{y}dy=dt$ 

$\Rightarrow \int \frac{dy}{y\log y}=\int \frac{dt}{t}$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log t$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log \left( \log y \right)$ 

Evaluating expression (1):

$\log \left( \log y \right)=\log x+\log C$ 

$\Rightarrow \log \left( \log y \right)=\log Cx$ 

$\Rightarrow \log y=Cx$ 

$\Rightarrow y={{e}^{Cx}}$ 

Thus the general solution of given differential equation is $\text{y=}{{\text{e}}^{\text{Cx}}}$.

8. Find the general solution for ${{\text{x}}^{\text{5}}}\frac{\text{dy}}{\text{dx}}\text{=-}{{\text{y}}^{\text{5}}}$.

Ans: The given differential equation is:

${{x}^{5}}\frac{dy}{dx}=-{{y}^{5}}$.

Simplify the expression:

$\frac{dy}{{{y}^{5}}}=-\frac{dx}{{{x}^{5}}}$ 

Integrate both side:

$\int \frac{dy}{{{y}^{5}}}=-\int \frac{dx}{{{x}^{-5}}}$ 

\[\Rightarrow \int {{y}^{-5}}dy=-\int {{x}^{-5}}dx\] 

$\Rightarrow \frac{{{y}^{-5+1}}}{-5+1}=-\frac{{{x}^{-5+1}}}{-5+1}+C$ 

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

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

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=A\,\,\,\,\left( A=-4C \right)$ 

Thus the general solution of given differential equation is ${{\text{x}}^{\text{-4}}}\text{+}{{\text{y}}^{\text{-4}}}\text{=A}$.

9. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=si}{{\text{n}}^{\text{-1}}}\text{x}$.

Ans: The given differential equation is:

$\frac{dy}{dx}={{\sin }^{-1}}x$.

Simplify the expression:

$dy={{\sin }^{-1}}xdx$ 

Integrate both side:

$\int dy=\int {{\sin }^{-1}}xdx$ 

\[\Rightarrow y=\int 1\times {{\sin }^{-1}}xdx\] 

Use product rule of integration:

$\int {{\sin }^{-1}}xdx={{\sin }^{-1}}x\int dx-\int \left( \frac{1}{\sqrt{1-{{x}^{2}}}}\int dx \right)dx$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x-\int \frac{x}{\sqrt{1-{{x}^{2}}}}dx$ 

Substitute $\text{1-}{{\text{x}}^{\text{2}}}\text{=}{{\text{t}}^{\text{2}}}$ 

$1-{{x}^{2}}={{t}^{2}}$ 

$\Rightarrow -2xdx=2tdt$ 

$\Rightarrow -xdx=tdt$ 

Evaluating the integral:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\int \frac{tdt}{\sqrt{{{t}^{2}}}}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+t+C$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

$\Rightarrow y=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

Thus the general solution of given differential equation is $\text{y=xsi}{{\text{n}}^{\text{-1}}}\text{x+}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{+C}$,

10. Find the general solution for ${{\text{e}}^{\text{x}}}\text{tanydx+}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)\text{se}{{\text{c}}^{\text{2}}}\text{ydy=0}$.

Ans: The given differential equation is:

${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$.

Simplify the expression:

$\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=-{{e}^{x}}\tan ydx$ 

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

Integrate both side:

$\int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ ……(1)

Substitute $\text{tan y=u}$ 

$\tan y=u$ 

\[\Rightarrow {{\sec }^{2}}y=du\] 

Evaluating the LHS integral of (1):

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log \left( \tan y \right)$ 

Substitute $\text{1-}{{\text{e}}^{x}}\text{=v}$ 

$1-{{e}^{x}}=v$ 

$\Rightarrow -{{e}^{x}}dx=dv$ 

Evaluating the RHS integral of (1):

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

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log v$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log \left( 1-{{e}^{x}} \right)$ 

Therefore the integral (1) will be:

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

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

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

Thus the general solution of given differential equation is $\text{tany=C}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)$,

11. Find the particular solution of $\left( {{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}\text{+x+1} \right)\frac{\text{dy}}{\text{dx}}\text{=2}{{\text{x}}^{\text{2}}}\text{+x;}\,\,\text{y=1,}\,\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)\frac{dy}{dx}=2{{x}^{2}}+x;\,\,y=1,\,x=0$.

Simplify the expression:

$\frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)}$ 

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

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

Integrate both side:

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

Use partial fraction method to simplify the RHS:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$

$\Rightarrow 2{{x}^{2}}+x=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right)$ 

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

By comparing coefficients:

$A+B=2$ 

$B+C=1$ 

$A+C=0$ 

Solving this we get:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{\left( \frac{1}{2} \right)}{x+1}+\frac{\left( \frac{3}{2} \right)x+\left( -\frac{1}{2} \right)}{{{x}^{2}}+1}$ 

$\Rightarrow \frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{1}{2}\left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)$ 

Rewriting the  integral(1):

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

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

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{2}\int \frac{x}{{{x}^{2}}+1}dx-\frac{1}{2}\int \frac{1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\int \frac{2x}{{{x}^{2}}+1}dx-\frac{1}{2}{{\tan }^{-1}}x\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=\frac{1}{2}\log \left( 0+1 \right)+\frac{3}{4}\log \left( 0+1 \right)-\frac{1}{2}{{\tan }^{-1}}0+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

$\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+1$.

12. Find the particular solution of $\text{x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)\frac{\text{dy}}{\text{dx}}\text{=1}$; $\text{y=0}$ when $\text{x=2}$ to satisfy the given condition.

Ans: The given differential equation is:

$x\left( {{x}^{2}}-1 \right)\frac{dy}{dx}=1$; $y=0$ when $x=2$ 

Simplify the expression:

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

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

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

Integrate both side:

$\int dy=\int \frac{dx}{x\left( x-1 \right)\left( x+1 \right)}$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

$\Rightarrow 1=A\left( {{x}^{2}}-1 \right)+Bx\left( x+1 \right)+Cx\left( x-1 \right)$ 

\[\Rightarrow 1=\left( A+B+C \right){{x}^{2}}+\left( B-C \right)x-A\] 

By comparing coefficients:

$A+B+C=0$ 

$B-C=0$ 

$-A=1$ 

Solving this we get:

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

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

Rewriting the  integral(1):

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

\[\Rightarrow y=-\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{1}{x+1}dx\] 

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

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

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

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

For $y=0$ when $\text{x=2}$. 

\[0=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{2}^{2}}-1 \right)}{{{2}^{2}}} \right]\] 

\[\Rightarrow 0=\log \left[ \frac{3{{C}^{2}}}{4} \right]\] 

$\Rightarrow \frac{3{{C}^{2}}}{4}=1$ 

$\Rightarrow {{C}^{2}}=\frac{4}{3}$ 

Thus the required particular solution is :

$y=\frac{1}{2}\log \left[ \frac{4\left( {{x}^{2}}-1 \right)}{3{{x}^{2}}} \right]$.

13. Find the particular solution of $\text{cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=a}\,\,\left( \text{a}\in \text{R} \right)$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\cos \left( \frac{dy}{dx} \right)=a\,\,\left( a\in R \right)$; $y=1$ when $x=0$

Simplify the expression:

$\cos \left( \frac{dy}{dx} \right)=a\,$ 

$\Rightarrow \frac{dy}{dx}={{\cos }^{-1}}a$ 

$\Rightarrow dy={{\cos }^{-1}}adx$ 

Integrate both side:

$\int dy=\int {{\cos }^{-1}}adx$ 

$\Rightarrow y={{\cos }^{-1}}a\int dx$ 

\[\Rightarrow y=x{{\cos }^{-1}}a+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=0{{\cos }^{-1}}a+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is:

\[y=x{{\cos }^{-1}}a+1\].

\[\Rightarrow \frac{y-1}{x}={{\cos }^{-1}}a\]

\[\Rightarrow \cos \left( \frac{y-1}{x} \right)= a\]

14. Find the particular solution of $\frac{\text{dy}}{\text{dx}}\text{=ytanx}$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\frac{dy}{dx}=y\tan x$; $y=1$ when $x=0$

Simplify the expression:

$\frac{dy}{dx}=y\tan x$

$\Rightarrow \frac{dy}{y}=\tan xdx$ 

Integrate both side:

$\int \frac{dy}{y}=\int \tan xdx$ 

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

\[\Rightarrow \log y=\log \left( C\sec x \right)\] 

$\Rightarrow y=C\sec x$ 

For $\text{y=1}$ when $\text{x=0}$. 

\[1=C\sec 0\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

\[y=\sec x\]. 

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\text{y }\!\!'\!\!\text{ =}{{\text{e}}^{\text{x}}}\text{sin x}$.

Ans: The given differential equation is:

$y'={{e}^{x}}\sin x$

The curve passes through $\left( 0,0 \right)$.

Simplify the expression:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\sin x$

$\Rightarrow dy={{e}^{x}}\sin xdx$ 

Integrate both side:

$\int dy=\int {{e}^{x}}\sin xdx$ 

Use product rules for integration of RHS. Let:

$I=\int {{e}^{x}}\sin xdx$ 

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

$\Rightarrow I={{e}^{x}}\sin x-\int {{e}^{x}}\cos xdx$ 

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

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

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-I$ 

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

Thus integral will be:

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

Thus as the curve passes through $\left( \text{0,0} \right)$ 

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

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

$\Rightarrow C=\frac{1}{2}$ 

Thus the equation of the curve will be:

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

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

16. For the differential equation $\text{xy}\frac{\text{dy}}{\text{dx}}\text{=}\left( \text{x+2} \right)\left( \text{y+2} \right)$ find the solution curve passing through the point $\left( \text{1,-1} \right)$.

Ans: The given differential equation is:

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

The curve passes through $\left( 1,-1 \right)$.

Simplify the expression:

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

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

Integrate both side:

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

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

$\Rightarrow y-2\log \left( y+2 \right)=x+2\log x+C$ 

$\Rightarrow y-x=2\log \left( y+2 \right)+2\log x+C$ 

$\Rightarrow \Rightarrow y-x=2\log \left[ x\left( y+2 \right) \right]+C$ 

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

Thus as the curve passes through $\left( \text{1,-1} \right)$ 

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

$\Rightarrow -2=\log 1+C$ 

$\Rightarrow C=-2$ 

Thus the equation of the curve will be:

$y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]-2$ 

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

17. Find the equation of a curve passing through the point $\left( \text{0,-2} \right)$ given that at any point $\left( \text{x,y} \right)$ on the curve, the product of the slope of its tangent and $\text{y}$-coordinate of the point is equal to the $\text{x}$ -coordinate of the point.

Ans: According to , the equation is given by:

$y\frac{dy}{dx}=x$ 

The curve passes through $\left( 0,-2 \right)$.

Simplify the expression:

\[\Rightarrow ydy=xdx\] 

Integrate both side:

$\int ydy=\int xdx$ 

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

$\Rightarrow {{y}^{2}}-{{x}^{2}}=2C$ 

Thus as the curve passes through $\left( \text{0,-2} \right)$ 

$\Rightarrow {{\left( -2 \right)}^{2}}-{{0}^{2}}=2C$

$\Rightarrow 4=2C$ 

$\Rightarrow C=2$ 

Thus the equation of the curve will be:

${{y}^{2}}-{{x}^{2}}=2\left( 2 \right)$ 

$\Rightarrow {{y}^{2}}-{{x}^{2}}=4$ .

18. At any point $\left( \text{x,y} \right)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point$\left( \text{-4,-3} \right)$ . Find the equation of the curve given that it passes through $\left( \text{-2,1} \right)$.

Ans: Let the point of contact of the tangent be $\left( \text{x,y} \right)$.Then the slope of the segment joining point of contact and $\left( \text{-4,-3} \right)$:

$m=\frac{y+3}{x+4}$ 

According to  the for the slope of tangent $\frac{\text{dy}}{\text{dx}}$ it follows:

$\frac{dy}{dx}=2m$ 

$\Rightarrow \frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$

Simplify the expression:

$\frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$ 

$\frac{dy}{y+3}=\frac{2}{x+4}dx$ 

Integrate both side:

$\int \frac{dy}{y+3}=\int \frac{2}{x+4}dx$ 

$\Rightarrow \log \left( y+3 \right)=2\log \left( x+4 \right)+\log C$ 

\[\Rightarrow \log \left( y+3 \right)=\log {{\left( x+4 \right)}^{2}}+\log C\] 

\[\Rightarrow \log \left( y+3 \right)=\log C{{\left( x+4 \right)}^{2}}\] 

\[\Rightarrow y+3=C{{\left( x+4 \right)}^{2}}\] 

Thus as the curve passes through $\left( \text{-2,1} \right)$ 

\[1+3=C{{\left( -2+4 \right)}^{2}}\]

$\Rightarrow 4=4C$ 

$\Rightarrow C=1$ 

Thus the equation of the curve will be:

\[y+3={{\left( x+4 \right)}^{2}}\] .

19. The volume of spherical balloons being inflated changes at a constant rate. If initially its radius is $\text{3}$ units and after \[\text{3}\] seconds it is $\text{6}$ units. Find the radius of balloon after $\text{t}$ seconds.

Ans: Let the volume of spherical balloon be $\text{V}$ and its radius $\text{r}$. Let the rate of change of volume be $\text{k}$.

$\frac{dV}{dt}=k$ 

$\Rightarrow \frac{d}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \frac{d}{dt}\left( {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \left( 3{{r}^{2}} \right)\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}dr=kdt$ 

Integrate both side:

$\int 4\pi {{r}^{2}}dr=\int kdt$ 

$\Rightarrow 4\pi \int {{r}^{2}}dr=kt+C$ 

$\Rightarrow \frac{4}{3}\pi {{r}^{3}}=kt+C$ 

At initial time, $\text{t=0}$ and $\text{r=3}$:

$\frac{4}{3}\pi {{3}^{3}}=k\left( 0 \right)+C$ 

$\Rightarrow C=36\pi $

At $\text{t=3}$ the radius $\text{r=6}$:

$\frac{4}{3}\pi \left( {{6}^{3}} \right)=k\left( 3 \right)+36\pi $ 

$\Rightarrow 3k=288\pi -36\pi $ 

$\Rightarrow k=84\pi $ 

Thus the radius-time relation can be given by:

$\frac{4}{3}\pi {{r}^{3}}=84\pi t+36\pi$ 

$\Rightarrow {{r}^{3}}=63t+27$ 

$\Rightarrow r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$

The radius of balloon after $\text{t}$ seconds is given by: $r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$ .

20. In a bank, principal increases continuously at the rate of $\text{r}$% per year. Find the value of $\text{r}$ if Rs. $\text{100}$doubles itself in $\text{10}$ years $\left( \text{lo}{{\text{g}}_{\text{e}}}\text{2=0}\text{.6931} \right)$.

Ans: Let the principal be $\text{p}$, according to :

$\frac{dp}{dt}=\left( \frac{r}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{p}=\left( \frac{r}{100} \right)dt$ 

Integrate both side:

$\int \frac{dp}{p}=\int \left( \frac{r}{100} \right)dt$ 

$\Rightarrow \log p=\frac{rt}{100}+c$ 

\[\Rightarrow p={{e}^{\frac{rt}{100}+c}}\] 

\[\Rightarrow p=A{{e}^{\frac{rt}{100}}}\,\,\,\left( A={{e}^{c}} \right)\] 

At $\text{t=0}$, $\text{p=100}$:

\[100=A{{e}^{\frac{r\left( 0 \right)}{100}}}\] 

$\Rightarrow A=100$ 

Thus the principle and rate of interest relation:

$p=100{{e}^{\frac{rt}{100}}}$ 

At $\text{t=10}$, $\text{p=2}\times \text{100=200}$:

\[200=100{{e}^{\frac{r\left( 10 \right)}{100}}}\] 

$\Rightarrow 2={{e}^{\frac{r}{10}}}$ 

Take logarithm on both side:

$\log \left( {{e}^{\frac{r}{10}}} \right)=\log \left( 2 \right)$ 

$\Rightarrow \frac{r}{10}=0.6931$ 

$\Rightarrow r=6.931$ 

Thus the rate of interest $\text{r=6}\text{.931 }\!\!%\!\!\text{ }$. 

21. In a bank, principal increases continuously at the rate of $\text{5}$% per year. An amount of Rs $\text{1000}$ is deposited with this bank, how much will it worth after $\text{10}$  years $\left( {{\text{e}}^{\text{0}\text{.5}}}\text{=1}\text{.648} \right)$.

Ans: Let the principal be $\text{p}$, according to  principle increases at the rate of $\text{5}$% per year.

$\frac{dp}{dt}=\left( \frac{5}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{dt}=\frac{p}{20}$ 

$\Rightarrow \frac{dp}{p}=\frac{1}{20}dt$ 

Integrate both side:

$\int \frac{dp}{p}=\int \frac{1}{20}dt$ 

$\Rightarrow \log p=\frac{t}{20}+C$ 

$\Rightarrow p={{e}^{\frac{t}{20}+C}}$ 

$\Rightarrow p=A{{e}^{\frac{t}{20}}}\,\,\left( A={{e}^{C}} \right)$ 

At $\text{t=0}$, $\text{p=1000}$: 

$1000=A{{e}^{\frac{0}{20}}}$ 

$\Rightarrow A=1000$ 

Thus the relation of principal and time relation:

$\Rightarrow p=1000{{e}^{\frac{t}{20}}}$ 

At $\text{t=10}$: 

$p=1000{{e}^{\frac{10}{20}}}$ 

$\Rightarrow p=1000{{e}^{0.5}}$

$\Rightarrow p=1000\times 1.648$ 

$\Rightarrow p=1648$  

Thus after $\text{10}$ this year the amount will become Rs. $\text{1648}$.

22. In a culture, the bacteria count is $\text{100000}$ . The number is increased by $\text{10 }\!\!%\!\!\text{ }$  in $\text{2}$ hours. In how many hours will the count reach $\text{200000}$, if the rate of growth of bacteria is proportional to the number present?

Ans: Let the number of bacteria be $\text{y}$ at time $\text{t}$. According to :

$\frac{dy}{dt}\propto y$ 

$\frac{dy}{dt}=cy$ 

Here $\text{c}$ is constant.

Simplify the expression:

$\frac{dy}{y}=cdt$ 

Integrate both side:

$\int \frac{dy}{y}=\int cdt$ 

$\Rightarrow \log y=ct+D$ 

$\Rightarrow y={{e}^{ct+D}}$ 

$\Rightarrow y=A{{e}^{ct}}\,\,\left( A={{e}^{D}} \right)$ 

At $\text{t=0}$, $\text{y=100000}$: 

$100000=A{{e}^{c\left( 0 \right)}}$ 

$\Rightarrow A=100000$ 

At $\text{t=2}$, $\text{y=}\frac{11}{10}\left( \text{100000} \right)=110000$: 

$y=100000{{e}^{ct}}$ 

$\Rightarrow 110000=100000{{e}^{c\left( 2 \right)}}$ 

$\Rightarrow {{e}^{2c}}=\frac{11}{10}$ 

$\Rightarrow 2c=\log \left( \frac{11}{10} \right)$ 

$\Rightarrow c=\frac{1}{2}\log \left( \frac{11}{10} \right)$ ……(1)

For $\text{y=200000}$:

$200000=100000{{e}^{ct}}$ 

$\Rightarrow {{e}^{ct}}=2$ 

$\Rightarrow ct=\log 2$ 

$\Rightarrow t=\frac{\log 2}{c}$ 

Back substituting using expression (1):

\[t=\frac{\log 2}{\frac{1}{2}\log \left( \frac{11}{10} \right)}\] 

$\Rightarrow t=\frac{2\log 2}{\log \left( \frac{11}{10} \right)}$ 

Thus time required for bacteria to reach $\text{200000}$ is \[\text{t=}\frac{\text{log2}}{\frac{\text{1}}{\text{2}}\text{log}\left( \frac{\text{11}}{\text{10}} \right)}\] hrs.

23. Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$.

(A)${{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-y}}}\text{=C}$ 

(B) ${{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{y}}}\text{=C}$

(C) ${{\text{e}}^{\text{-x}}}\text{+}{{\text{e}}^{\text{y}}}\text{=C}$

(D) ${{\text{e}}^{\text{-x}}}\text{+}{{\text{e}}^{\text{-y}}}\text{=C}$

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$. Simplify the expression:

$\frac{dy}{dx}={{e}^{x}}{{e}^{y}}$ 

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

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

Integrate both side:

$\int {{e}^{-y}}dy=\int {{e}^{x}}dx$ 

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

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

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

Thus the general solution of given differential equation is $ {{e}^{x}}+{{e}^{-y}}=C$ 

Thus the correct option is (A).

Conclusion

NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations Exercise 9.3 provides a comprehensive approach to solving first-order, first-degree differential equations. With a focus on key methods such as separating variables, these solutions offer clear, step-by-step guidance to help students grasp and apply these techniques effectively. By practising these methods, students can build a solid foundation in differential equations, enhancing their understanding and performance in exams.

Class 12 Maths Chapter 9: Exercises Breakdown

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