Exercise 9.5 (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths
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NCERT Class 12 Maths Solutions: Chapter 9 - Differential Equations
In each of the following differential equations given in each Questions 1 to 4, find the general solution:
Ex 9.5 Question 1.
$\frac{d y}{d x}+2 y=\sin x$
Answer.
Given: Differential equation $\frac{d y}{d x}+2 y=\sin x$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have $\mathrm{P}=2$ and $\mathrm{Q}=\sin \mathrm{x}$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int 2 d x=2 \int 1 d x=2 x \\
& \Rightarrow \text { I.F. }=e^{\int P d x}=e^{2 x}
\end{aligned}
$
Solution is $y$ (I.F.) $=\int \mathrm{Q}($ I.F.) $d x+c$
$
\begin{aligned}
& \Rightarrow y e^{2 x}=\int e^{2 x} \sin x d x+c \\
& \Rightarrow y e^{2 x}=I+c
\end{aligned}
$
Applying product rule,
$
\begin{aligned}
& \Rightarrow \mathrm{I}=e^{2 x}(-\cos x)-\int 2 e^{2 x}(-\cos x) d x \\
& \Rightarrow-e^{2 x} \cos x+2 \int e^{2 x} \cos x d x
\end{aligned}
$
Again applying product rule,
$
\begin{aligned}
& \Rightarrow \mathrm{I}=-e^{2 x} \cos x+2\left[e^{2 x} \sin x-\int 2 e^{2 x} \sin x d x\right] \\
& \Rightarrow \mathrm{I}=-e^{2 x} \cos x+2 e^{2 x} \sin x-4 \int e^{2 x} \sin x d x \\
& \Rightarrow \mathrm{I}=e^{2 x}(-\cos x+2 \sin x)-4 \mathrm{I} \\
& \Rightarrow 5 \mathrm{I}=e^{2 x}(-\cos x+2 \sin x) \\
& \Rightarrow \mathrm{I}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)
\end{aligned}
$
Putting the value of I in eq. (i),
$
\begin{aligned}
& \Rightarrow y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c \\
& \Rightarrow y=\frac{1}{5}(2 \sin x-\cos x)+\frac{c}{e^{2 x}} \\
& \Rightarrow y=\frac{1}{5}(2 \sin x-\cos x)+c e^{-2 x}
\end{aligned}
$
Ex 9.5 Question 2.
$\frac{d y}{d x}+3 y=e^{-2 x}$
Answer.
Given: Differential equation $\frac{d y}{d x}+3 y=e^{-2 x}$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=2$ and $\mathrm{Q}=e^{-2 x}$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int 3 d x=3 \int 1 d x=3 x \\
& \Rightarrow \text { I.F. }=e^{\int P d x}=e^{3 x}
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int Q(\text { I.F.) } d x+c \\
& \Rightarrow y e^{3 x}=\int e^{-2 x} e^{3 x} d x+c \\
& \Rightarrow \int e^{-2 x+3 x} d x+c=\int e^x d x+c \\
& \Rightarrow y e^{3 x}=e^x+c \\
& \Rightarrow y=\frac{e^x}{e^{3 x}}+\frac{c}{e^{3 x}} \\
& \Rightarrow y=e^{-2 x}+c e^{-3 x}
\end{aligned}
$
Ex 9.5 Question 3.
$\frac{d y}{d x}+\frac{y}{x}=x^2$
Answer.
Given: Differential equation $\frac{d y}{d x}+\frac{y}{x}=x^2$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{1}{x}$ and $\mathrm{Q}=x^2$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \frac{1}{x} d x=\log x \\
& \Rightarrow \text { I.F. }=e^{\int P d x}=e^{\log x=x}
\end{aligned}
$
Solution is
$
\Rightarrow y \text { (I.F.) }=\int \mathrm{Q}(\text { I.F.) } d x+c
$
$\begin{aligned}
& \Rightarrow y x=\int x^2 x d x+c \\
& \Rightarrow y x=\int x^3 d x+c \\
& \Rightarrow x y=\frac{x^4}{4}+c
\end{aligned}$
Ex 9.5 Question 4.
$\frac{d y}{d x}+(\sec x) y=\tan x\left(0 \leq y<\frac{\pi}{2}\right)$
Answer.
Given: Differential equation $\frac{d y}{d x}+(\sec x) y=\tan x$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have $\mathrm{P}=\sec x$ and $\mathrm{Q}=\tan x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \sec x d x=\log (\sec x+\tan x) \\
& \text { I.F. }=e^{\int P d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int \mathrm{Q}(\text { I.F. }) d x+c \\
& \Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+c \\
& \Rightarrow y(\sec x+\tan x)=\int\left(\sec x \tan x+\tan ^2 x\right) d x+c \\
& =>y(\sec x+\tan x)=\int\left(\sec x \cdot \tan x+\sec ^2 x-1\right) d x+c \\
& \Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+c
\end{aligned}
$
For each of the following differential equations given in Question 5 to 8, find the general solution:
Ex 9.5 Question 5.
$\cos ^2 x \frac{d y}{d x}+y=\tan x\left(0 \leq y<\frac{\pi}{2}\right)$
Answer.
Given: Differential equation $\cos ^2 x \frac{d y}{d x}+y=\tan x$
$
\begin{aligned}
& =>\frac{d y}{d x}+\frac{y}{\cos ^2 x}=\frac{\tan x}{\cos ^2 x} \\
& \Rightarrow \frac{d y}{d x}+\left(\sec ^2 x\right) y=\sec ^2 x \tan x
\end{aligned}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\sec ^2 x$ and $\mathrm{Q}=\sec ^2 x \tan x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \sec ^2 x d x=\tan x \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{Pdx}}=e^{\tan x}
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int Q(\text { I.F. }) d x+c \\
& \Rightarrow y e^{\tan x}=\int \sec ^2 x \tan x e^{\tan x} d x+c
\end{aligned}
$
Putting $\tan x=t$ and differentiating $\sec ^2 x d x=d t$
$
\Rightarrow \int \sec ^2 x \tan x e^{\tan x} d x=\int t e^t d t
$
Applying product rule,
$
\Rightarrow \int \sec ^2 x \tan x e^{\tan x} d x=t \cdot e^t-\int 1 \cdot e^t d t=t \cdot e^t-e^t=(t-1) e^t=(\tan x-1) e^{\tan x}
$
Putting this value in eq. (i),
$
\Rightarrow y e^{\tan x}=(\tan x-1) e^{\tan x}+c
$
divide by $\mathrm{e}^{\tan \mathrm{x}}$, we get
$
\Rightarrow \mathrm{y}=(\tan \mathrm{x}-1)+\mathrm{ce}^{-\tan \mathrm{x}}
$
Ex 9.5 Question 6.
$x \frac{d y}{d x}+2 y=x^2 \log x$
Answer.
Given: Differential equation $x \frac{d y}{d x}+2 y=x^2 \log x$
$
\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{2}{x}$ and $\mathrm{Q}=x \log x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=2 \int \frac{1}{x} d x=2 \log x \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{P} d x}=e^{2 \log x}=e^{\log x^2}=x^2
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int \mathrm{Q}(\text { I.F. }) d x+c \\
& \Rightarrow y x^2=\int(x \log x) x^2 d x+c \\
& \Rightarrow y x^2=\int \log x x^3 d x+c \\
& \Rightarrow y x^2=\log x \cdot \frac{x^4}{4}-\int \frac{1}{x} \cdot \frac{x^4}{4} d x+c \\
& \Rightarrow y x^2=\frac{x^4}{4} \log x-\frac{1}{4} \int x^3 d x+c
\end{aligned}
$
$\begin{aligned}
& \Rightarrow y x^2=\frac{x^4}{4} \log x-\frac{x^4}{16}+c \\
& \Rightarrow y=\frac{x^2}{4} \log x-\frac{x^2}{16}+\frac{c}{x^2} \\
& \Rightarrow y=\frac{x^2}{16}(4 \log x-1)+\frac{c}{x^2}
\end{aligned}$
Ex 9.5 Question 7.
$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$
Answer.
Given: Differential equation $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$
$
\Rightarrow \frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2}{x^2}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{1}{x \log x}$ and $\mathrm{Q}=\frac{2}{x^2}$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \frac{1}{x \log x} d x=\int \frac{1 / x}{\log x} d x=\log (\log x) \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{P} \alpha x}=e^{\log (\log x)}=\log x
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int \mathrm{Q}(\text { I.F. }) d x+c \\
& \Rightarrow y \log x=\int \frac{2}{x^2} \log x d x=2 \int(\log x) x^{-2} d x+c
\end{aligned}
$
Applying Product rule of Integration,
$
\begin{aligned}
& \Rightarrow y \log x=2\left[(\log x) \frac{x^{-1}}{-1}-\int \frac{1}{x} \cdot \frac{x^{-1}}{-1} d x\right]+c \\
& \Rightarrow y \log x=2\left[\frac{-\log x}{x}+\int x^{-2} d x\right]+c \\
& \Rightarrow y \log x=2\left[\frac{-\log x}{x}+\frac{x^{-1}}{-1}\right]+c \\
& \Rightarrow y \log x=\frac{-2}{x}(1+\log x)+c
\end{aligned}
$
Ex 9.5 Question 8.
$\left(1+x^2\right) d y+2 x y d x=\cot x d x(x \neq 0)$
Answer.
Given: Differential equation $\left(1+x^2\right) d y+2 x y d x=\cot x d x$
$
\begin{aligned}
& \Rightarrow\left(1+x^2\right) \frac{d y}{d x}+2 x y=\cot x \\
& \Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{\cot x}{1+x^2}
\end{aligned}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{2 x}{1+x^2}$ and $\mathrm{Q}=\frac{\cot x}{1+x^2}$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \frac{2 x}{1+x^2} d x=\log \left|1+x^2\right|=\log \left(1+x^2\right) \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{P} d x}=e^{\log \left(1+x^2\right)}=1+x^2
\end{aligned}
$
Solution is
$\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int \mathrm{Q}(\text { I.F. }) d x+c \\
& \Rightarrow y\left(1+x^2\right)=\int \frac{\cot x}{1+x^2}\left(1+x^2\right) d x+c \\
& \Rightarrow y\left(1+x^2\right)=\int \cot x d x+c \\
& \Rightarrow y\left(1+x^2\right)=\log |\sin x|+c \\
& \Rightarrow y=\frac{\log |\sin x|}{1+x^2}+\frac{c}{1+x^2} \\
& \Rightarrow y=\left(1+x^2\right)^{-1} \log |\sin x|+c\left(1+x^2\right)^{-1}
\end{aligned}$
For each of the following differential equations given in Question 9 to 12, find the general solution:
Ex 9.5 Question 9.
$x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)$
Answer.
Given: Differential equation $x \frac{d y}{d x}+y-x+x y \cot x=0$
$
\begin{aligned}
& \Rightarrow x \frac{d y}{d x}+y+x y \cot x=x \\
& \Rightarrow x \frac{d y}{d x}+(1+x \cot x) y=x \\
& \Rightarrow \frac{d y}{d x}+\frac{(1+x \cot x)}{x} y=1
\end{aligned}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{1+x \cot x}{x}$ and $\mathrm{Q}=1$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \frac{1+x \cot x}{x} d x=\int\left(\frac{1}{x}+\cot x\right) d x \\
& =\log x+\log \sin x=\log (x \sin x) \\
& =\text { I.F. }=e^{\int \mathrm{P} d x}=e^{\log (x \sin x)}=x \sin x
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow y \text { (I.F.) }=\int \mathrm{Q}(\text { I.F.) } d x+c \\
& \Rightarrow y(x \sin x)=\int 1 x \sin x d x+c
\end{aligned}
$
Applying product rule of Integration,
$
\begin{aligned}
& \Rightarrow y(x \sin x)=x(-\cos x)-\int 1(-\cos x) d x+c=-x \cos x+\int \cos x d x+c \\
& \Rightarrow y(x \sin x)=-x \cos x+\sin x+c \\
& \Rightarrow y=\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{c}{x \sin x} \\
& \Rightarrow y=-\cot x+\frac{1}{x}+\frac{c}{x \sin x}
\end{aligned}
$
Ex 9.5 Question 10.
$(x+y) \frac{d y}{d x}=1$
Answer.
Given: Differential equation $(x+y) \frac{d y}{d x}=1$
$
\begin{aligned}
& \Rightarrow \frac{d x}{d y}=x+y \\
& \Rightarrow \frac{d x}{d y}-x=y
\end{aligned}
$
Comparing with $\frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}$, we have $\mathrm{P}=-1$ and $\mathrm{Q}=y$.
$
\therefore \int \mathrm{P} d y=\int-1 d y=-\int 1 d t=-y \quad \text { I.F. }=e^{\int \mathrm{P} d y}=e^{-y}
$
Solution is
$
\begin{aligned}
& \Rightarrow \mathrm{x} \text { (I.F.) }=\int Q(\text { I.F. }) d y+c \\
& \Rightarrow x e^{-y}=\int y e^{-y} d y+c
\end{aligned}
$
Applying product rule of Integration,
$
\begin{aligned}
& \Rightarrow x e^{-y}=y \frac{e^{-y}}{-1}-\int 1 \frac{e^{-y}}{-1} d y+c \\
& \Rightarrow \mathrm{xe}^{-\mathrm{y}}=-y e^{-y}+\int e^{-y} d y+c \\
& \Rightarrow \mathrm{xe}^{-\mathrm{y}}=-y e^{-y}+\frac{e^{-y}}{-1}+c \\
& \Rightarrow x e^{-y}=-y e^{-y}-e^{-y}+c \\
& \Rightarrow x=-y-1+\frac{c}{e^{-y}} \\
& \Rightarrow \mathrm{x}+\mathrm{y}+1=\mathrm{ce}^{\mathrm{y}}
\end{aligned}
$
Ex 9.5 Question 11.
$y d x+\left(x-y^2\right) d y=0$
Answer.
Given: Differential equation $y d x+\left(x-y^2\right) d y=0$
$
\Rightarrow y \frac{d x}{d y}+x-y^2=0
$
$\begin{aligned}
& \Rightarrow y \frac{d x}{d y}+x=y^2 \\
& \Rightarrow \frac{d x}{d y}+\frac{1}{y} x=y \\
& \text { Comparing with } \frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}, \mathrm{w} \\
& \text { e have } \mathrm{P}=\frac{1}{y} \text { and } \mathrm{Q}=y . \\
& \therefore \int \mathrm{P} \text { dy }=\int \frac{1}{y} d y=\log y \\
& \text { I.F. }=e^{\int \mathrm{P}}=e^{\mathrm{log} y}=y \\
& \text { Solution is x(I.F.) }=\int Q(\text { I.F. }) d y+c \\
& \Rightarrow x \cdot y=\int y y d y+c \\
& \Rightarrow x y=\int y^2 d y+c \\
& \Rightarrow x y=\frac{y^3}{3}+c \\
& \Rightarrow x=\frac{y^2}{3}+\frac{c}{y}
\end{aligned}$
Ex 9.5 Question 12.
$\left(x+3 y^2\right) \frac{d y}{d x}=y(y>0)$
Answer.
Given: Differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$
$
\Rightarrow y \frac{d x}{d y}=x+3 y^2
$
$
\begin{aligned}
& \Rightarrow y \frac{d x}{d y}-x=3 y^2 \\
& \Rightarrow \frac{d x}{d y}-\frac{1}{y} x=3 y
\end{aligned}
$
Comparing with $\frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}$,
we have $\mathrm{P}=\frac{-1}{y}$ and $\mathrm{Q}=3 y$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d y=-\int \frac{1}{y} d y=-\log y=\log y^{-1} \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{P} a y}=e^{\log y^{-1}}=y^{-1}=\frac{1}{y}
\end{aligned}
$
Solution is $\mathrm{x}$ (I.F.) $=\int Q($ I.F. $) d y+c$
$
\begin{aligned}
& \Rightarrow x \cdot \frac{1}{y}=\int 3 y \cdot \frac{1}{y} d y+c \\
& \Rightarrow \frac{x}{y}=3 \int 1 d y+c=3 y+c \\
& \Rightarrow x=3 y^2+c y
\end{aligned}
$
For each of the differential equations given in Questions 13 to 15 , find a particular solution satisfying the given condition:
Ex 9.5 Question 13.
$\frac{d y}{d x}+2 y \tan x=\sin x, y=0$ when $x=\frac{\pi}{3}$
Answer.
Given: Differential equation $\frac{d y}{d x}+2 y \tan x=\sin x, y=0$ when $x=\frac{\pi}{3}$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=2 \tan x$ and $\mathrm{Q}=\sin x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=2 \int \tan x d x=2 \log \sec x=\log (\sec x)^2 \\
& \Rightarrow \text { I.F.= } e^{\int \mathrm{Pdx}}=e^{\log (* x)^2}=(\sec x)^2=\sec ^2 x
\end{aligned}
$
Solution is $y$ (I.F.) $=\int \mathrm{Q}($ I.F.) $d x+c$
$
\begin{aligned}
& \Rightarrow y \sec ^2 x=\int \sin x \cdot \sec ^2 x d x+c \\
& \Rightarrow y \sec ^2 x=\int \frac{\sin x}{\cos ^2 x} d x+c
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow y \sec ^2 x=\int \frac{\sin x}{\cos x \cos x} d x+c \\
& \Rightarrow y \sec ^2 x=\int \tan x \cdot \sec x d x+c=\sec x+c \\
& \Rightarrow \frac{y}{\cos ^2 x}=\frac{1}{\cos x}+c
\end{aligned}
$
put value of $x$ and $y$, we get
$
\begin{aligned}
& =>\frac{0}{\cos ^2 \frac{\pi}{3}}=\frac{1}{\cos \frac{\pi}{3}}+c \\
& =>0=2+c \\
& =>c=-2 \text { put value of } \mathrm{c} \text { in (1), we get } \\
& =>\frac{y}{\cos ^2 x}=\frac{1}{\cos x}-2 \\
& =>y=\cos x-2 \cos ^2 x
\end{aligned}
$
Ex 9.5 Question 14.
$\left(1+x^2\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^2} ; y=0$ when $x=1$
Answer.
Given: Differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^2} ; y=0$ when $x=1$
$
\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{2 x}{1+x^2}$ and $\mathrm{Q}=\frac{1}{\left(1+x^2\right)^2}$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=\int \frac{2 x}{1+x^2} d x=\log \left(1+x^2\right) \\
& \Rightarrow \text { I.F. }=e^{\int P a x}=e^{\log \left(1+x^2\right)}=1+x^2
\end{aligned}
$
Solution is $y$ (I.F.) $=\int \mathrm{Q}($ I.F.) $d x+c$
$
\begin{aligned}
& \Rightarrow y\left(1+x^2\right)=\int \frac{1}{\left(1+x^2\right)^2}\left(1+x^2\right) d x+c \\
& \Rightarrow y\left(1+x^2\right)=\int \frac{1}{\left(1+x^2\right)} d x+c=\tan ^{-1} x+c
\end{aligned}
$
Now putting $y=0, x=1 \quad 0=\tan ^{-1} x+c$
$
\Rightarrow 0=\frac{\pi}{4}+c
$
$\Rightarrow c=-\frac{\pi}{4}$
Putting the value of $c$ in eq. (i),
$
\Rightarrow y\left(1+x^2\right)=\tan ^{-1} x-\frac{\pi}{4}
$
Ex 9.5 Question 15.
$\frac{d y}{d x}-3 y \cot x=\sin 2 x, y=2$ when $x=\frac{\pi}{2}$
Answer.
Given: Differential equation $\frac{d y}{d x}-3 y \cot x=\sin 2 x$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=-3 \cot x$ and $\mathrm{Q}=\sin 2 x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=-3 \int \cot x d x=-3 \log \sin x=\log (\sin x)^{-3} \\
& \text { I.F. }=e^{\int \mathrm{P} \alpha x}=e^{\log (\sin x)^{-3}}=(\sin x)^{-3}=\frac{1}{\sin ^3 x}
\end{aligned}
$
Solution is $y$ (I.F.) $=\int Q($ I.F.) $d x+c$
$
\begin{aligned}
& \Rightarrow y \frac{1}{\sin ^3 x}=\int \sin 2 x \cdot \frac{1}{\sin ^3 x} d x+c \\
& \Rightarrow \frac{y}{\sin ^3 x}=\int \frac{2 \sin x \cos x}{\sin ^3 x} d x+c=2 \int \frac{\cos x}{\sin ^2 x} d x+c \\
& \Rightarrow \frac{y}{\sin ^3 x}=2 \int \frac{\cos x}{\sin x \cdot \sin x} d x+c \\
& \Rightarrow \frac{y}{\sin ^3 x}=2 \int \operatorname{cosec} x \cot x d x+c \\
& \Rightarrow \frac{y}{\sin ^3 x}=-2 \operatorname{cosec} x+c
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \frac{y}{\sin ^3 x}=-\frac{2}{\sin x}+c \\
& \Rightarrow y=-2 \sin ^2 x+c \sin ^3 x
\end{aligned}
$
Now putting $y=2, x=\frac{\pi}{2}$ in eq. (i),
$
\begin{aligned}
& 2=-2 \sin ^2 \frac{\pi}{2}+c \sin ^3 \frac{\pi}{2} \\
& \Rightarrow 2=-2+c \\
& \Rightarrow c=4
\end{aligned}
$
Putting $c=4$ in eq. (i),
$
\begin{aligned}
& \Rightarrow y=-2 \sin ^2 x+4 \sin ^3 x \\
& \Rightarrow y=4 \sin ^3 x-2 \sin ^2 x
\end{aligned}
$
Ex 9.5 Question 16.
Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point ( $x, y$ ) is equal to the sum of coordinates of that point.
Answer.
Slope of the tangent to the curve at any point $(x, y)=$ Sum of coordinates of the point
$
\begin{aligned}
& (x, y) \\
& \Rightarrow \frac{d y}{d x}=x+y \\
& \Rightarrow \frac{d y}{d x}-y=x
\end{aligned}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=-1$ and $\mathrm{Q}=x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=-\int 1 d x=-x \\
& \Rightarrow \text { I.F. }=e^{\int P d x}=e^{-x}
\end{aligned}
$
Solution is
$
\begin{aligned}
& =>y(\text { I.F. })=\int \mathrm{Q}(\text { I.F. }) d x+c \\
& \Rightarrow y e^{-x}=\int x \cdot e^{-x} d x+c
\end{aligned}
$
Applying Product rule of Integration,
$
\begin{aligned}
& \Rightarrow y e^{-x}=x \frac{e^{-x}}{-1}-\int 1 \frac{e^{-x}}{-1} d x+c \\
& \Rightarrow y e^{-x}=-x e^{-x}+\frac{e^{-x}}{-1}+c \\
& \Rightarrow y e^{-x}=-x e^{-x}-e^{-x}+c \\
& \Rightarrow \frac{y}{e^x}=-\frac{x}{e^x}-\frac{1}{e^x}+c \\
& \Rightarrow y=-x-1+c e^x
\end{aligned}
$
Now, since curve (i) passes through the origin $(0,0)$, therefore putting $x=0, y=0$ in eq. (i)
$
\begin{aligned}
& \Rightarrow-c=-1 \\
& \Rightarrow c=1
\end{aligned}
$
Putting $c=1$ in eq. (i),
$\begin{aligned}
& y=-x-1+e^x \\
& \Rightarrow y+x+1=e^x
\end{aligned}$
Ex 9.5 Question 17.
Find the equation of the curve passing through the point $(0,2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5 .
Answer.
According to the question, Sum of the coordinates of any point say $(x, y)$ on the curve
$
\begin{aligned}
& =\text { Magnitude of the sl } \\
& \Rightarrow x+y=\frac{d y}{d x}+5 \\
& \Rightarrow \frac{d y}{d x}-y=x-5
\end{aligned}
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=-1$ and $\mathrm{Q}=x-5$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=-\int 1 d x=-x \\
& \Rightarrow \text { I.F.= } e^{\int \mathrm{P} d x}=e^{-x}
\end{aligned}
$
Solution is
$
\begin{aligned}
& \Rightarrow>\text { (I.F.) }=\int \mathrm{Q}(\text { I.F. }) d x+c \\
& \Rightarrow y e^{-x}=\int(x-5) \cdot e^{-x} d x+c
\end{aligned}
$
Applying Product rule of Integration,
$
\Rightarrow y e^{-x}=(x-5) \frac{e^{-x}}{-1}-\int 1 \frac{e^{-x}}{-1} d x+c
$
$\Rightarrow y e^{-x}=-(x-5) e^{-x}+\int e^{-x} d x+c$
$
\begin{aligned}
& \Rightarrow y e^{-x}=-(x-5) e^{-x}+\frac{e^{-x}}{-1}+c \\
& \Rightarrow \frac{y}{e^x}=-\frac{x-5}{e^x}-\frac{1}{e^x}+c \\
& \Rightarrow y=-x+5-1+c e^x \\
& \Rightarrow x+y=4+c e^x
\end{aligned}
$
Now, since curve (i) passes through the point $(0,2)$, therefore putting $x=0, y=2$ in eq. (i)
$
\begin{aligned}
& \Rightarrow 0+2=4+c e^0 \\
& \Rightarrow c=-2
\end{aligned}
$
Putting $c=-2$ in eq. (i),
$
\begin{aligned}
& \Rightarrow x+y=4-2 e^x \\
& \Rightarrow y=4-x-2 e^x
\end{aligned}
$
Ex 9.5 Question 18.
Choose the correct answer:
The integrating factor of the differential equation $x \frac{d y}{d x}-y=2 x^2$ is:
(A) $e^{-x}$
(B) $e^{-y}$
(C) $\frac{1}{x}$
(D) $x$
Answer.
Given: Differential equation $x \frac{d y}{d x}-y=2 x^2$
$
\Rightarrow \frac{d y}{d x}-\frac{1}{x} y=2 x
$
Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
we have $\mathrm{P}=\frac{-1}{x}$ and $\mathrm{Q}=2 x$.
$
\begin{aligned}
& \therefore \int \mathrm{P} d x=-\int \frac{1}{x} d x=-\log x=\log x^{-1} \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{P} d x}=e^{\log x^{-1}}=x^{-1}=\frac{1}{x}
\end{aligned}
$
Therefore, option (C) is correct.
Ex 9.5 Question 19.
Choose the correct answer:
The integrating factor of the differential equation $\left(1-y^2\right) \frac{d x}{d y}+y x=a y(-1<y<1) i s$
(A) $\frac{1}{y^2-1}$
(B) $\frac{1}{\sqrt{y^2-1}}$
(C) $\frac{1}{1-y^2}$
(D) $\frac{1}{\sqrt{1-y^2}}$
Answer.
Given: Differential equation $\left(1-y^2\right) \frac{d x}{d y}+y x=a y(-1<y<1)$
$
\begin{aligned}
& \Rightarrow \frac{d x}{d y}+\frac{y}{1-y^2} x=\frac{a y}{1-y^2} \\
& \text { Comparing with } \frac{d x}{d y}+\mathrm{P} x=\mathrm{Q},
\end{aligned}
$
we have $\mathrm{P}=\frac{y}{1-y^2}$ and $\mathrm{Q}=\frac{a y}{1-y^2}$
$
\begin{aligned}
& \therefore \int \mathrm{P} d y=\int \frac{y}{1-y^2} d y=\frac{-1}{2} \int \frac{-2 y}{1-y^2} d y \\
& =\frac{-1}{2} \log \left(1-y^2\right)=\log \left(1-y^2\right)^{-1 / 2} \\
& \Rightarrow \text { I.F. }=e^{\int P \alpha y}=e^{\log \left(1-y^2\right)^{-1 / 2}}=\left(1-y^2\right)^{-1 / 2}=\frac{1}{\sqrt{1-y^2}}
\end{aligned}
$
Therefore, option (D) is correct.
