RD Sharma Class 12 Exercise 21.3 Differential Equation Solutions Maths

RD Sharma Class 12 Exercise 21.3 Differential Equation Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 04:19 PM IST

The students of class 12 have a packed schedule for doing their homework, preparing for public exams and other tests & examinations too. For subjects like mathematics, the RD Sharma Class 12th exercise 21.3 book will be extremely helpful.

RD Sharma class 12 solutions Differentials Equations ex 21.3 ought to be benefited of by all students who have maths in their board exams. Chapter 21 of the NCERT is named Differential Equations and the ideas covered are fundamental thoughts on differential conditions, general and specific solutions of a differential condition, development of a differential condition, and how to address first request, first degree differential conditions. Exercise 21.3 has 25 inquiries that cover ideas from the whole chapter.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
Differential Equations Excercise: 21.3
RD Sharma Chapter wise Solutions

Most of the CBSE board and the other schools recommend their students to own a set of RD Sharma solutions reference guides to help them with their homework. These books also help them practice numerous sums to test their knowledge.

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.3

Differential equation exercise 21.3 question 1

Answer:
$y=b e^{x}+c e^{2 x}$ is a solution of the given differential equation.
Hint:
Differentiate the given solution of differential equation on both sides with respect to $x$
Given:
$y=b e^{x}+c e^{2 x}$
Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(b e^{x}+c e^{2 x}\right) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$\begin{aligned} &\frac{d y}{d x}=\left[b \frac{d\left(e^{x}\right)}{d x}+e^{x} \frac{d b}{d x}\right]+\left[c \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d c}{d x}\right] \\\\ &\frac{d y}{d x}=\left[b e^{x}+e^{x}(0)\right]+\left[2 c e^{2 x}+e^{2 x}(0)\right] \\\\ &\frac{d y}{d x}=b e^{x}+2 c e^{2 x} \end{aligned}$ ......(i)
Now to obtain the second order derivative, differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(b e^{x}+2 c e^{2 x}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=\left[b \frac{d\left(e^{x}\right)}{d x}+e^{x} \frac{d b}{d x}\right]+2\left[c \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d c}{d x}\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\left[b e^{x}+e^{x}(0)\right]+2\left[2 c e^{2 x}+e^{2 x}(0)\right] \\\\ &\frac{d^{2} y}{d x^{2}}=b e^{x}+4 c e^{2 x} \end{aligned}$ ........(ii)
Now put both equation (i) and (ii) in given differential equation as follows
$\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0$
LHS
$\begin{aligned} &=\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y \\\\ &=\left(b e^{x}+4 c e^{2 x}\right)-3\left(b e^{x}+2 c e^{2 x}\right)+2\left(b e^{x}+c e^{2 x}\right) \end{aligned}$
$\begin{aligned} &=b e^{x}+4 c e^{2 x}-3 b e^{x}-6 c e^{2 x}+2 b e^{x}+2 c e^{2 x} \\\\ &=0 \end{aligned}$
$=$ RHS
Thus, $y=b e^{x}+c e^{2 x}$ is a solution of the given differential equation.

Differential equation exercise 21.3 question 2

Answer:
$y=4 \sin 3 x$ is a solution of $\frac{d^{2} y}{d x^{2}}+9 y=0$
Hint:
Differentiate the given solution of differential equation on both sides with respect to $x$
Given: $y=4 \sin 3 x$ is a solution

Solution:
Differentiating on both sides,
$\frac{d y}{d x}=4 \cos 3 x(3)$
$\frac{d y}{d x}=12 \cos 3 x$ ........(i)
Now again differentiating equation (i)
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(12 \cos 3 x) \\\\ &\frac{d^{2} y}{d x^{2}}=3(-12 \sin 3 x) \\\\ &\frac{d^{2} y}{d x^{2}}=-36 \sin 3 x \end{aligned}$ ........(ii)
Put value of equation (ii) in given problem
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}+9 y=0 \\\\ &L H S=-36 \sin 3 x+9(4 \sin 3 x) \end{aligned}$
$\begin{aligned} &=-36 \sin 3 x+9(4 \sin 3 x) \\\\ &=-36 \sin 3 x+36 \sin 3 x \\\\ &=0 \\\\ &=R H S \end{aligned}$
Thus, $y=4 \sin 3 x$ is a solution of $\frac{d^{2} y}{d x^{2}}+9 y=0$ .

Differential equation exercise 21.3 question 3

Answer:
$y=a e^{2 x}+b e^{-x}$ is a solution of $\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0$
Hint:
Differentiate the given solution of differential equation on both sides with respect to
Given:
$y=a e^{2 x}+b e^{-x}$ is a solution.

Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(a e^{2 x}+b e^{-x}\right) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$\frac{d y}{d x}=\left[a \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d a}{d x}\right]+\left[b \frac{d\left(e^{-x}\right)}{d x}+e^{-x} \frac{d b}{d x}\right]$
$\begin{aligned} &\frac{d y}{d x}=\left[2 a e^{2 x}+e^{2 x}(0)\right]+\left[-b e^{-x}+e^{-x}(0)\right] \\\\ &\frac{d y}{d x}=2 a e^{2 x}-b e^{-x} \end{aligned}$ ........(i)
Now to obtain the second order derivative, differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(2 a e^{2 x}-b e^{-x}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=2\left[a \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d a}{d x}\right]-\left[b \frac{d\left(e^{-x}\right)}{d x}+e^{-x} \frac{d b}{d x}\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=2\left[2 a e^{2 x}+e^{2 x}(0)\right]-\left[-b e^{-x}+e^{-x}(0)\right] \\\\ &\frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+b e^{-x} \end{aligned}$ .........(ii)
Now put both equation (i) and (ii) in given differential equation as follows
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0 \\\\ &\left(4 a e^{2 x}+b e^{-x}\right)-\left(2 a e^{2 x}-b e^{-x}\right)-2\left(a e^{2 x}+b e^{-x}\right)=0 \\\\ &4 a e^{2 x}+b e^{-x}-2 a e^{2 x}+b e^{-x}-2 a e^{2 x}-2 b e^{-x}=0 \end{aligned}$
Thus, _.

Differential equation exercise 21.3 question 4

Answer:
$y=A \cos x+B \sin x$ is a solution of differential equation.
Hint:
Differentiate the given solution of differential equation on both sides with respect to
Given:
$y=A \cos x+B \sin x$ is a solution.

Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}(A \cos x+B \sin x) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$\frac{d y}{d x}=\left[A \frac{d(\cos x)}{d x}+\cos x \frac{d A}{d x}\right]+\left[B \frac{d(\sin x)}{d x}+\sin x \frac{d B}{d x}\right]$
$\begin{aligned} &\frac{d y}{d x}=[-A \sin x+0]+[B \cos x+0] \\\\ &\frac{d y}{d x}=-A \sin x+B \cos x \end{aligned}$ ...........(i)
Now to obtain the second order derivative, differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(-A \sin x+B \cos x) \\\\ &\frac{d^{2} y}{d x^{2}}=\left[-A \frac{d(\sin x)}{d x}+\sin x \frac{d(-A)}{d x}\right]+\left[B \frac{d(\cos x)}{d x}+\cos x \frac{d B}{d x}\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=[-A \cos x+0]+[-B \sin x] \\\\ &\frac{d^{2} y}{d x^{2}}=-A \cos x-B \sin x \end{aligned}$ .................(ii)
Now put equation (ii) in given differential equation as follows
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}+y=0 \\\\ &L H S=\frac{d^{2} y}{d x^{2}}+y \end{aligned}$
$\begin{aligned} &=(-A \cos x-B \sin x)+(A \cos x+B \sin x) \\\\ &=0 \\\\ &=R H S \end{aligned}$

Thus, $y=A \cos x+B \sin x$ is a solution of differential equation.

Differential equation exercise 21.3 question 5

Answer:
$y=A \cos 2 x-B \sin 2 x$ is a solution of differential equation.
Hint:
Differentiate the given solution of differential equation on both sides with respect to
Given:
$y=A \cos 2 x-B \sin 2 x$ is a solution of differential equation.

Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}(A \cos 2 x-B \sin 2 x) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$\frac{d y}{d x}=\left[A \frac{d(\cos 2 x)}{d x}+\cos 2 x \frac{d A}{d x}\right]-\left[B \frac{d(\sin 2 x)}{d x}+\sin 2 x \frac{d B}{d x}\right]$
$\begin{aligned} &\frac{d y}{d x}=[-2 A \sin 2 x+\cos 2 x(0)]-[2 B \cos 2 x+\sin 2 x(0)] \\\\ &\frac{d y}{d x}=-2 A \sin 2 x-2 B \cos 2 x \end{aligned}$ .......(i)
Now to obtain the second order derivative, differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(-2 A \sin 2 x-2 B \cos 2 x) \\\\ &\frac{d^{2} y}{d x^{2}}=-2\left[A \frac{d(\sin 2 x)}{d x}+\sin 2 x \frac{d(A)}{d x}\right]-2\left[B \frac{d(\cos 2 x)}{d x}+\cos 2 x \frac{d B}{d x}\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-2[2 A \cos 2 x+\sin 2 x(0)]-2[-2 B \sin 2 x+\cos 2 x(0)] \\\\ &\frac{d^{2} y}{d x^{2}}=-4 A \cos 2 x+4 B \sin 2 x \end{aligned}$ ..............(ii)
Now put equation (ii) in given differential equation as follows
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}+4 y=0 \\\\ &L H S=\frac{d^{2} y}{d x^{2}}+4 y \end{aligned}$
$\begin{aligned} &=(-4 A \cos 2 x+4 B \sin 2 x)+4(A \cos 2 x-B \sin 2 x) \\\\ &=0 \\\\ &=R H S \end{aligned}$
Thus, $y=A \cos 2 x-B \sin 2 x$ is a solution of differential equation.

Differential equation exercise 21.3 question 6

Answer:
$y=A e^{B x}$ is a solution of differential equation
Hint:
Differentiate the given solution of differential equation on both sides with respect to
Given:
$y=A e^{B x}$ is a solution.

Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(A e^{B x}\right) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$\begin{aligned} &\frac{d y}{d x}=\left[A \frac{d\left(e^{B x}\right)}{d x}+e^{B x} \frac{d A}{d x}\right] \\\\ &\frac{d y}{d x}=B A e^{B x}+0 \\\\ &\frac{d y}{d x}=B A e^{B x} \end{aligned}$ ..............(i)
Now to obtain the second order derivative, differentiate equation (i) with respect to $x$
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{d}{d x}\left(B A e^{B x}\right) \\\\ \frac{d^{2} y}{d x^{2}} &=\left[B A \frac{d\left(e^{B x}\right)}{d x}+e^{B x} \frac{d(B A)}{d x}\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=B^{2} A e^{B x}+0 \\\\ &\frac{d^{2} y}{d x^{2}}=B^{2} A e^{B x} \end{aligned}$ .......(ii)
Now put both equation (i) and (ii) in given differential equation as follows
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{1}{y}\left(\frac{d y}{d x}\right)^{2} \\\\ &B^{2} A e^{B x}=\frac{1}{A e^{B x}}\left(B A e^{B x}\right)^{2} \end{aligned}$
$R H S=\frac{1}{A e^{B x}}\left(B A e^{B x}\right)^{2}$
$\begin{aligned} &=\frac{B^{2}\left(A e^{B x}\right)\left(A e^{B x}\right)}{A e^{B x}} \\\\ &=B^{2} A e^{B x} \\\\ &=L H S \end{aligned}$
Thus, $y=A e^{B x}$ is a solution of differential equation.

Differential equation exercise 21.3 question 7

Answer:
$y=\frac{a}{x}+b$ is a solution of differential equation
Hint:
Differentiate the given solution of differential equation on both sides with respect to
Given:
$y=\frac{a}{x}+b$ is a solution of the equation

Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{a}{x}+b\right) \quad\left[\because \frac{d(u+v)}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{a}{x}\right)+\frac{d}{d x}(b) \\\\ &\frac{d y}{d x}=a \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{d}{d x}(b) \end{aligned}$
$\frac{d y}{d x}=a \frac{d}{d x}\left(x^{-1}\right)+\frac{d}{d x}(b) \quad\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$
$\begin{aligned} &\frac{d y}{d x}=a(-1) x^{-1-1}+0 \\\\ &\frac{d y}{d x}=-a x^{-2}+0 \end{aligned}$
$\frac{d y}{d x}=\frac{-a}{x^{2}}$ ..............(i)
Differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{-a}{x^{2}}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=-a \frac{d}{d x}\left(\frac{1}{x^{2}}\right) \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-a \frac{d}{d x}\left(x^{-2}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=-a\left((-2) x^{-2-1}\right) \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-a\left((-2) x^{-3}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=2 a\left(x^{-3}\right) \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=\frac{2 a}{x^{3}}$ ............(ii)
Now put both equation (i) and (ii) in given differential equation as follows
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}+\frac{2}{x}\left(\frac{d y}{d x}\right)=0 \\\\ &L H S=\frac{d^{2} y}{d x^{2}}+\frac{2}{x}\left(\frac{d y}{d x}\right) \end{aligned}$
$\begin{aligned} &=\frac{2 a}{x^{3}}+\frac{2}{x}\left(\frac{-a}{x^{2}}\right) \\\\ &=\frac{2 a}{x^{3}}-\frac{2 a}{x^{3}} \\\\ &=0 \\\\ &=R H S \end{aligned}$
Thus, $y=\frac{a}{x}+b$ is a solution of differential equation.

Differential equation exercise 21.3 question 8

Answer:
$y^{2}=4 a x$ is a solution of differential equation
Hint:
Differentiate the given solution of differential equation on both sides with respect toand once with respect to $y$
Given:
$y^{2}=4 a x$ is a solution of the equation

Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &2 y \frac{d y}{d x}=4 a \\\\ &\frac{d y}{d x}=\frac{4 a}{2 y} \\\\ &\frac{d y}{d x}=\frac{2 a}{y} \end{aligned}$ .........(i)
Now differentiating $y^{2}=4 a x$ with respect to $y$
$\begin{aligned} &2 y=4 a \frac{d x}{d y} \\\\ &\frac{d x}{d y}=\frac{2 y}{4 a} \\\\ &\frac{d x}{d y}=\frac{y}{2 a} \end{aligned}$ .........(ii)
Now put both equation (i) and (ii) in given differential equation as follows
$\begin{aligned} &y=x \frac{d y}{d x}+a \frac{d x}{d y} \\\\ &R H S=x \frac{d y}{d x}+a \frac{d x}{d y} \end{aligned}$
$\begin{aligned} &=x\left(\frac{2 a}{y}\right)+a\left(\frac{y}{2 a}\right) \\\\ &=\frac{y^{2}}{4 a}\left(\frac{2 a}{y}\right)+a\left(\frac{y}{2 a}\right) \quad\left[y^{2}=4 a x \Rightarrow \frac{y^{2}}{4 a}=x\right] \end{aligned}$
$\begin{aligned} &=\frac{y}{2}+\frac{y}{2} \\\\ &=\frac{2 y}{2} \\\\ &=y \\\\ &=L H S \end{aligned}$
Thus, $y^{2}=4 a x$ is a solution of differential equation.
$\begin{aligned} &y=x \frac{d y}{d x}+a \frac{d x}{d y} \end{aligned}$

Differential equation exercise 21.3 question 9

Answer:
$A x^{2}+B y^{2}=1$ is a solution of differential equation
Hint:
Differentiate and substitute values of equation to obtain differential equation.
Given:
$A x^{2}+B y^{2}=1$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d\left(A x^{2}+B y^{2}\right)}{d x}=\frac{d}{d x}(1) \\\\ &\frac{d\left(A x^{2}\right)}{d x}+\frac{d\left(B y^{2}\right)}{d x}=\frac{d}{d x}(1) \end{aligned}$
$2 A x+2 B y \frac{d y}{d x}=0$ ..........(i)
Differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d}{d x}\left(2 A x+2 B y \frac{d y}{d x}\right)=\frac{d}{d x}(0) \\\\ &2 A \frac{d}{d x}(x)+2 B \frac{d}{d x}\left(y \frac{d y}{d x}\right)=0 \end{aligned}$
$\begin{aligned} &2 A+2 B\left[\frac{d y}{d x} \frac{d y}{d x}+y \frac{d}{d x}\left(\frac{d y}{d x}\right)\right]=0 \\\\ &2 B\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=-2 A \end{aligned}$
$\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=\frac{-2 A}{2 B}$ .........(ii)
Using equation (i), we can find the values of $\frac{-2 A}{2 B}$
$\begin{aligned} &2 A x+2 B y \frac{d y}{d x}=0 \\\\ &2 B y \frac{d y}{d x}=-2 A x \\\\ &\frac{y}{x} \frac{d y}{d x}=\frac{-2 A}{2 B} \end{aligned}$ ..........(iii)
Now put equation (ii) in (iii) ,
$\begin{aligned} &{\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=\frac{y d y}{x d x}} \end{aligned}$
$x\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=y \frac{d y}{d x}$
Hence proved.
Thus, $A x^{2}+B y^{2}=1$ is a solution of differential equation.

Differential equation exercise 21.3 question 10

Answer:
$y=a x^{3}+b x^{2}+c$ is a solution
Hint:
Differentiate the given solution_.
Given:
$y=a x^{3}+b x^{2}+c$

Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(a x^{3}+b x^{2}+c\right) \\\\ &\frac{d y}{d x}=3 a x^{2}+2 b x+0 \\\\ &\frac{d y}{d x}=3 a x^{2}+2 b x \end{aligned}$ .................(i)
Differentiate equation (i) with respect to $x$ ,
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(3 a x^{2}+2 b x\right) \\\\ &\frac{d^{2} y}{d x^{2}}=6 a x+2 b \end{aligned}$ ..............(ii)
Differentiate equation (ii) with respect to $x$ ,
$\frac{d^{3} y}{d x^{3}}=6 a$
Thus, $y=a x^{3}+b x^{2}+c$ is a solution of differential equation.

Differential equation exercise 21.3 question 11

Answer:
$y=\frac{c-x}{(1+c x)}$ is a solution of differential equation
Hint:
Differentiate the given solution and substitute in the differential equation.
Given:
$y=\frac{c-x}{(1+c x)}$ is a solution of the equation

Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{c-x}{(1+c x)}\right) \\\\ &\frac{d y}{d x}=\frac{\left[(1+c x) \frac{d}{d x}(c-x)\right]-\left[(c-x) \frac{d}{d x}(1+c x)\right]}{(1+c x)^{2}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{[(1+c x)(0-1)]-[(c-x)(0+c)]}{(1+c x)^{2}} \\\\ &\frac{d y}{d x}=\frac{[(1+c x)(-1)]-[(c-x)(c)]}{(1+c x)^{2}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{-1-c x-c^{2}+c x}{(1+c x)^{2}} \\\\ &\frac{d y}{d x}=\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}} \end{aligned}$ .............(i)
Put equation (i) in differential equation as follows
$\begin{aligned} &\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0 \\\\ &L H S=\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right) \end{aligned}$
$\begin{aligned} &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\left(1+y^{2}\right) \\\\ &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\left[1+\left(\frac{(c-x)}{(1+c x)}\right)^{2}\right] \end{aligned}$
$\begin{aligned} &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+1+\frac{(c-x)^{2}}{(1+c x)^{2}} \\\\ &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\frac{(1+c x)^{2}+(c-x)^{2}}{(1+c x)^{2}} \end{aligned}$
$=\left[\frac{-\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\frac{1+2 c x+c^{2} x^{2}+c^{2}-2 c x+x^{2}}{(1+c x)^{2}}$
$\begin{aligned} &=\left[\frac{-\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\frac{1+c^{2} x^{2}+c^{2}+x^{2}}{(1+c x)^{2}} \\\\ &=\left[\frac{-\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\left[\frac{\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right] \\\\ &=0 \\\\ &=R H S \end{aligned}$

Differential equation exercise 21.3 question 12

Answer:
$y=e^{x}(A \cos x+B \sin x)$ is a solution of given equation
Hint:
Differentiate the given solution and substitute in the differential equation.
Given:
$y=e^{x}(A \cos x+B \sin x)$ is a solution of the equation

Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[e^{x}(A \cos x+B \sin x)\right] \\\\ &\frac{d y}{d x}=\frac{d}{d x}\left[e^{x} A \cos x\right]+\frac{d}{d x}\left[e^{x} B \sin x\right] \end{aligned}$
$\frac{d y}{d x}=A\left[e^{x} \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}\left(e^{x}\right)\right]+B\left[e^{x} \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x}\left(e^{x}\right)\right]$
$\begin{aligned} &\frac{d y}{d x}=A\left[e^{x}(-\sin x)+\cos x\left(e^{x}\right)\right]+B\left[e^{x}(\cos x)+\sin x\left(e^{x}\right)\right] \\\\ &\frac{d y}{d x}=e^{x}[-A \sin x+A \cos x]+e^{x}[B \cos x+B \sin x] \end{aligned}$ ............(i)
Differentiate equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left[e^{x}[-A \sin x+A \cos x]+e^{x}[B \cos x+B \sin x]\right] \\\\ &\frac{d^{2} y}{d x^{2}}=A \frac{d}{d x}\left[-e^{x} \sin x+e^{x} \cos x\right]+B \frac{d}{d x}\left[e^{x} \cos x+e^{x} \sin x\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=A\left[-\left(e^{x} \cos x+e^{x} \sin x\right)+e^{x}(-\sin x)+e^{x} \cos x\right]+B\left[\left(e^{x}(-\sin x)+e^{x} \cos x\right)+e^{x}(\cos x)+e^{x} \sin x\right] \\\\ &\frac{d^{2} y}{d x^{2}}=A\left[e^{x}(-\cos x-\sin x)+e^{x}(\cos x-\sin x)\right]+B\left[e^{x}(\cos x-\sin x)+e^{x}(\cos x+\sin x)\right] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=e^{x}[-A \sin x-A \cos x+A \cos x-A \sin x]+e^{x}[B \cos x-B \sin x+B \cos x+B \sin x] \\\\ &\frac{d^{2} y}{d x^{2}}=e^{x}[-2 A \sin x]+e^{x}[2 B \cos x] \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=e^{x}[2 B \cos x-2 A \sin x] \\\\ &\frac{d^{2} y}{d x^{2}}=2 e^{x}[-A \sin x+B \cos x]+2 e^{x}[A \cos x+B \sin x]-2 e^{x}[A \cos x+B \sin x] \end{aligned}$
[on Adding and subtracting the value of $y$ in the above step ]
$\frac{d^{2} y}{d x^{2}}=2\left[e^{x}(-A \sin x+B \cos x)+e^{x}(A \cos x+B \sin x)\right]-2 e^{x}\left ( A\cos x+B\sin x \right )$
Put value of $y$ as per question and value of equation (i) in above equation
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=2 \frac{d y}{d x}-2 y \\\\ &\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0 \end{aligned}$
Hence it is proved that $y=e^{x}(A \cos x+B \sin x)$ is solution of given equation.

Differential equation exercise 21.3 question 13

Answer:
$y=c x+2 c^{2}$ is a solution of differential equation
Hint:
Just differentiate once and put value in given differential problem
Given:
$y=c x+2 c^{2}$ is a solution of the equation

Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d(c x)}{d x}+\frac{d\left(2 c^{2}\right)}{d x} \\\\ &\frac{d y}{d x}=c \end{aligned}$ ................(i)
Put value in given problem as follows
$\begin{aligned} &2\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}-y=0 \\\\ &L H S=2\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}-y \end{aligned}$
$\begin{aligned} &=2 c^{2}+x c-y \\\\ &=2 c^{2}+x c-\left(c x+2 c^{2}\right) \\\\ &=2 c^{2}+x c-c x-2 c^{2} \\\\ &=0 \\\\ &=R H S \end{aligned}$
Thus, $y=c x+2 c^{2}$ is a solution of differential equation.

Differential equation exercise 21.3 question 14

Answer:
$y=-x-1$ is a solution of differential equation
Hint:
Differentiate the solution and modify the given differential equation
Given:
$y=-x-1$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=-\frac{d(x)}{d x}-\frac{d(1)}{d x} \\\\ &\frac{d y}{d x}=-1 \end{aligned}$ ...............(i)
Now the given equation has to be modified
$\begin{aligned} &(y-x) d y-\left(y^{2}-x^{2}\right) d x=0 \\\\ &(y-x) d y=\left(y^{2}-x^{2}\right) d x \\\\ &\frac{d y}{d x}=\frac{y-x}{y^{2}-x^{2}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{y-x}{(y-x)(y+x)} \\\\ &\frac{d y}{d x}=\frac{1}{y+x} \end{aligned}$ ..............(ii)
Comparing equation (i) and (ii)
$\begin{aligned} &\frac{1}{y+x}=-1 \\\\ &1=-(y+x) \\\\ &y=-x-1 \end{aligned}$
Thus,it is proved that $y=-x-1$ is a solution of differential equation.

Differential equation exercise 21.3 question 15

Answer:
$y^{2}=4 a(x+a)$ is a solution of differential equation
Hint:
Differentiate the solution and modify the given differential equation
Given:
$y^{2}=4 a(x+a)$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(4 a(x+a)) \\\\ &2 y \frac{d y}{d x}=4 a \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{4 a}{2 y} \\\\ &\frac{d y}{d x}=\frac{2 a}{y} \end{aligned}$ ...............(i)
Put value of equation (i) in given problem as follows
$\begin{aligned} &y\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=2 x \frac{d y}{d x} \\\\ &L H S=y\left[1-\left(\frac{d y}{d x}\right)^{2}\right] \end{aligned}$
$\begin{aligned} &=y\left[1-\left(\frac{2 a}{y}\right)^{2}\right] \\\\ &=y-\frac{4 a^{2} y}{y^{2}} \\\\ &=y-\frac{4 a^{2}}{y} \end{aligned}$ ..............(ii)
$\begin{aligned} &R H S=2 x \frac{d y}{d x} \\\\ &=2 x\left(\frac{2 a}{y}\right) \\\\ &=\frac{4 a x}{y} \end{aligned}$ ..............(iii)
Now, $y^{2}=4 a(x+a)$
$\begin{aligned} &y^{2}=4 a x+4 a^{2} \\\\ &\frac{y^{2}-4 a^{2}}{4 a}=x \end{aligned}$ ...........(iv)
Put value of equation (iv) in equation (iii)
$\begin{aligned} &=\frac{4 a}{y}\left(\frac{y^{2}-4 a^{2}}{4 a}\right) \\\\ &=\frac{y^{2}}{y}-\frac{4 a^{2}}{y} \\\\ &=y-\frac{4 a^{2}}{y} \end{aligned}$ ............(v)
From (ii) and (v)
$L H S=R H S$
Thus, $y^{2}=4 a(x+a)$ is a solution of differential equation.

Differential equation exercise 21.3 question 16

Answer:
$y=c e^{\tan ^{-1} x}$ is a solution of differential equation
Hint:
Differentiate with respect to
Use formula $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}$
Given:
$y=c e^{\tan ^{-1} x}$

Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(c e^{\tan ^{-1} x}\right) \\\\ &\frac{d y}{d x}=c e^{\tan ^{-1} x}\left(\frac{1}{1+x^{2}}\right) \end{aligned}$ ................(i)
Differentiating equation (i) with respect to $x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=c \frac{d}{d x}\left(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\right) \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=c\left[\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(e^{\tan ^{-1} x}\right)-e^{\tan ^{-1} x} \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=c\left[\frac{\left(1+x^{2}\right) \frac{e^{\tan ^{-1} x}}{1+x^{2}}-e^{\tan ^{-1} x}(2 x)}{\left(1+x^{2}\right)^{2}}\right] \\\\ &\frac{d^{2} y}{d x^{2}}=c\left[\frac{e^{\tan ^{-1} x}-2 x e^{\tan ^{-1} x}}{\left(1+x^{2}\right)^{2}}\right] \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=\left[\frac{c(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^{2}\right)^{2}}\right]$ .............(ii)
Put (i) and (ii) in given equation
$\begin{aligned} &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0 \\\\ &L H S=\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x} \end{aligned}$
$\begin{aligned} &=\left(1+x^{2}\right) \frac{c(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^{2}\right)^{2}}+(2 x-1) \frac{c e^{\tan ^{-1} x}}{\left(1+x^{2}\right)} \\\\ &=\frac{c(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^{2}\right)}-(1-2 x) \frac{c e^{\tan ^{-1} x}}{\left(1+x^{2}\right)} \\\\ &=0 \\\\ &=R H S \end{aligned}$
Thus, $y=c e^{\tan ^{-1} x}$ is a solution of the differential equation.

Differential equation exercise 21.3 question 17

Answer:
$y=e^{m \cos ^{-1} x}$ is a solution of differential equation
Hint:
Differentiate with respect to $x$ and put values in given equation.
Given:
$y=e^{m \cos ^{-1} x}$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(e^{m \cos ^{-1} x}\right) \\\\ &\frac{d y}{d x}=m e^{m \cos ^{-1} x}\left(\frac{-1}{\sqrt{1+x^{2}}}\right) \\\\ &\frac{d y}{d x}=\left(\frac{-m e^{m \cos ^{-1} x}}{\sqrt{1+x^{2}}}\right) \end{aligned}$ ............(i)
Differentiating equation (i) with respect to $x$ ,
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-m\left[\sqrt{1+x^{2}} \frac{d}{d x}\left(e^{m \cos ^{-1} x}\right)-e^{m \cos ^{-1} x} \frac{d}{d x}\left(\sqrt{1+x^{2}}\right)\right]}{\left(\sqrt{1+x^{2}}\right)^{2}} \\\\ &\frac{d^{2} y}{d x^{2}}=\frac{-m\left[\sqrt{1+x^{2}}\left(\frac{m e^{m \cos ^{-1} x}}{\sqrt{1+x^{2}}}\right)-e^{m \cos ^{-1} x}\left(\frac{-x}{\sqrt{1+x^{2}}}\right)\right]}{\left(\sqrt{1+x^{2}}\right)^{2}} \end{aligned}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-m\left[-m e^{m \cos ^{-1} x}+\left(\frac{x e^{m \cos ^{-1} x}}{\sqrt{1+x^{2}}}\right)\right]}{\left(\sqrt{1+x^{2}}\right)^{2}} \\\\ &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}=m^{2} e^{m \cos ^{-1} x}+\left(\frac{-m x e^{m \cos ^{-1} x}}{\sqrt{1+x^{2}}}\right) \end{aligned}$
$\begin{aligned} &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}=m^{2} y+x \frac{d y}{d x} \\\\ &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-m^{2} y=0 \end{aligned}$
Hence proved, the given function is the solution to given differential equation

Differential equation exercise 21.3 question 18

Answer:

Answer:
$y=\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{2}$ is a solution of differential equation
Hint:
Differentiate and substitute the values
Given:
$y=\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{2}$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{2}\right] \\\\ &\frac{d y}{d x}=\frac{d}{d x}\left[2 \log \left(x+\sqrt{x^{2}+a^{2}}\right)\right] \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{2\left(1+\frac{x}{\sqrt{x^{2}+a^{2}}}\right)}{\left(x+\sqrt{x^{2}+a^{2}}\right)} \\\\ &\frac{d y}{d x}=\frac{2\left(\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}}\right)}{\left(x+\sqrt{x^{2}+a^{2}}\right)} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{2\left(x+\sqrt{x^{2}+a^{2}}\right)}{\sqrt{x^{2}+a^{2}}\left(x+\sqrt{x^{2}+a^{2}}\right)} \\\\ &\frac{d y}{d x}=\frac{2}{\sqrt{x^{2}+a^{2}}} \end{aligned}$ $\begin{aligned} &\frac{d y}{d x}=\frac{2\left(x+\sqrt{x^{2}+a^{2}}\right)}{\sqrt{x^{2}+a^{2}}\left(x+\sqrt{x^{2}+a^{2}}\right)} \\ &\frac{d y}{d x}=\frac{2}{\sqrt{x^{2}+a^{2}}} \end{aligned}$ ...............(i)
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=2\left(\frac{-1}{2}\right)\left(\frac{2 x}{\left(x^{2}+a^{2}\right) \sqrt{x^{2}+a^{2}}}\right) \\\\ &\left(x^{2}+a^{2}\right) \frac{d^{2} y}{d x^{2}}=\frac{-2 x}{\sqrt{x^{2}+a^{2}}} \end{aligned}$
$\begin{aligned} &\left(a^{2}+x^{2}\right) \frac{d^{2} y}{d x^{2}}=-x \frac{d y}{d x} \\\\ &\left(a^{2}+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=0 \end{aligned}$
Hence the given function is the solution to given differential equation.

Differential equation exercise 21.3 question 19

Answer:
$\begin{aligned} &y=2\left(x^{2}-1\right)+c e^{-x^{2}}\\ \end{aligned}$ is a solution of differential equation
Hint:
Differentiate both sides and add $2x^{2}$ and $-2x^{2}$
Given:
$\begin{aligned} &y=2\left(x^{2}-1\right)+c e^{-x^{2}}\\ \end{aligned}$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[2 x^{2}-2+c e^{-x^{2}}\right] \\\\ &\frac{d y}{d x}=4 x+c\left(e^{-x^{2}}(-2 x)\right) \\\\\ &\frac{d y}{d x}=4 x-2 x c e^{-x^{2}} \end{aligned}$ ............(i)
Now taking common and then adding $2x^{2}$ and $-2x^{2}$
$\begin{aligned} &\frac{d y}{d x}=2 x\left(2-c e^{-x^{2}}\right) \\\\ &\frac{d y}{d x}=-2 x\left[c e^{-x^{2}}-2+2 x^{2}-2 x^{2}\right] \\\\ &\frac{d y}{d x}=-2 x\left[2 x^{2}+c e^{-x^{2}}-2-2 x^{2}\right] \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=-2 x\left[2\left(x^{2}-1\right)+c e^{-x^{2}}-2 x^{2}\right] \\\\ &\frac{d y}{d x}=-2 x\left[y-2 x^{2}\right] \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=-2 x y+4 x^{3} \\\\ &\frac{d y}{d x}+2 x y=4 x^{3} \end{aligned}$
Thus, $\begin{aligned} &y=2\left(x^{2}-1\right)+c e^{-x^{2}}\\ \end{aligned}$ is a solution to the given differential equation.

Differential equation exercise 21.3 question 20

Answer:
$y=e^{-x}+a x+b$ is a solution of differential equation
Hint:
Just differentiate two times to obtain values.
Given:
$y=e^{-x}+a x+b$
Solution:
$y=e^{-x}+a x+b$
Differentiating on both sides with respect to $x$
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(e^{-x}+a x+b\right) \\\\ &\Rightarrow \frac{d y}{d x}=-e^{-x}+a \end{aligned}$ ................(i)
Differentiating equation (i)
$\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=-\left(-e^{-x}\right) \\\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=\left(e^{-x}\right) \\\\ &\Rightarrow e^{x} \cdot \frac{d^{2} y}{d x^{2}}=1 \end{aligned}$
Thus, $\Rightarrow y=e^{-x}+a x+b$ is a solution.

Differential equation exercise 21.3 question 21(i)

Answer:
The given function is the solution to the given differential equation.
Hint:
Just differentiate the given function to obtain differential equation.
Given:
$y=ax$
Solution:
$\Rightarrow y=a x$
Differentiating above equation
$\Rightarrow \frac{d y}{d x}=a$ ...........(i)
Now from $y=ax$
$a=\frac{y}{x}$
Substitute (ii) in (i)
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{y}{x} \\\\ &\Rightarrow x \cdot \frac{d y}{d x}=y \end{aligned}$
Hence, the given function is the solution to the given differential equation.

Differential equation exercise 21.3 question 21(ii)

Answer:
Function satisfies the given differential equation
Hint:
Modify the function by taking squares on both sides.
Given:
$y=\pm \sqrt{a^{2}-x^{2}}$
Solution:
Given, $y=\pm \sqrt{a^{2}-x^{2}}$ as a function
Taking squares on both sides
$\begin{aligned} &\Rightarrow y^{2}=\left(\sqrt{a^{2}-x^{2}}\right)^{2} \\\\ &\Rightarrow y^{2}=a^{2}-x^{2} \end{aligned}$
Now differentiating with respect to x
$\Rightarrow 2 y \frac{d y}{d x}=-2 x$
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=-\frac{x}{y} \\\\ &\Rightarrow y \frac{d y}{d x}=-x \\\\ &\Rightarrow x+y \frac{d y}{d x}=0 \end{aligned}$
Thus, the function satisfies the differential equation.

Differential Equation exercise 21.3 question 21(iii)

Answer:
$y=\frac{a}{x+a}$ is the function which satisfies the differential equation
Hint:
Just differentiate the function and substitute in given equation
Given:
$y=\frac{a}{x+a}$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{(x+a)(0)-a(1)}{(x+a)^{2}} \\\\ &\frac{d y}{d x}=\frac{-a}{(x+a)^{2}} \end{aligned}$ .............(i)
Substitute equation (i) in differential equation
$\begin{aligned} &x \frac{d y}{d x}+y=y^{2} \\\\ &L H S=x \frac{d y}{d x}+y \end{aligned}$
$\begin{aligned} &=x\left(\frac{-a}{(x+a)^{2}}\right)+y \\\\ &=\frac{-x a}{(x+a)^{2}}+\frac{a}{x+a} \\\\ &=\frac{-x a+a(x+a)}{(x+a)^{2}} \end{aligned}$
$\begin{aligned} &=\frac{-x a+x a+a^{2}}{(x+a)^{2}} \\\\ &=\frac{a^{2}}{(x+a)^{2}} \\\\ &=\left(\frac{a}{(x+a)}\right)^{2} \end{aligned}$
$=y^{2}$ ..................(ii)
$LHS = RHS$
Thus, $y=\frac{a}{x+a}$ is the function which satisfies the differential equation.

Differential equation exercise 21.3 question 21(v)

Answer:
$y=\frac{1}{4}(x \pm a)^{2}$ is a solution of differential equation
Hint:
Differentiate the function
Given:
$y=\frac{1}{4}(x \pm a)^{2}$
Solution:
Differentiating on both sides with respect to $x$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{4} \times 2(x \pm a)(1) \\\\ &\frac{d y}{d x}=\frac{1}{2}(x \pm a) \end{aligned}$ ..............(i)
Squaring on both sides,
$\begin{aligned} &\left(\frac{d y}{d x}\right)^{2}=\frac{1}{4}(x \pm a)^{2} \\\\ &\left(\frac{d y}{d x}\right)^{2}=y \\\\ &y=\left(\frac{d y}{d x}\right)^{2} \end{aligned}$
Hence, the given function satisfies the equation.

Since RD Sharma class 12th exercise 21.3 is such a well known name among numerous NCERT solutions, there has to be some good reasons why they are so popular. Here are a list of reasons which make RD Sharma class 12 chapter 21 exercise 21.3 such a great book:-

In RD Sharma class 12th exercise 21.3 the answers are created by experts who have immense knowledge in maths. They use some advanced calculations and techniques which help students solve questions faster.
Hundreds of students place their trust in RD Sharma class 12th exercise 21.3 every year and have benefitted from the answers. They have even found common questions in their board exams.
Teachers also trust RD Sharma class 12 solutions chapter 21 ex 21.3 which is why they often give out homework questions from the book. Students can seek help from the answers when they are unable to complete complex questions.
The RD Sharma class 12th exercise 21.3 always comes with an updated and latest syllabus. The free copy of the book can be readily downloaded from the Career360 website.

RD Sharma Chapter wise Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Frequently Asked Questions (FAQs)

1. What are the ideas shrouded in chapter 21 of the class 12 maths book?

The 19th chapter of the class 12 maths book in CBSE contains the chapter Differential Equations.

2. How are MCQ, VSA, FBQ, and RE segments significant for exam arrangements?

The MCQ, VSA, FBQ, and RE areas are fundamental for all exams. These parts contain inquiries from the whole chapter.

3. Who makes the appropriate responses in the RD Sharma solutions?

Specialists and gifted experts are liable for making the appropriate responses in the RD Sharma solutions.

4. How might I download the RD Sharma class 12th exercise 21.3 pdf?

Students will actually want to download the RD Sharma class 12th exercise 21.3 pdf free of charge from the Career360 site. It is the one-stop objective for all RD Sharma solutions.

5. Would i be able to utilize class 12 RD Sharma chapter 21 exercise 21.3 solution for exam arrangements?

Students can utilize the class 12 RD Sharma chapter 21 exercise 21.3 solution for their exam arrangements. The prospectus followed by these books covers exams like school tests, sheets, and JEE mains.