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

---

Share this to Friend on WhatsApp

NCERT Class 12 Maths Solutions: Chapter 9 - Differential Equations

Miscellaneous Exercise Question 1.

For each of the differential equations given below, indicate its order and degree (if defined):
(i) $\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2-6 y=\log x$
(ii) $\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2+7 y=\sin x$
(iii) $\frac{d^4 y}{d x^4}-\sin \left(\frac{d^3 y}{d x^3}\right)=0$

Answer.

(i) Given: Differential equation $\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2-6 y=\log x$
The highest order derivative present in this differential equation is $\frac{d^2 y}{d x^2}$ and hence order of this differential equation if 2 .

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d^2 y}{d x^2}$ is 1 .

Therefore, Order $=2$, Degree $=1$
(ii) Given: Differential equation $\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2+7 y=\sin x$

The highest order derivative present in this differential equation is $\frac{d y}{d x}$ and hence order of this differential equation if 1 .

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d y}{d x}$ is 3 .

Therefore, Order $=1$, Degree $=3$
(iii) Given: Differential equation $\frac{d^4 y}{d x^4}-\sin \left(\frac{d^3 y}{d x^3}\right)=0$

The highest order derivative present in this differential equation is $\frac{d^4 y}{d x^4}$ and hence order of this differential equation if 4 .

The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.

Therefore, Order = 4, Degree not defined

Miscellaneous Exercise Question 2.

For each of the exercises given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:
(i) $x y=a e^x+b e^{-x}+x^2: x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0$
(ii) $y=e^x(a \cos x+b \sin x): \frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0$
(iii) $y=x \sin 3 x: \frac{d^2 y}{d x^2}+9 y-6 \cos 3 x=0$
(iv) $x^2=2 y^2 \log y:\left(x^2+y^2\right) \frac{d y}{d x}-x y=0$

Answer.

(i) The given function is $x y=a e^x+b e^{-x}+x^2$
To verify: Function (i) is a solution of D.E. $\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0$
Differentiating both sides of eq. (i) w.r.t. $x, x \frac{d y}{d x}+y .1=a e^x+b e^{-x}+2 x$
Again differentiating both sides w.r.t. $x$,
$
\begin{aligned}
& x \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot 1+\frac{d y}{d x}=a e^x+b e^{-x}+2 \\
& \Rightarrow x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}=a e^x+b e^{-x}+2
\end{aligned}
$

Putting $a e^x+b e^{-x}=x y-x^2$ from eq. (i), we have,
$
\begin{aligned}
& x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}=x y-x^2+2 \\
& \Rightarrow \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0
\end{aligned}
$

Therefore, Function given by eq. (i) is a solution of D.E. (ii).
(ii) The given function is $y=e^x(a \cos x+b \sin x)$

To verify: Function given by (i) is a solution of D.E. $\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0$
From (i), $\frac{d y}{d x}=\frac{d}{d x} e^x(a \cos x+b \sin x)+e^x \frac{d}{d x}(a \cos x+b \sin x)$

$\Rightarrow \frac{d y}{d x}=e^x(a \cos x+b \sin x)+e^x(-a \sin x+b \cos x)$

$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=y+e^x(-a \sin x+b \cos x) \text { [By eq. (i)] .........(iii) } \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x}+e^x(-a \sin x+b \cos x)+e^x(-a \cos x-b \sin x) \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x}+e^x(-a \sin x+b \cos x)-e^x(a \cos x+b \sin x) \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x}+\left(\frac{d y}{d x}-y\right)-y[\text { Using eq. (iii) and (i)] } \\
& \Rightarrow \frac{d^2 y}{d x^2}=2 \frac{d y}{d x}-2 y \\
& \Rightarrow \frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0
\end{aligned}
$

Therefore, Function given by eq. (i) is a solution of D.E. (ii).
(iii) The given function is $y=x \sin 3 x$ $\qquad$
To verify: Function given by eq. (i) is a solution of D.E. $\frac{d^2 y}{d x^2}+9 y-6 \cos 3 x=0$
From eq. (i), $\frac{d y}{d x}=x \cos 3 x \cdot 3+\sin 3 x .1$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=3 x \cos 3 x+\sin 3 x \\
& \Rightarrow \frac{d^2 y}{d x^2}=3[x(-\sin 3 x) 3+\cos 3 x \cdot 1]+(\cos 3 x) 3
\end{aligned}
$

$\Rightarrow \frac{d^2 y}{d x^2}=-9 x \sin 3 x+6 \cos 3 x=-9 y+6 \cos 3 x \quad \text { [Using eq. (i)] }$

$
\Rightarrow \frac{d^2 y}{d x^2}+9 y-6 \cos 3 x=0
$

Therefore, Function given by eq. (i) is a solution of D.E. (ii).
(iv) The given function is $x^2=2 y^2 \log y$

To verify: Function given by eq. (i) is a solution of D.E. $\left(x^2+y^2\right) \frac{d y}{d x}-x y=0$
Differentiating both sides of eq. (i) w.r.t. $x$,
$
\begin{aligned}
& 2 x=2\left[y^2 \cdot \frac{1}{y} \frac{d y}{d x}+(\log y) 2 y \frac{d y}{d x}\right] \\
& \Rightarrow x=\frac{d y}{d x}(y+2 y \log y) \\
& \Rightarrow \frac{d y}{d x}=\frac{x}{y+2 y \log y}=\frac{x}{y(1+2 \log y)}
\end{aligned}
$

Putting $2 \log y=\frac{x^2}{y^2}$ from eq. (i), we get
$
\begin{aligned}
& \frac{d y}{d x}=\frac{x}{y\left(1+\frac{x^2}{y^2}\right)}=\frac{x}{y\left(\frac{y^2+x^2}{y^2}\right)}=\frac{x y^2}{y\left(x^2+y^2\right)} \\
& \Rightarrow \frac{d y}{d x}=\frac{x y}{x^2+y^2} \\
& \Rightarrow\left(x^2+y^2\right) \frac{d y}{d x}=x y
\end{aligned}
$

$\Rightarrow\left(x^2+y^2\right) \frac{d y}{d x}-x y=0$

$\text { Therefore, Function given by eq. (i) is a solution of D.E. (ii). }$

Miscellaneous Exercise Question 3.

Form the differential equation representing the family of curves $(x-a)^2+2 y^2=a^2$, where $a$ ia an arbitrary constant.

Answer.

Equation of the given family of curves is $(x-a)^2+2 y^2=a^2$
$
\begin{aligned}
& \Rightarrow x^2+a^2-2 a x+2 y^2=a^2 \\
& \Rightarrow x^2-2 a x+2 y^2=0 \\
& \Rightarrow x^2+2 y^2=2 a x
\end{aligned}
$

Here number of arbitrary constants is one only $(a)$.
So, we will differentiate both sides of equation only once, w.r.t. $x$,
$
\begin{aligned}
& \Rightarrow 2 x+2.2 y \frac{d y}{d x}=2 a \\
& \Rightarrow 2 x+4 y \frac{d y}{d x}=2 a
\end{aligned}
$

Dividing eq. (i) by eq. (ii), we have
$
\begin{aligned}
& \Rightarrow \frac{x^2+2 y^2}{2 x+4 y \frac{d y}{d x}}=\frac{2 a x}{2 a} \\
& \Rightarrow \frac{x^2+2 y^2}{2 x+4 y \frac{d y}{d x}}=x \\
& \Rightarrow x\left(2 x+4 y \frac{d y}{d x}\right)=x^2+2 y^2
\end{aligned}
$

$\begin{aligned}
& \Rightarrow 2 x^2+4 x y \frac{d y}{d x}=x^2+2 y^2 \\
& \Rightarrow 4 x y \frac{d y}{d x}=2 y^2-x^2 \\
& \Rightarrow \frac{d y}{d x}=\frac{2 y^2-x^2}{4 x y}
\end{aligned}$

Miscellaneous Exercise Question 4.

Prove that $x^2-y^2=c\left(x^2+y^2\right)^2$ is the general equation of the differential equation $\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y$, where $c$ is a parameter.

Answer.

Given: Differential equation $\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y$
Here each coefficient of $d x$ and $d y$ is of same degree, i.e., 3, therefore differential equation looks to be homogeneous.
$
\begin{aligned}
& \therefore \frac{d y}{d x}=\frac{\left(x^3-3 x y^2\right)}{\left(y^3-3 x^2 y\right)} \\
& \Rightarrow \frac{d y}{d x}=\frac{1-3\left(\frac{y}{x}\right)^2}{\left(\frac{y}{x}\right)^3-3\left(\frac{y}{x}\right)}=f\left(\frac{y}{x}\right)
\end{aligned}
$

Therefore, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$

Putting these values in eq. (ii),
$
\begin{aligned}
& \Rightarrow v+x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}-v=\frac{1-3 v^2-v^4+3 v^2}{v^3-3 v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{1-v^4}{v^3-3 v} \\
& \Rightarrow x\left(v^3-3 v\right) d v=\left(1-v^4\right) d x \\
& \Rightarrow \frac{v^3-3 v}{1-v^4} d v=\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$

Integrating both sides
$
\int \frac{v^3-3 v}{1-v^4} d v=\int \frac{1}{x} d x=\log x+\log c
$

Now forming partial fraction of $\frac{v^3-3 v}{1-v^4}$
$
\begin{aligned}
& \frac{v^3-3 v}{1-v^4}=\frac{v^3-3 v}{\left(1-v^2\right)\left(1+v^2\right)} \\
& \Rightarrow \frac{v^3-3 v}{1-v^4}=\frac{v^3-3 v}{(1-v)(1+v)\left(1+v^2\right)} \\
& \Rightarrow \frac{v^3-3 v}{1-v^4}=\frac{\mathrm{A}}{(1-v)}=\frac{\mathrm{B}}{(1+v)}=\frac{\mathrm{C} v+\mathrm{D}}{\left(1+v^2\right)}
\end{aligned}
$

$\begin{aligned}
& \Rightarrow v^3-3 v=\mathrm{A}(1+v)\left(1+v^2\right)+\mathrm{B}(1-v)\left(1+v^2\right)+(\mathrm{C} v+\mathrm{D})\left(1-v^2\right) \\
& \Rightarrow v^3-3 v=\mathrm{A}\left(1+v^2+v+v^3\right)+\mathrm{B}\left(1+v^2-v-v^3\right)+\mathrm{C} v-\mathrm{C} v^3+\mathrm{D}-\mathrm{D} v^2
\end{aligned}$

Comparing coefficients of like powers of
$
\begin{aligned}
& v^3 \mathrm{~A}-\mathrm{B}-\mathrm{C}=1 \\
& v^2 \mathrm{~A}+\mathrm{B}-\mathrm{D}=0 \\
& v \mathrm{~A}-\mathrm{B}+\mathrm{C}=-3
\end{aligned}
$

Constants $\mathrm{A}+\mathrm{B}+\mathrm{D}=0$ (viii)

Now eq. (v) - eq. (vii)
$
\begin{aligned}
& \Rightarrow-2 C=4 \\
& \Rightarrow C=-2
\end{aligned}
$

Eq. (vi) - eq. (viii)
$
\Rightarrow-2 \mathrm{D}=0 \Rightarrow \mathrm{D}=0
$

Putting $\mathrm{C}=-2$ in eq. (v), $\mathrm{A}-\mathrm{B}+2=1$
$
\Rightarrow \mathrm{A}-\mathrm{B}=-1
$

Putting $\mathrm{D}=0$ in eq. (vi) $\mathrm{A}+\mathrm{B}=0$
Adding eq. (ix) and (x) $2 A=-1$
$
\Rightarrow \mathrm{A}=\frac{-1}{2}
$

From eq. (x), $B=-A=\frac{1}{2}$
Putting the values of A, B, C and D in eq. (iv), we have

$\frac{v^3-3 v}{1-v^4}=\frac{\frac{-1}{2}}{1-v}+\frac{\frac{1}{2}}{1+v}-\frac{2 v}{1+v^2}$

$
\begin{aligned}
& \Rightarrow \int \frac{v^3-3 v}{1-v^4} d v=\frac{-1}{2} \frac{\log (1-v)}{-1}+\frac{1}{2} \log (1+v)-\log \left(1+v^2\right) \\
& \Rightarrow \int \frac{v^3-3 v}{1-v^4} d v=\frac{1}{2} \log (1-v)+\frac{1}{2} \log (1+v)-\log \left(1+v^2\right) \\
& \Rightarrow \int \frac{v^3-3 v}{1-v^4} d v=\frac{1}{2}[\log (1-v)+\log (1+v)]-\log \left(1+v^2\right) \\
& \Rightarrow \int \frac{v^3-3 v}{1-v^4} d v=\frac{1}{2}[\log (1-v)(1+v)]-\log \left(1+v^2\right) \\
& \Rightarrow \int \frac{v^3-3 v}{1-v^4} d v=\log \left(1-v^2\right)^{\frac{1}{2}}-\log \left(1+v^2\right) \\
& =\log \left(\frac{\sqrt{1-v^2}}{1+v^2}\right)
\end{aligned}
$

Putting this value in eq. (iii),
$
\begin{aligned}
& \Rightarrow \log \left(\frac{\sqrt{1-v^2}}{1+v^2}\right)=\log x c \\
& \Rightarrow \frac{\sqrt{1-v^2}}{1+v^2}=x c
\end{aligned}
$

Squaring both sides and cross-multiplying,
$
\Rightarrow 1-v^2=c^2 x^2\left(1+v^2\right)^2
$

Putting $v=\frac{y}{x}$,

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

$\begin{aligned}
& \Rightarrow \frac{x^2-y^2}{x^2}=c^2 x^2 \frac{\left(x^2+y^2\right)^2}{x^4} \\
& \Rightarrow \frac{x^2-y^2}{x^2}=c^2 \frac{\left(x^2+y^2\right)^2}{x^2} \\
& \Rightarrow x^2-y^2=\mathrm{C}\left(x^2+y^2\right)^2 \text { where } c^2=\mathrm{C}
\end{aligned}$

Miscellaneous Exercise Question 6.

Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$

Answer.

Given: Differential Equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}} \\
& \Rightarrow \sqrt{1-x^2} d y=-\sqrt{1-y^2} d x \\
& \Rightarrow \frac{d y}{\sqrt{1-y^2}}=\frac{-d x}{\sqrt{1-x^2}}
\end{aligned}
$

Integrating both sides
$
\begin{aligned}
& \Rightarrow \int \frac{1}{\sqrt{1-y^2}} d y=-\int \frac{1}{\sqrt{1-x^2}} d x \\
& \Rightarrow \sin ^{-1} \mathrm{y}=-\sin ^{-1} \mathrm{x}+\mathrm{c} \\
& \Rightarrow \sin ^{-1} x+\sin ^{-1} y=c
\end{aligned}
$
Miscellaneous Exercise Question 7.

Show that the general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ 

given by $(x+y+1)=A(1-x-y-2 x y)$, where $A$ is parameter.
Answer.

Given: Differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=-\left(\frac{y^2+y+1}{x^2+x+1}\right) \\
& \Rightarrow \frac{d y}{y^2+y+1}=\frac{-d x}{x^2+x+1} \\
& \Rightarrow \frac{d y}{y^2+y+1}+\frac{d x}{x^2+x+1}=0
\end{aligned}
$

Integrating both sides,
$
\Rightarrow \int \frac{1}{y^2+y+1} d y+\int \frac{1}{x^2+x+1} d x=0
$

Now $y^2+y+1=y^2+y+\frac{1}{4}-\frac{1}{4}+1 \quad$ [Completing the squares]
$
\Rightarrow y^2+y+1=\left(y+\frac{1}{2}\right)^2+\frac{3}{4}=\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2
$

Therefore, $\int \frac{1}{y^2+y+1} d y=\int \frac{1}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d y$

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

Similarly, $\int \frac{1}{x^2+x+1} d x=\frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}$
Putting these values in eq. (i),
$
\begin{aligned}
& \frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 y+1}{\sqrt{3}}+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}=c \\
& \Rightarrow \tan ^{-1} \frac{2 y+1}{\sqrt{3}}+\tan ^{-1} \frac{2 x+1}{\sqrt{3}}=\frac{\sqrt{3}}{2} c \text { [Multiplying by } \frac{\sqrt{3}}{2} \text { ] } \\
& \Rightarrow \tan ^{-1} \frac{\frac{2 x+1}{\sqrt{3}}+\frac{2 y+1}{\sqrt{3}}}{1-\frac{2 x+1}{\sqrt{3}} \cdot \frac{2 y+1}{\sqrt{3}}}=\tan ^{-1} c^{\prime} \text { where } c^{\prime}=\frac{\sqrt{3}}{2} c \\
& \Rightarrow \frac{\sqrt{3}(2 x+2 y+2)}{3-(4 x y+2 x+2 y+1)}=c^{\prime} \\
& \Rightarrow \sqrt{3}(2 x+2 y+2)=c^{\prime}(2-2 x-2 y-4 x y) \\
& \Rightarrow 2 \sqrt{3}(x+y+1)=2 c^{\prime}(1-x-y-2 x y)
\end{aligned}
$
dividing by $2 \sqrt{3}$
$
\begin{aligned}
& x+y+1=\frac{c^{\prime}}{\sqrt{3}}(1-x-y-2 x y) \\
& \Rightarrow x+y+1=\mathrm{A}(1-x-y-2 x y) \text { where } \mathrm{A}=\frac{c^{\prime}}{\sqrt{3}}
\end{aligned}
$

Miscellaneous Exercise Question 8.

Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is $\sin x \cos y d x+\operatorname{cox} \sin y d y=0$

Answer.

Given: Differential equation $\sin x \cos y d x+\cos x \sin y d y=0$

$
\begin{aligned}
& \Rightarrow \sin x \cos y d x=-\cos x \sin y d y \\
& \Rightarrow \frac{\sin x}{\cos x} d x=\frac{-\sin y}{\cos y} d y \\
& \Rightarrow \tan x d x=-\tan y d y
\end{aligned}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \tan x d x=-\int \tan y d y \\
& \Rightarrow \log |\sec x|=-\log |\sec y|+\log |c| \\
& \Rightarrow \log |\sec x|+\log |\sec y|=\log |c| \\
& \Rightarrow \log |\sec x \sec y|=\log |c| \\
& \Rightarrow \sec x \sec y=c
\end{aligned}
$

Now, curve (i) passes through $\left(0, \frac{\pi}{4}\right)$.
Therefore, putting $x=0, y=\frac{\pi}{4}$ in eq. (i),
$
\begin{aligned}
& \Rightarrow \sec 0 \sec \frac{\pi}{4}=c \\
& \Rightarrow c=\sqrt{2}
\end{aligned}
$

Putting $c=\sqrt{2}$ in eq. (i),
$
\Rightarrow \sec x \sec y=\sqrt{2}
$

$\Rightarrow \frac{\sec x}{\cos y}=\sqrt{2}$

$\Rightarrow \cos y=\frac{\sec x}{\sqrt{2}}$

Miscellaneous Exercise Question 9.

Find the particular solution of the differential equation $\left(1+e^{2 x}\right) d y+\left(1+y^2\right) e^x d x=0$ given that $y=1$ when $x=0$.

Answer.

Given: Differential equation $\left(1+e^{2 x}\right) d y+\left(1+y^2\right) e^x d x=0$
Dividing every term by $\left(1+y^2\right)\left(1+e^{2 x}\right)$, we have
$
\Rightarrow \frac{d y}{1+y^2}+\frac{e^x}{1+e^{2 x}} d x=0
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{1}{1+y^2} d y+\int \frac{e^x}{1+e^{2 x}} d x=c \\
& \Rightarrow \tan ^{-1} y+\int \frac{e^x}{1+e^{2 x}} d x=c
\end{aligned}
$

Now to evaluate $\int \frac{e^x}{1+e^{2 x}} d x$,
putting $e^x=t$
$
\begin{aligned}
& \Rightarrow e^x=\frac{d t}{d x} \Rightarrow e^x d x=d t \\
& \therefore \int \frac{e^x}{1+e^{2 x}} d x=\int \frac{d t}{1+t^2}=\tan ^{-1} t=\tan ^{-1} e^x
\end{aligned}
$

Putting this value in eq. (i), $\tan ^{-1} y+\tan ^{-1} e^x=c$
Now putting $y=1, x=0$ in eq. (ii),

$
\begin{aligned}
& \Rightarrow \tan ^{-1} 1+\tan ^{-1} e^0=c \\
& \Rightarrow \tan ^{-1} 1+\tan ^{-1} 1=c \\
& \Rightarrow \frac{\pi}{4}+\frac{\pi}{4}=c \\
& \Rightarrow c=\frac{\pi}{2}
\end{aligned}
$

Putting $c=\frac{\pi}{2}$ in eq. (ii),
$
\Rightarrow \tan ^{-1} y+\tan ^{-1} e^x=\frac{\pi}{2}
$

Miscellaneous Exercise Question 10.

Solve the differential equation: $y e^{x / y} d x=\left(x e^{x / y}+y^2\right) d y(y \neq 0)$

Answer.

Given: Differential equation $y \cdot e^{x / y} d x=\left(x \cdot e^{x / y}+y^2\right) d y, y \neq 0$
$
\begin{aligned}
& \Rightarrow \frac{d x}{d y}=\frac{x \cdot e^{x / y}+y^2}{y \cdot e^{x / y}}=\frac{x \cdot e^{x / y}}{y \cdot e^{x / y}}+\frac{y^2}{y \cdot e^{x / y}} \\
& \Rightarrow \frac{d x}{d y}=\frac{x}{y}+y \cdot e^{x / y} \ldots \ldots \ldots . \text { (i) }
\end{aligned}
$

It is not a homogeneous differential equation because of presence of only $y$ as a factor, yet it can be solved by putting $\frac{x}{y}=v$, i.e., $x=v y$.
$
\Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}
$

Putting these values in eq. (i), we get

$
\begin{aligned}
& \Rightarrow v+y \frac{d v}{d y}=v+y e^{-v} \\
& \Rightarrow y \frac{d v}{d y}=y \cdot e^{-v} \\
& \Rightarrow y \frac{d v}{d y}=\frac{y}{e^v} \\
& \Rightarrow e^v d v=d y
\end{aligned}
$

Integrate both sides, we get
$
\begin{aligned}
& \Rightarrow e^v=y+c \\
& \Rightarrow e^{x / y}=y+c
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow-\frac{d y}{d x}=\frac{d t}{d x}-1 \\
& \Rightarrow \frac{d y}{d x}=\frac{-d t}{d x}+1
\end{aligned}
$

Putting this value in eq. (i),
$
\begin{aligned}
& \Rightarrow \frac{-d t}{d x}+1=-\left(\frac{t-1}{t+1}\right) \\
& \Rightarrow \frac{-d t}{d x}=-1-\left(\frac{t-1}{t+1}\right) \\
& \Rightarrow \frac{d t}{d x}=1+\left(\frac{t-1}{t+1}\right)=\frac{t+1+t-1}{t+1} \\
& \Rightarrow \frac{d t}{d x}=\frac{2 t}{t+1} \\
& \Rightarrow(t+1) d t=2 t d x \\
& \Rightarrow \frac{t+1}{t} d t=2 d x
\end{aligned}
$

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

$\Rightarrow t+\log |t|=2 x+c$

Putting $x-y=t$,
$
\begin{aligned}
& \Rightarrow x-y+\log |x-y|=2 x+c \\
& \Rightarrow \log |x-y|=x+y+c \ldots . . . .
\end{aligned}
$

Now putting $y=-1, x=0$ in eq. (ii),
$
\begin{aligned}
& \Rightarrow \log 1=0-1+c \\
& \Rightarrow 0=-1+c \\
& \Rightarrow c=1
\end{aligned}
$

Putting $c=1$ in eq. (ii),
$
\Rightarrow \log |x-y|=x+y+1
$

Miscellaneous Exercise Question 12.

Solve the differential equation: $\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}=1(x \neq 0)$

Answer.

Given: Differential equation $\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}=1$
$
\begin{aligned}
& \Rightarrow \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}=\frac{d y}{d x} \\
& \Rightarrow \frac{d y}{d x}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}
\end{aligned}
$

Comparing this equation with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
$
\mathrm{P}=\frac{1}{\sqrt{x}} \text { and } \mathrm{Q}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}
$

$
\begin{aligned}
& \int \mathrm{P} d x=\int \frac{1}{\sqrt{x}} d x=\int x^{-1 / 2} d x=\frac{x^{1 / 2}}{1 / 2}=2 \sqrt{x} \\
& \Rightarrow \text { I.F. }=e^{\lceil\mathrm{P} \alpha x}=e^{2 \sqrt{x}}
\end{aligned}
$

The general solution is
$
\begin{aligned}
& \Rightarrow \mathrm{y} \text { (I.F.) }=\int \mathrm{Q}(\text { I.F.) }) d x+c \\
& \Rightarrow y e^{2 \sqrt{x}}=\int \frac{e^{-2 \sqrt{x}}}{\sqrt{x}} e^{2 \sqrt{x}} d x+c \\
& \Rightarrow y e^{2 \sqrt{x}}=\int \frac{1}{\sqrt{x}} d x+c \\
& \Rightarrow y e^{2 \sqrt{x}}=2 \sqrt{x}+c \\
& \Rightarrow y=e^{-2 \sqrt{x}}(2 \sqrt{x}+c)
\end{aligned}
$

Miscellaneous Exercise Question 13.

Find the particular solution of the differential equation $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x(x \neq 0)$ given that $y=0$ when $x=\frac{\pi}{2}$.

Answer.

Given: Differential equation $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x$
Comparing this equation with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$,
$
\begin{aligned}
& \mathrm{P}=\cot x \text { and } \mathrm{Q}=4 x \operatorname{cosec} x \\
& \int \mathrm{P} d x=\int \cot x d x=\log \sin x \\
& \Rightarrow \text { I.F. }=e^{\int \mathrm{P} d x}=e^{\mathrm{bg} \sin x}=\sin x
\end{aligned}
$

The general solution is
$
\Rightarrow \mathrm{y} \text { (I.F.) }=\int \mathrm{Q}(\text { I.F. }) d x+c
$

$
\begin{aligned}
& \Rightarrow y(\sin x)=\int 4 x \operatorname{cosec} x \sin x d x+c \\
& \Rightarrow y(\sin x)=4 \int x \frac{1}{\sin x} \sin x d x+c \\
& \Rightarrow y(\sin x)=4 \int x d x+c=4 \cdot \frac{x^2}{2}+c \\
& \Rightarrow y \sin x=2 x^2+c
\end{aligned}
$

Now putting $y=0, x=\frac{\pi}{2}$ in eq. (i),
$
\begin{aligned}
& \Rightarrow 0=2 \cdot \frac{\pi^2}{4}+c \\
& \Rightarrow c=\frac{-\pi^2}{2}
\end{aligned}
$

Putting $c=\frac{-\pi^2}{2}$ in eq. (i),
$
\Rightarrow y \sin x=2 x^2-\frac{\pi^2}{2}
$

Miscellaneous Exercise Question 14.

Find the particular solution of the differential equation $(x+1) \frac{d y}{d x}=2 e^{-y}-1$ given that $y=0$ when $x=0$.

Answer.

Given: Differential equation $(x+1) \frac{d y}{d x}=2 e^{-y}-1$
$
\begin{aligned}
& \Rightarrow(x+1) \frac{d y}{d x}=\frac{2}{e^y}-1=\frac{2-e^y}{e^y} \\
& \Rightarrow(x+1) e^y d y=\left(2-e^y\right) d x
\end{aligned}
$

$
\Rightarrow \frac{e^y d y}{2-e^y}=\frac{d x}{x+1}
$

Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{e^y}{2-e^y} d y=\int \frac{1}{x+1} d x \\
& \text { Putting } e^y=t \\
& \Rightarrow e^y=\frac{d t}{d y} \\
& \Rightarrow e^y d y=d t \\
& \therefore \int \frac{d t}{2-t}=\log |x+1| \\
& \Rightarrow \frac{\log |2-t|}{-1}=\log |x+1|+c
\end{aligned}
$

Putting $e^y=t$,
$
\begin{aligned}
& \Rightarrow-\log \left|2-e^y\right|=\log |x+1|+c \\
& \Rightarrow \log |x+1|+\log \left|2-e^y\right|=-c \\
& \Rightarrow \log |x+1|\left|2-e^y\right|=-c \\
& \Rightarrow|x+1|\left|2-e^y\right|=e^{-c} \\
& \Rightarrow(x+1)\left(2-e^y\right)= \pm e^{-c} \\
& \Rightarrow(x+1)\left(2-e^y\right)=\mathrm{C} \text { where } \mathrm{C}= \pm e^{-c}
\end{aligned}
$

$\text { Putting } x=0, y=0 \text { in eq. (i), }$

$
\begin{aligned}
& \Rightarrow(1)(2-1)=c \\
& \Rightarrow c=1
\end{aligned}
$

Putting $C=1$ in eq. (i),
$
\Rightarrow(x+1)\left(2-e^y\right)=1
$

This solution may be written as
$
\begin{aligned}
& \Rightarrow 2-e^y=\frac{1}{x+1} \\
& \Rightarrow e^y=2-\frac{1}{x+1}=\frac{2 x+1}{x+1} \\
& \Rightarrow \log e^y=\log \left(\frac{2 x+1}{x+1}\right) \\
& \Rightarrow y=\log \left(\frac{2 x+1}{x+1}\right)
\end{aligned}
$
where expresses $y$ as an explicit function of $x$.

Choose the correct answer:
Miscellaneous Exercise Question 16.

The general solution of the differential equation $\frac{y d x-x d y}{y}=0$ is:
(A) $x y=\mathrm{C}$
(B) $x=\mathrm{C} y^2$
(C) $y=\mathrm{C} x$
(D) $y=\mathrm{C} x^2$

Answer.

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

Integrating both sides, $\log |x|=\log |y|+\log |c|$
$
\begin{aligned}
& \Rightarrow \log |x|=\log |c y| \\
& \Rightarrow x= \pm c y \\
& \Rightarrow y= \pm \frac{1}{c} x
\end{aligned}
$

$
\Rightarrow y=\mathrm{C} x \text { where } \mathrm{C}= \pm \frac{1}{\mathrm{c}}
$

Therefore, option (C) is correct.

Miscellaneous Exercise Question 17.

  $\text { The general equation of a differential equation of the type } \frac{d y}{d x}+P_1 x=Q_1 \text { is: }$

Answer.

$\text {We know that general solution of differential equation of the type } \frac{d y}{d x}=\mathrm{P}_1 x=\mathrm{Q}_1 \text { is }$

Therefore, option (C) is correct. 

Miscellaneous Exercise Question 18.

The general solution of the differential equation $e^x d y+\left(y e^x+2 x\right) d x=0$ is:
(A) $x e^y+x^2=\mathrm{C}$
(B) $x e^y+y^2=\mathrm{C}$
(C) $y e^x+x^2=\mathrm{C}$

(D) $y e^x+x^2=\mathrm{C}$

Answer.

Given: Differential equation $e^x d y+\left(y e^x+2 x\right) d x=0$
$
\begin{aligned}
& \Rightarrow e^x \frac{d y}{d x}+y e^x+2 x=0 \\
& \Rightarrow e^x \frac{d y}{d x}+y e^x=-2 x \\
& \Rightarrow \frac{d y}{d x}+y=\frac{-2 x}{e^x}
\end{aligned}
$

Comparing with $\frac{d y}{d x}=\mathrm{P} x=\mathrm{Q} \quad \mathrm{P}=1$ and $\mathrm{Q}=\frac{\frac{-2 x}{y}}{\frac{x}{x}}$
$
\int \mathrm{P} d x=\int 1 d x=x \quad \text { I.F. }=e^{\int \mathrm{P} d x}=e^x
$

Solution is $y($ I.F. $)=\int \mathrm{Q}($ I.F. $) d x+\mathrm{C}$
$
\begin{aligned}
& \Rightarrow y e^x=\int \frac{-2 x}{e^x} e^x d x+\mathrm{C} \\
& \Rightarrow y e^x=-2 \int x d x+\mathrm{C} \\
& \Rightarrow y e^x=-2 \frac{x^2}{2}+\mathrm{C} \\
& \Rightarrow y e^x=-x^2+\mathrm{C} \\
& \Rightarrow y e^x+x^2=\mathrm{C}
\end{aligned}
$

Therefore, option (C) is correct.