Exercise 9.3 (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 9 - Differential Equations

For each of the differential equations in Questions 1 to 4, find the general solution:
Ex 9.3 Question 1.

$\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

Answer.

Given: Differential equation $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$
$\Rightarrow d y=\frac{1-\cos x}{1+\cos x} d x$
Integrating both sides, $\int d y=\int \frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x$
$
\begin{aligned}
& \Rightarrow y=\int \tan ^2 \frac{x}{2} d x \\
& \Rightarrow \int\left(\sec ^2 \frac{x}{2}-1\right) d x \\
& \Rightarrow \frac{\tan \frac{x}{2}}{\frac{1}{2}}-x+c \\
& \Rightarrow y=2 \tan \frac{x}{2}-x+c
\end{aligned}
$

Ex 9.3 Question 2.

$\frac{d y}{d x}=\sqrt{4-y^2}(-2)$

Answer.

Given: Differential equation $\frac{d y}{d x}=\sqrt{4-y^2}$
$
\begin{aligned}
& \Rightarrow d y=\sqrt{4-y^2} d x \\
& \Rightarrow \frac{d y}{\sqrt{4-y^2}}=d x
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{d y}{\sqrt{4-y^2}} d y=\int 1 d x \\
& \Rightarrow \sin ^{-1} \frac{y}{2}=x+c\left[\because \int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \frac{x}{a}\right] \\
& \Rightarrow \frac{y}{2}=\sin (x+c) \\
& \Rightarrow y=2 \sin (x+c)
\end{aligned}
$

Ex 9.3 Question 3.

$\frac{d y}{d x}+y=1(y \neq 1)$

Answer.

Given: Differential equation $\frac{d y}{d x}+y=1$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=1-y \\
& \Rightarrow d y=(1-y) d x \\
& \Rightarrow d y=-(y-1) d x
\end{aligned}
$

$
\Rightarrow \frac{d y}{y-1}=-d x
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{d y}{y-1} d y=-\int 1 d x \\
& \Rightarrow \log |y-1|=-x+c \\
& \Rightarrow|y-1|=e^{-x+c}\left[\because \text { if } \log x=t \text {, then } x=e^x\right] \\
& \Rightarrow y-1= \pm e^{-x+c} \\
& \Rightarrow y=1 \pm e^{-x} e^c \\
& \Rightarrow y=1 \pm e^c e^{-x} \\
& \Rightarrow y=1+\mathrm{A} e^{-x}, \text { where } \mathrm{A}= \pm e^c
\end{aligned}
$

Ex 9.3 Question 4.

$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$

Answer.

Given: Differential equation $\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
Dividing by $\tan x \tan y$, we have
$
\begin{aligned}
& \frac{\sec ^2 x}{\tan x} d x+\frac{\sec ^2 y}{\tan y} d y=0 \\
& \Rightarrow \log |\tan x|+\log |\tan y|=\log c\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right] \\
& \Rightarrow \log |\tan x \tan y|=\log c \\
& \Rightarrow|\tan x \tan y|=c
\end{aligned}
$

$\Rightarrow \tan x \tan y= \pm c=\mathrm{C}[\because|t|=a(a \geq 0) \Rightarrow t= \pm a]$

Ex 9.3 Question 5.

$\left(e^x+e^{-x}\right) d y-\left(e^x-e^{-x}\right) d x=0$

Answer.

Given: Differential equation $\left(e^x+e^{-x}\right) d y-\left(e^x-e^{-x}\right) d x=0$
$
\begin{aligned}
& \Rightarrow\left(e^x+e^{-x}\right) d y=\left(e^x-e^{-x}\right) d x \\
& \Rightarrow d y=\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) d x
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int d y=\int\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) d x \\
& \Rightarrow y=\log \left|e^x+e^{-x}\right|+c\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right]
\end{aligned}
$

Ex 9.3 Question 6.

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

Answer.

Given: Differential equation $\frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right)$
$\Rightarrow d y=\left(1+x^2\right)\left(1+y^2\right) d x$
$\Rightarrow \frac{d y}{1+y^2}=\left(1+x^2\right) d x$ [Separating variables]
Integrating both sides,
$\Rightarrow \int \frac{1}{y^2+1} d y=\int\left(x^2+1\right) d x$

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

Ex 9.3 Question 7.

$y \log y d x-x d y=0$

Answer.

Given: Differential equation $y \log y d x-x d y=0$
$
\begin{aligned}
& \Rightarrow-x d y=-y \log y d x \\
& \Rightarrow \frac{d y}{y \log y}=\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$

Integrating both sides,
$
\Rightarrow \int \frac{d y}{y \log y}=\int \frac{d x}{x}
$

Putting $\log y=t$ on L.H.S., we get
$
\begin{aligned}
& \frac{1}{y}=\frac{d t}{d y} \\
& \Rightarrow \frac{d y}{y}=d t
\end{aligned}
$

Now eq. (i) becomes
$
\begin{aligned}
& \int \frac{d t}{t}=\int \frac{d x}{x} \\
& \Rightarrow \log |t|=\log |x|+\log |c|=\log |x c|
\end{aligned}
$
[If all the terms in the solution of a differential equation involve $\log$, it is better to use $\log c$ or $\log |c|$ instead of $c$ in the solution.]

$\begin{aligned}
& \Rightarrow|t|=|x c| \\
& \Rightarrow t= \pm x c
\end{aligned}$

$\begin{aligned}
& \Rightarrow \log y= \pm x c=a x \text { where } a= \pm c \\
& \Rightarrow y=e^{a x}
\end{aligned}$

Ex 9.3 Question 8.

$x^5 \frac{d y}{d x}=-y^5$

Answer.

Given: Differential equation $x^5 \frac{d y}{d x}=-y^5$
$
\begin{aligned}
& \Rightarrow x^5 d y=-y^5 d x \\
& \Rightarrow \frac{d y}{y^5}=-\frac{d x}{x^5} \text { [Separating variables] } \\
& \Rightarrow y^{-5} d y=-x^{-5} d x
\end{aligned}
$

Integrating both sides
$
\begin{aligned}
& \Rightarrow \int y^{-5} d y=-\int x^{-5} d x \\
& \frac{y^{-4}}{-4}=-\frac{x^{-4}}{-4}+c \\
& \Rightarrow y^{-4}=-x^{-4}-4 c \\
& \Rightarrow x^{-4}+y^{-4}=-4 c \\
& \Rightarrow y^{-4}=x^{-4}+y^{-4}=\mathrm{C}, \text { where } \mathrm{C}=-4 c
\end{aligned}
$

Ex 9.3 Question 9.

$\frac{d y}{d x}=\sin ^{-1} x$

Answer.

Given: Differential equation $\frac{d y}{d x}=\sin ^{-1} x$
$
\Rightarrow d y=\sin ^{-1} x d x
$

Integrating both sides, $\int 1 d y=\int \sin ^{-1} x d x$
$
\Rightarrow y=\int \sin ^{-1} x 1 d x
$

Applying product rule,
$
\begin{aligned}
& y=\left(\sin ^{-1} x\right) \int 1 d x-\int \frac{d}{d x}\left(\sin ^{-1} x\right) \int 1 d x d x \\
& =x \sin ^{-1} x-\int \frac{1}{\sqrt{1-x^2}} x d x \ldots \ldots . . \text { (i) }
\end{aligned}
$

To evaluate $\int \frac{x}{\sqrt{1-x^2}} d x=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x$
Putting, $1-x^2=t$, differentiate $-2 x d x=d t$
$
\Rightarrow \int \frac{x}{\sqrt{1-x^2}} d x=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\frac{1}{2} \int t^{-1 / 2} d t=-\frac{1}{2} \cdot \frac{t^{1 / 2}}{1 / 2}=-\sqrt{t}=-\sqrt{1-x^2}
$

Putting this value in eq. (i), the required general solution is
$
y=x \sin ^{-1} x+\sqrt{1-x^2}+c
$

Ex 9.3 Question 10.

$e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0$

Answer.

Given: Differential equation $e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0$
Dividing each term by $\left(1-e^x\right) \tan y$, we get
$
\frac{e^x}{1-e^x} d x+\frac{\sec ^2 y}{\tan y} d y=0 \text { [Separating variables] }
$

Integrating both sides,

$\begin{aligned}
& \int \frac{e^x}{1-e^x} d x+\int \frac{\sec ^2 y}{\tan y} d y=c \\
& \Rightarrow-\int \frac{-e^x}{1-e^x} d x+\log |\tan y|=c \\
& \Rightarrow-\log \left|1-e^x\right|+\log |\tan y|=c \\
& \Rightarrow \log \frac{|\tan y|}{\left|1-e^x\right|}=\log c^{\prime} \\
& \Rightarrow \frac{|\tan y|}{\left|1-e^x\right|}=c^{\prime} \\
& \Rightarrow \tan y=\mathrm{C}\left(1-e^x\right)\left[\because|t|=c^{\prime} \Rightarrow t= \pm c^{\prime}=\mathrm{C}(\mathrm{say})\right]
\end{aligned}$

For each of the differential equations in Question 11 to 14, find a particular solution satisfying the given condition:
Ex 9.3 Question 11.

$\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x, y=1$, when $x=0$

Answer.

Given: $\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x$
$
\begin{aligned}
& \Rightarrow\left(x^3+x^2+x+1\right) d y=\left(2 x^2+x\right) d x \\
& \Rightarrow d y=\frac{2 x^2+x}{x^3+x^2+x+1} d x \text { [Separating variables] } \\
& \Rightarrow d y=\frac{2 x^2+x}{x^2(x+1)+(x+1)} d x
\end{aligned}
$

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

Integrating both sides,
$
\begin{aligned}
& \int 1 d y=\int \frac{2 x^2+x}{(x+1)\left(x^2+1\right)} d x \\
& \Rightarrow y=\int \frac{2 x^2+x}{(x+1)\left(x^2+1\right)} d x
\end{aligned}
$

Let $\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}$ [Partial fraction] $\qquad$ (ii)
$
\Rightarrow 2 x^2+x=\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(x+1)
$

Comparing the coefficients of $x^2$ on both sides, $\mathrm{A}+\mathrm{B}=2$
Comparing the coefficients of $x$ on both sides, $\mathrm{B}+\mathrm{C}=1$ $\qquad$ .(iv)

Comparing constants on both sides, $\mathrm{A}+\mathrm{C}=0$ $\qquad$ .(v)

From eq. (iii) - (iv), we haveA - $\mathrm{C}=1$ $\qquad$ (vi)

Adding eq. (v) and (vi), we have $2 \mathrm{~A}=1$
$
\Rightarrow \mathrm{A}=\frac{1}{2}
$

From eq. (v), we have $\mathrm{C}=-\mathrm{A}=\frac{-1}{2}$
Putting the value of $\mathrm{C}$ in eq. (iv), $\mathrm{B}-\frac{1}{2}=1$

$\Rightarrow \mathrm{B}=1+\frac{1}{2}=\frac{3}{2}$

Putting the values of A, B, and C in eq. (ii), we have
$
\begin{aligned}
& \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{\frac{1}{2}}{x+1}+\frac{\frac{3}{2} x-\frac{1}{2}}{x^2+1} \\
& =\frac{1}{2} \cdot \frac{1}{x+1}+\frac{3}{2} \cdot \frac{x}{x^2+1}-\frac{1}{2} \cdot \frac{1}{x^2+1} \\
& =\frac{1}{2} \cdot \frac{1}{x+1}+\frac{3}{4} \cdot \frac{2 x}{x^2+1}-\frac{1}{2} \cdot \frac{1}{x^2+1}
\end{aligned}
$

Putting this value in eq. (i),
$
\begin{aligned}
& y=\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{4} \int \frac{2 x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x \\
& \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+c \\
& {\left[\because \int \frac{2 x}{x^2+1} d x=\int \frac{f^{\prime}(x)}{f(x)} d x=\log f(x)\right]}
\end{aligned}
$

Now, when $x=0, y=1$, putting these values in eq. (vii),
$
\begin{aligned}
& 1=\frac{1}{2} \log 1+\frac{3}{4} \log 1-\frac{1}{2} \tan ^{-1} 0+c \\
& \Rightarrow 1=c\left[\log 1=0, \tan ^{-1} 0=0\right]
\end{aligned}
$

Putting value of $c$ in eq. (vii), the required general solution is
$
y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+1
$

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

$\begin{aligned}
& \Rightarrow y=\frac{1}{4}\left[\log (x+1)^2+\log \left(x^2+1\right)^3\right]-\frac{1}{2} \tan ^{-1} x+1 \\
& \Rightarrow y=\frac{1}{4}\left[\log (x+1)^2\left(x^2+1\right)^3\right]-\frac{1}{2} \tan ^{-1} x+1
\end{aligned}$

Ex 9.3 Question 12.

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

Answer.

Given: Differential equation $x\left(x^2-1\right) \frac{d y}{d x}=1$
$
\Rightarrow d y=\frac{d x}{x\left(x^2-1\right)}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int 1 d y=\int \frac{1}{x\left(x^2-1\right)} d x \\
& \Rightarrow y=\int \frac{1}{x(x+1)(x-1)} d x+c \ldots \ldots \ldots .(i) \\
& \text { Let } \frac{1}{x(x+1)(x-1)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{x-1} \ldots \ldots \ldots .(\mathrm{ii}) \\
& \Rightarrow 1=\mathrm{A}(x+1)(x-1)+\mathrm{B} x(x-1)+\mathrm{C} x(x+1) \\
& \Rightarrow 1=\mathrm{A}\left(x^2-1\right)+\mathrm{B}\left(x^2-x\right)+\mathrm{C}\left(x^2+x\right) \\
& \Rightarrow 1=\mathrm{A} x^2-\mathrm{A}+\mathrm{B} x^2-\mathrm{B} x+\mathrm{C} x^2+\mathrm{C} x
\end{aligned}
$

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

Comparing the coefficients of $x^2$ on both sides, $\mathrm{A}+\mathrm{B}+\mathrm{C}=0$
Comparing the coefficients of $x$ on both sides, $-\mathrm{B}+\mathrm{C}=0$
$
\Rightarrow C=B
$

Comparing constants on both sides, $-\mathrm{A}=1$
$
\Rightarrow \mathrm{A}=-1
$

Putting $A=-1$ and $C=B$ in eq. (iii),
$
\begin{aligned}
& -1+B+B=0 \\
& \Rightarrow 2 B=1 \\
& \Rightarrow B=\frac{1}{2}
\end{aligned}
$

From eq. (iv), $C=B=\frac{1}{2}$
Putting the values of $A, B$ and $C$ in eq. (ii), we get
$
\begin{aligned}
& \frac{1}{x(x+1)(x-1)}=\frac{-1}{x}+\frac{\frac{1}{2}}{x+1}+\frac{\frac{1}{2}}{x-1} \\
& \Rightarrow \int \frac{1}{x(x+1)(x-1)} d x=-\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x \\
& \Rightarrow \int \frac{1}{x(x+1)(x-1)} d x=-\log |x|+\frac{1}{2} \log |x+1|+\frac{1}{2} \log |x-1|
\end{aligned}
$

$\begin{aligned}
& \Rightarrow \int \frac{1}{x(x+1)(x-1)} d x=\frac{1}{2}[2-\log |x|+\log |x+1|+\log |x-1|] \\
& \Rightarrow \int \frac{1}{x(x+1)(x-1)} d x=\frac{1}{2}\left[-\log |x|^2+\log |x+1||x-1|\right] \\
& \Rightarrow \int \frac{1}{x(x+1)(x-1)} d x=\frac{1}{2}\left[\log \frac{\left|x^2-1\right|}{|x|^2}=\frac{1}{2} \log \left|\frac{x^2-1}{x^2}\right|\right]
\end{aligned}$

Putting this value in eq. (i),
$
y=\frac{1}{2} \log \left|\frac{x^2-1}{x^2}\right|+c
$

Now, putting $y=0$, when $x=2$ in eq. (v), we get
$
0=\frac{1}{2} \log \frac{3}{4}+c \Rightarrow c=\frac{-1}{2} \log \frac{3}{4}
$

Putting the value of $c$ in eq. (v), the required general solution is
$
y=\frac{1}{2} \log \left|\frac{x^2-1}{x^2}\right|-\frac{1}{2} \log \frac{3}{4}
$
[NOTE: You can also do, to evaluate $\int \frac{1}{x\left(x^2-1\right)} d x=\int \frac{x}{x^2\left(x^2-1\right)} d x=$ $\frac{1}{2} \int \frac{2 x}{x^2\left(x^2-1\right)} d x$ Put $x^2=t$

Ex 9.3 Question 13.

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

Answer.

Given: Differential equation $\cos \frac{d y}{d x}=a(a \in \mathrm{R}) ; y=1$ when $x=0$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\cos ^{-1} a \\
& \Rightarrow d y=\left(\cos ^{-1} a\right) d x
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int 1 d y=\int\left(\cos ^{-1} a\right) d x \\
& \Rightarrow y=\left(\cos ^{-1} a\right) \int 1 d x
\end{aligned}
$

$
\Rightarrow y=\left(\cos ^{-1} a\right) x+c
$

Now putting $y=1$ when $x=0$ in eq. (i), we get $c=1$
Putting $c=1$ in eq. (i), $y=\left(\cos ^{-1} a\right) x+1$
$
\begin{aligned}
& \Rightarrow \frac{y-1}{x}=\cos ^{-1} a \\
& \Rightarrow \cos \left(\frac{y-1}{x}\right)=a
\end{aligned}
$

Ex 9.3 Question 14.

$\frac{d y}{d x}=y \tan x, y=1$ when $x=0$

Answer.

Given: Differential equation $\frac{d y}{d x}=y \tan x$
$
\begin{aligned}
& \Rightarrow d y=y \tan x d x \\
& \Rightarrow \frac{d y}{y}=\tan x d x \text { [Separating variables] }
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& =\int \frac{1}{y} d y=\int \tan x d x \\
& \Rightarrow \log |y|=\log |\sec x|+\log |c| \\
& \Rightarrow \log |y|=\log |c \sec x| \\
& \Rightarrow|y|=|c \sec x| \\
& \Rightarrow y= \pm c \sec x \\
& \Rightarrow y=\mathrm{C} \sec x \text { where } \pm c=\mathrm{C}
\end{aligned}
$

Now putting $y=1$ and $x=0$ in eq. (i), we get $1=\mathrm{C} \sec 0=\mathrm{C}$
Putting $\mathrm{C}=1 \mathrm{in}$ eq. (i), we get the required general solution
$
\Rightarrow y=\sec x
$

Ex 9.3 Question 15.

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

Answer.

Given: Differential equation $y^{\prime}=e^x \sin x$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=e^x \sin x \\
& \Rightarrow d y=e^x \sin x d x
\end{aligned}
$

Integrating both sides
$
\begin{aligned}
& \Rightarrow \int 1 d y=\int e^x \sin x d x \\
& \Rightarrow y=I+C \ldots \ldots . .(i)
\end{aligned}
$
where $\mathrm{I}=\int e^x \sin x d x$
Applying product rule,
$
\begin{aligned}
& \mathrm{I}=e^x(-\cos x)-\int e^x(-\cos x) d x \\
& \Rightarrow \mathrm{I}=-e^x \cos x+\int e^x \cos x d x
\end{aligned}
$

Again applying product rule,
$
\begin{aligned}
& \mathrm{I}=-e^x \cos x+e^x \sin x-\int e^x \sin x d x \\
& \Rightarrow \mathrm{I}=e^x(-\cos x+\sin x)-\mathrm{I} \text { [By eq. (ii)] } \\
& \Rightarrow 2 \mathrm{I}=e^x(\sin x-\cos x)
\end{aligned}
$

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

Putting this value of I in eq. (i), we get
$
y=\frac{1}{2} e^x(\sin x-\cos x)+c
$

Now putting $x=0$ and $y=0$ in eq. (iii)
$
\begin{aligned}
& 0=\frac{1}{2}(-1)+c \\
& \Rightarrow c=\frac{1}{2}
\end{aligned}
$

Putting the value of $c$ in eq. (iii), we get the required general solution
$
\begin{aligned}
& y=\frac{1}{2} e^x(\sin x-\cos x)+\frac{1}{2} \\
& \Rightarrow 2 y=e^x(\sin x-\cos x)+1 \\
& \Rightarrow 2 y-1=e^x(\sin x-\cos x)
\end{aligned}
$

Ex 9.3 Question 16.

For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$, find the solution curve passing through the point $(1,-1)$.

Answer.

Given: Differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$
$
\begin{aligned}
& \Rightarrow x y d y=(x+2)(y+2) d x \\
& \Rightarrow \frac{y}{y+2} d y=\frac{x+2}{x} d x \text { [Separating both sides] }
\end{aligned}
$

Integrating both sides

$
\begin{aligned}
& \Rightarrow \int \frac{y}{y+2} d y=\int \frac{x+2}{x} d x \\
& =>\int \frac{y+2-2}{y+2} d y=\int\left(\frac{x}{x}+\frac{2}{x}\right) d x \\
& =>\int \frac{y+2}{y+2}-\frac{2}{y+2} d y=\int\left(1+\frac{2}{x}\right) d x \\
& =>\int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x \\
& \Rightarrow y-2 \log |y+2|=x+2 \log |x|+c \\
& \Rightarrow y-x=\log (y+2)^2+\log (x)^2+c \\
& \Rightarrow y-x=\log (y+2)^2 \cdot(x)^2+c \ldots \ldots \ldots . .(i)
\end{aligned}
$

Now putting $x=1, y=-1$ in eq. (i),
$
\begin{aligned}
& -1-1=\log (1)+c \\
& \Rightarrow c=-2
\end{aligned}
$

Putting this value of $c$ in eq. (i) to get the required solution curve
$
\begin{aligned}
& y-x=\log (y+2)^2 \cdot(x)^2-2 \\
& \Rightarrow y-x+2=\log (y+2)^2 \cdot(x)^2
\end{aligned}
$

Ex 9.3 Question 17.

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

Answer.

Let $\mathrm{P}(x, y)$ be any point on the required curve.
According to the question, Slope of the tangent to the curve at $\mathrm{P}(x, y) \times y=x$

$
\begin{aligned}
& \Rightarrow \frac{d y}{d x} \cdot y=x \\
& \Rightarrow y d y=x d x
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int y d y=\int x d x \\
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+c \\
& \Rightarrow y^2=x^2+2 c \\
& \Rightarrow y^2=x^2+\mathrm{C} \text { where } \mathrm{C}=2 c
\end{aligned}
$

Now it is given that curve $y^2=x^2+\mathrm{C}$ passes through the point $(0,-2)$.
Therefore, putting $x=0$ and $y=-2$ in this equation, we get $\mathrm{C}=4$
Putting the value of $\mathrm{C}$ in the equation $y^2=x^2+\mathrm{C}$,
$
\begin{aligned}
& y^2=x^2+4 \\
& \Rightarrow y^2-x^2=4
\end{aligned}
$

Ex 9.3 Question 18.

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

Answer.

According to the question, slope of the tangent at any point $\mathrm{P}(x, y)$ of the required curve
=2. Slope of the line joining the point of contact $\mathrm{P}(x, y)$ to the given point $\mathrm{A}(-4,-3)$

$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=2\left(\frac{y-(-3)}{x-(-4)}\right) \\
& \Rightarrow \frac{d y}{d x}=2\left(\frac{y+3}{x+4}\right) \\
& \Rightarrow(x+4) d y=2(y+3) d x \\
& \Rightarrow \frac{d y}{y+3}=\frac{2}{x+4} d x \text { [Separating variables] }
\end{aligned}
$

Integrating both side,
$
\begin{aligned}
& \Rightarrow \int \frac{1}{y+3} d y=2 \int \frac{1}{x+4} d x \\
& \Rightarrow \log |y+3|=2 \log |x+4|+\log |c| \\
& \Rightarrow \log |y+3|=\log |x+4|^2+\log |c|=\log |c|(x+4)^2 \\
& \Rightarrow|y+3|= \pm|c|(x+4)^2 \\
& \Rightarrow y+3= \pm|c|(x+4)^2 \\
& \Rightarrow y+3=\mathrm{C}(x+4)^2 \text { where } \pm|c|=\mathrm{C} \ldots \ldots . . . \text { (i) }
\end{aligned}
$

Now it is given that curve (i) passes through the point $(-2,1)$.
Therefore, putting $x=-2$ and $y=1$ in eq. (i),
$
\begin{aligned}
& \Rightarrow 1+3=C(-2+4)^2 \\
& \Rightarrow 4=4 C
\end{aligned}
$

$\Rightarrow C=1$

Putting $\mathrm{C}=1$ in eq. (i), we get the required solution,
$
\begin{aligned}
& y+3=(x+4)^2 \\
& \Rightarrow(x+4)^2=y+3
\end{aligned}
$

Ex 9.3 Question 19.

The volume of the spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after $t$ seconds.

Answer.

Let $x$ be the radius of the spherical balloon at time $t$.
Given: Rate of change of volume of spherical balloon is constant $=k$ (say)
$
\begin{aligned}
& \Rightarrow \frac{d}{d t}\left(\frac{4 \pi}{3} x^3\right)=k \\
& \Rightarrow \frac{4 \pi}{3} 3 x^2 \cdot \frac{d x}{d t}=k \\
& \Rightarrow 4 \pi x^2 \frac{d x}{d t}=k \\
& \Rightarrow 4 \pi x^2 d x=k d t \quad \text { [Separating variables] }
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow 4 \pi \int x^2 d x=k \int 1 d t \\
& \Rightarrow 4 \pi \frac{x^3}{3}=k t+c \ldots \ldots . .(i)
\end{aligned}
$

Now it is given that initially radius is 3 units, when $t=0, x=3$.
Therefore, putting $t=0, x=3$ in eq. (i),
$
=>\frac{4 \pi \times 27}{3}=c
$

$
\Rightarrow c=36 \pi
$
$\qquad$
Again when $t=3 \mathrm{sec}$, then $x=6$ units
Therefore, putting $t=3$ and $x=6$ in eq. (i),
$
\begin{aligned}
& \Rightarrow \frac{4 \pi}{3}(6)^3=3 k+c \\
& \Rightarrow \frac{4 \pi}{3}(216)=3 k+36 \pi \text { [From eq. (ii)] } \\
& \Rightarrow 4 \pi(72)-36 \pi=36 \\
& \Rightarrow 288 \pi-36 \pi=3 k \Rightarrow 3 k=252 \pi \\
& \Rightarrow k=84 \pi \ldots . . . . .(\text { iii) }
\end{aligned}
$

Putting the value of $c$ and $k$ in eq. (i), we get
$
\begin{aligned}
& \frac{4 \pi}{3} x^3=84 \pi t+36 \pi \\
& \Rightarrow \frac{x^3}{3}=21 t+9 \\
& \Rightarrow x^3=63 t+27 \\
& \Rightarrow x=(63 t+27)^{\frac{1}{3}}
\end{aligned}
$

Ex 9.3 Question 20.

In a bank principal increases continuously at the rate of $r \%$ per year. Find the value of $r$ if Rs $\mathbf{1 0 0}$ double itself in $\mathbf{1 0}$ years. $\left(\log _\epsilon 2=0.6931\right)$

Answer.

Let $\mathrm{P}$ be the principal (amount) at the end of $t$ years.
According to the given condition, rate of increase of principal per year $=r \%$ (of principal)

$
\begin{aligned}
& \Rightarrow \frac{d \mathrm{P}}{d t}=\frac{r}{100} \times \mathrm{P} \\
& \Rightarrow \frac{d \mathrm{P}}{\mathrm{P}}=\frac{r}{100} d t \text { [Separating variables] }
\end{aligned}
$

Integrating both sides,
$
\log \mathrm{P}=\frac{r}{100} t+c
$
[Since P being principal $>0$, hence $\log |P|=\log P$ ]
Now initial principal $=\mathrm{Rs} 100$ (given), i.e., when $t=0$, then $\mathrm{P}=100$
Therefore, putting $t=0, \mathrm{P}=100$ in eq. (i), $\log 100=c$
$
\begin{aligned}
& \text { Putting } \log 100=c \text { in eq. (i), } \log \mathrm{P}=\frac{r}{100} t+\log 100 \\
& \Rightarrow \log \mathrm{P}-\log 100=\frac{r}{100} t \\
& \Rightarrow \log \frac{\mathrm{P}}{100}=\frac{r}{100} t \ldots . . . . . . \text { (ii) }
\end{aligned}
$

Now putting $\mathrm{P}=$ double of itself $=2 \times 100=\mathrm{Rs} 200$, when $t=10$ years (given)
$
\begin{aligned}
& \log \frac{200}{100}=\frac{r}{100} \times 10 \\
& \Rightarrow \log 2=\frac{r}{10} \\
& \Rightarrow r=10 \log 2
\end{aligned}
$

$
\Rightarrow 10 \times 0.6931=6.931 \%
$
Ex 9.3 Question 21.

In a bank principal increases at the rate of $5 \%$ per year. An amount of Rs 1000 is 

deposited with this bank, how much will it worth after 10 years? $\left(e^{0.5}=1.648\right)$
Answer.

Let $\mathrm{P}$ be the principal (amount) at the end of $t$ years.
According to the given condition, rate of increase of principal per year $=5 \%$ (of principal)
$
\begin{aligned}
& \Rightarrow \frac{d \mathrm{P}}{d t}=\frac{5}{100} \times \mathrm{P} \\
& \Rightarrow \frac{d \mathrm{P}}{d t}=\frac{\mathrm{P}}{20} \\
& \Rightarrow \frac{d \mathrm{P}}{\mathrm{P}}=\frac{d t}{20} \quad \text { [Separating variables] }
\end{aligned}
$

Integrating both sides,
$
\log \mathrm{P}=\frac{1}{20} t+c
$
[Since $P$ being principal $>0$, hence $\log |P|=\log P$ ]
Now initial principal $=$ Rs 1000 (given), i.e., when $t=0$, then $\mathrm{P}=1000$
Therefore, putting $t=0, \mathrm{P}=1000$ in eq. (i), $\log 1000=c$
Putting $\log 1000=c$ in eq. (i), $\log \mathrm{P}=\frac{1}{20} t+\log 1000$
$
\begin{aligned}
& \Rightarrow \log \mathrm{P}-\log 1000=\frac{1}{20} t \\
& \Rightarrow \log \frac{\mathrm{P}}{1000}=\frac{1}{20} t \ldots \ldots \ldots .(\text { ii) }
\end{aligned}
$

Now putting $t=10$ years (given)
$
\log \frac{P}{100}=\frac{1}{20} \times 10=\frac{1}{2}=0.5
$

$\begin{aligned}
& \Rightarrow \frac{\mathrm{P}}{1000}=e^{0.5}\left[\because \text { if } x=t \text {, then } x=e^t\right] \\
& \Rightarrow \mathrm{P}=1000 \times 1.648=\operatorname{Rs} 1648
\end{aligned}$

Ex 9.3 Question 22.

In a culture the bacteria count is $1,00,000$. The number is increased by $10 \%$ in 2 hours. In how many hours will the count reach $2,00,000$, if the rate of growth of bacteria is proportional to the number present.

Answer.

Let $x$ be the bacteria present in the culture at time $t$ hours.
According to the question,
Rate of growth of bacteria is proportional to the number present
$
\begin{aligned}
& \Rightarrow \frac{d x}{d t} \text { is proportional to } x \\
& \Rightarrow \frac{d x}{d t}=k x \text { where } k \text { is the constant of proportionality } \\
& \Rightarrow d x=k x d t \\
& \Rightarrow \frac{d x}{x}=k d t
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{1}{x} d x=k \int 1 d t \\
& \Rightarrow \log x=k t+c . .
\end{aligned}
$

Now it is given that initially the bacteria count is $x_0$ (say) $=1,00,000$
$\Rightarrow$ when $t=0$, then $x=x_0$
Putting these values in eq. (i)

$
\log x_0=c
$

Putting $\log x_0=c$ in eq. (i), we get
$
\begin{aligned}
& \Rightarrow \log x=k t+\log x_0 \\
& \Rightarrow \log x-\log x_0=k t \\
& \Rightarrow \log \frac{x}{x_0}=k t \ldots . . . . . . \text { (ii) }
\end{aligned}
$

Now it is given also that the number of bacteria increased by $10 \%$ in 2 hours.
Therefore, increase in bacteria in 2 hours $=\frac{10}{100} \times 100000=10,000$
$\therefore \mathrm{x}$, the amount of bacteria at $t=2=1,00,000+10,000=1,10,000=x_1$ (say)
Putting $x=x_1$ and $t=2$ in eq. (ii),
$
\begin{aligned}
& \log \frac{x_1}{x_0}=2 k \\
& \Rightarrow k=\frac{1}{2} \log \frac{x_1}{x_0} \\
& \Rightarrow k=\frac{1}{2} \log \frac{110000}{100000} \\
& \Rightarrow k=\frac{1}{2} \log \frac{11}{10}
\end{aligned}
$

Putting this value of $k$ in eq. (ii), we get,

$\log \frac{x}{x_0}=\frac{1}{2}\left(\log \frac{11}{10}\right) t$

$\begin{aligned}
& \Rightarrow \log \frac{200000}{100000}=\frac{1}{2}\left(\log \frac{11}{10}\right) t[\text { when } x=200000] \\
& \Rightarrow \log 2=\frac{1}{2}\left(\log \frac{11}{10}\right) t \\
& \Rightarrow 2 \log 2=\left(\log \frac{11}{10}\right) t \\
& \Rightarrow t=\frac{2 \log 2}{\log \frac{11}{10}} \text { hours }
\end{aligned}$

Ex 9.3 Question 23.

The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is:
(A) $e^x+e^{-y}=c$
(B) $e^x+e^y=c$
(C) $e^{-x}+e^y=c$
(D) $e^{-x}+e^{-y}=c$

Answer.

Given: Differential equation $\frac{d y}{d x}=e^{x+y}$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=e^x \cdot e^y \\
& \Rightarrow d y=e^x \cdot e^y d x \\
& \Rightarrow \frac{d y}{e^y}=e^x d x \text { [Separating variables] } \\
& \Rightarrow e^{-y} d y=e^x d x
\end{aligned}
$

Integrating both sides,

$
\begin{aligned}
& \int e^{-y} d y=\int e^x d x \\
& \Rightarrow \frac{e^{-y}}{-1}=e^x+c \\
& \Rightarrow-e^{-y}-e^x=c \\
& \Rightarrow e^{-y}+e^x=-c \\
& \Rightarrow e^{-y}+e^x=\mathrm{C} \text { where } \mathrm{C}=-c \text { which is required solution }
\end{aligned}
$

Therefore, option (A) is correct.