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

# 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.

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.

## 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$x$
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, $LHS =RHS$.

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$x$
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$x$
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$x$
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$x$
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 to$x$and 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$x$
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.

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