Exercise 9.3 (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
You can Download the Exercise 9.3 (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends
Share this to Friend on WhatsApp
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.
