RD Sharma Class 12 Exercise 21.10 Differential Equation Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 21.10 Differential Equation Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 04:19 PM IST

Most of the class 12 students worry about the board exams. However, there is nothing to be worried about; all they need is proper guidance to score marks in the public exam. Most importantly, students require an excellent solution book to refer to the sums given in the RD Sharma Class 12th Exercise 21.10 part. Even though the differential equation is a challenging portion, this book will lend a helping hand to the class 12 students.

## 21.10

Differential Equation Exercise 21.10 Question 1

Answer: $y=\frac{1}{5} e^{3 x}+C e^{-2 x}$
Hint: To solve this equation we use formula
Give:$\frac{d y}{d x}+2 y=e^{3 x}$
Solution: \begin{aligned} & \frac{d y}{d x}+P y=Q \\ & \end{aligned}
$p=2, Q=e^{3 x} \\$
$\text { If } e^{\int p d x}$
\begin{aligned} &=e^{\operatorname{\int xdx}} \\ & \end{aligned}
$=e^{2 x} \\$
$y \cdot e^{2 x}=\int e^{3 x} \cdot e^{2 x} d x+C$
$y \times I f=\int Q \times I f d x+C \\$
\begin{aligned} &=\int e^{5 x} \\ & \end{aligned}
$=y e^{2 x}=\frac{e^{5 x}}{5}+C \\$
$=y=\frac{1}{e^{2 x}}\left[\frac{e^{5 x}}{5}+C\right] \\$
$=y=\frac{e^{3 x}}{5}+C e^{-2 x}$

Differential Equation Exercise 21.10 Question 2

Answer: $y=\frac{-5}{4} e^{-3 x}+C e^{-2 x}$
Hint: To solve this equation we use $e\int fx dx$ formula
Give:$4 \frac{d y}{d x}+8 y=5 e^{-3 x}$
Solution:$\frac{d y}{d x}+2 y=\frac{5}{4} e^{-3 x}$
\begin{aligned} &P(x)=2, Q(x)=\frac{5}{4} e^{-3 x} \\ & \end{aligned}
$I\! f=e^{\int f(x) d x} \\$
$=e^{\int 2 d x} \\$
$=e^{2 x} \\$
$=y e^{2 x}=\int e^{2 x} \times \frac{5}{4} e^{-3 x} d x+C \\$
$=y e^{2 x}=\frac{-5}{4} e^{-x} d x+C \\$
$=y=\frac{-5}{4} e^{-3 x}+C e^{-2 x}$

Differential Equation Exercise 21.10 Question 3

Answer: $y e^{2 x}=2 e^{3 x}+C$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ formula
Give:$\frac{d y}{d x}+2 y=6 e^{x}$
Solution: $\frac{d y}{d x}+2 y=6 e^{x} \cdot \cdot \cdot \left ( i \right )$
$\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=2, Q=6 e^{x} \\ \end{aligned}
$I \! f=e^{\int f d x} \\$
$=e^{2 \int d x} \\$
$=e^{2 x} \\$
$=y \times I f=\int(I f \times Q) d x \\$
$=y \times e^{2 x}=\int e^{2 x} \times 6 e^{x} d x \\$
$=y e^{2 x}=6 \int e^{3 x} d x \\$
$=y e^{2 x}=\frac{6}{3} e^{x}+C \\$
$=y e^{2 x}=2 e^{x}+C$

Differential Equation Excercise 21.10 Question 4

Answer: $y=-e^{-2 x}+C e^{-x}$
Hint: To solve this equation we use $ylf$ formula
Give: $\frac{d y}{d x}+y=e^{-2 x}$
Solution: $\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=1, Q=e^{-2 x} \\ \end{aligned}
$I\! f=e^{\int P d x}=e^{\int d x}=e^{x} \\$
$y \times I f=\int Q \times I\! f d x+C \\$
$y \times e^{x}=\int e^{-2 x} e^{x} d x+C$
\begin{aligned} &y e^{x}=\int e^{-x} d x+C \\ \end{aligned}
$y e^{x}=-e^{-x}+C \\$
$y=-e^{-x} e^{x}+C e^{-x} \\$
$y=-e^{2 x}+C e^{-x}$

Differential Equation Excercise 21.10 Question 5

Answer: $\frac{y}{x}=\log |x|+C$
Hint: To solve this equation we use
Give: $x \frac{d y}{d x}=x+y$
Solution: $\frac{d y}{d x}=1+\frac{y}{x}$
\begin{aligned} &\frac{d y}{d x}-\frac{1}{x} y=1 \\ \end{aligned}
$\frac{d y}{d x}-\frac{1}{x} y=1 \\$
$\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=\frac{1}{x^{\prime}}\, Q=1 \\ & \end{aligned}
$\text { If }=e^{\int P d x} \\$
$=e^{-\int\frac{1}{x}dx }$
$=e^{-\log x} \\$
$=x^{-1} \\$
$=\frac{1}{x}$
\begin{aligned} &y \times \frac{1}{x}=\int 1 \times \frac{1}{x} d x+C \\ & \end{aligned}
$\frac{y}{x}=\log x+C \\$
$\frac{y}{x}=\log |x|+C$

Differential Equation Excercise 21.10 Question 6

Answer: $y=(2 x-1)+C e^{-2 x}$
Hint: To solve this equation we will use differentiate different.
Give: $\frac{d y}{d x}+2 y=4 x$
Solution: $\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=2, Q=4 x \\ & \end{aligned}
$I\! f=e^{\int P d x} \\$
$=e^{\int 2 d x} \\$
$=e^{-2 x} \\$
\begin{aligned} &y e^{2 x}=\int 4 x e^{2 x} d x+C \\ \end{aligned}
$y e^{2 x}=4 \int x e^{2 x} d x+C \\$
$y e^{2 x}=4\left[\frac{x e^{2 x}}{2}\right]-\left[\frac{e^{2 x}}{2} d x\right]+C \\$
$y e^{2 x}=2 x e^{2}-2 \frac{e^{2 x}}{2}+C \\$
$y=(2 x-1)+C e^{-2 x}$
$y \times I f=\int Q \times \operatorname{If} d x+C$

Differential Equation Exercise 21.10 Question 7

Answer: $y=\left(\frac{x-1}{x}\right) e^{x}+\frac{C}{x}$
Hint: To solve this equation we will use differentiate different.
Give: $x \frac{d y}{d x}+y=x e^{x}$
Solution: $\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=\frac{1}{x^{\prime}} f=e^{x} \\ & \end{aligned}
$I\! f=e^{\int P d x}$
\begin{aligned} &=e^{\int \frac{1}{x} d x} \\ & \end{aligned}
$=e^{\log e^{x}} \\$
$=x \\$
$y I\! f=\int f I\! f$
\begin{aligned} &y x=\int e^{x} x \\ & \end{aligned}
$y x=x e^{x}+\int e^{x}+C \\$
$y x=(x+1) e^{x}+C \\$
$y=y \frac{(x-1)}{x} e^{x}+\frac{C}{x}$

Differential Equation Exercise 21.10 Question 8

Answer: $y\left(x^{2}+1\right)^{2}=-x+C$
Hint: To solve this equation we will use differentiate both terms.
Give: $\frac{d y}{d x}+\frac{4 x}{x^{2}+1} y+\frac{1}{\left(x^{2}+1\right)^{2}}=0$
Solution: $\frac{d y}{d x}+\frac{4 x}{x^{2}+1} y=\frac{-1}{\left(x^{2}+1\right)^{2}}$
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ & \end{aligned}
$I\! f=e^{\int P d x} \\$
$=e^{\int \frac{4 x}{x^{2}+1} d x}$ $\quad\left[\text { Let } x^{2}+1=u, 2 x d x=d u\right.$
\begin{aligned} &=2 e^{\int \frac{d u}{u}} \\ & \end{aligned}
$=e^{2} \ln \left|x^{2}+1\right| \\$
$=\left(x^{2}+1\right)^{2} \\$
$y\, I\! f=\int Q \, I\! f d x \\$
$y\left(x^{2}+1\right)^{2}=\int-\frac{1}{\left(x^{2}+1\right)^{2}}\left(x^{2}+1\right)^{2} d x \\$
$y=\int-d x \\$
$y\left(x^{2}+1\right)^{2}=-x+C$

Differential Equation Exercise 21.10 Question 9

Answer: $4 x y=2 x^{2} \log |x|-x^{2}+C$
Hint: To solve this equation we will use differentiation method.
Give: $\frac{d y}{d x}+\frac{y}{x}=\log x$
Solution: $\frac{d y}{d x}+\frac{y}{x}=\log x$
$\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=\frac{1}{x^{\prime}} Q=\log x \\ & \end{aligned}
$I\! f=e^{\int \frac{1}{x} d x} \\$
$=e^{\log x} \\$
$=x \\$
$y I\! f=\int Q I\! f d x \\$
$y x=\int x \log x d x+C$
\begin{aligned} &=\log x \frac{x^{2}}{2}-\int \frac{1}{x} \frac{x^{2}}{2} d x+C \\ & \end{aligned}
$x y=\frac{x^{2} \log x}{2}-\int \frac{x}{2} d x+C \\$
$=\frac{x^{2} \log x}{2}-\frac{x^{2}}{4}+C \\$
$=y=\frac{x \log x}{2}-\frac{x}{4}+\frac{C}{x} \\$
$=4 x y=2 x^{2} \log x-x+C$

Differential Equation exercise 21.10 question 10

Answer: $y=e^{x}+C x$
Hint: To solve this equation we will use differentiate separately.
Give: \begin{aligned} &x \frac{d y}{d x}-y=(x-1) e^{x} \\ & \end{aligned}
Solution: $\frac{d y}{d x}-\frac{1}{x} y \frac{(x-1) e^{x}}{x}$
\begin{aligned} &\quad \frac{d y}{d x}+P y=Q \\ & \end{aligned}
$P=\frac{-1}{x}, Q=\frac{(x-1) e^{x}}{x} \\$
$I \! f=e^{\int P d x} \\$
$=e^{-\int \frac{1}{x} d x}$
\begin{aligned} &=e^{\log x^{-1}} \\ & \end{aligned}
$=x^{-1} \\$
$y I \! f=\int Q I\! f+C \\$
$=y \frac{1}{x}=\int(x-1) e^{x} \frac{1}{x} d x+C \\$
$=\frac{y}{x}=\int \frac{x e^{x}}{x^{2}} d x-\int \frac{e^{x}}{x^{2}} d x+C$
\begin{aligned} &=\frac{1}{x} e^{x}-\int \frac{1}{x^{2}} e^{x} d x-\int \frac{1}{x^{2}} e^{x} d x \\ \end{aligned}
$=\frac{1}{x} e^{x} \\$
$=\frac{y}{x}=\frac{e^{x}}{x}+C \\$
$=y=e^{x}+C x$

Differential Equation exercise 21.10 question 11

Answer: $5 x y=e^{5}+C$
Hint: To solve this equation we use $e^{\int P dx}$ formula.
Give: \begin{aligned} &x \frac{d y}{d x}+\frac{y}{x}=x^{3} \\ & \end{aligned}
Solution: $\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=\frac{1}{x^{\prime}} Q=x^{3} \\ & \end{aligned}
$I\! f=e^{\int P d x} \\$
$=e^{\int \frac{1}{x} d x} \\$
$=e^{\log x} \\$
$=x \\$
$y I \! f=\int Q I\! f+C$
\begin{aligned} &=y x=\int x^{3} x d x+C \\ & \end{aligned}
$=y x=\int x^{4} d x+C \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\$
$=x y=\frac{x^{5}}{5}+C \\$
$=y=\frac{x^{4}}{5}+\frac{C}{x}$
$=5 x y=x^{5}+C$

Differential Equation exercise 21.10 question 12

Answer: $y=C e^{-x}+\frac{1}{2}(\sin x-\cos x)$
Hint: To solve this equation we use $\int u v dx$ formula.
Give: \begin{aligned} &\frac{d y}{d x}+y=\sin x \\ & \end{aligned}
Solution: $\frac{d y}{d x}+P y=Q \\$
$P=1, Q=\sin x \\$
$\text { If }=e^{\int P d x} \\$
$=e^{\int 1 d x} \\$
$=e^{x}$
\begin{aligned} &y I\! f=\int Q I\! f d x+C \\ & \end{aligned}
$y e^{x}=\int \sin x e^{x} d x+C \\$
$=I=\sin \int e^{x}-\int \frac{d}{d x} \sin x \int e^{x} d x d x \\$
$=I=\sin e^{x}-\int \cos x e^{x} d x$
\begin{aligned} &=I=\sin e^{x}-\int \cos x-e^{x} d x+\int \sin x e^{x} d x \\ & \end{aligned}
$=I=e^{x}[\sin x-\cos x]-I \\$
$=2 I=e^{x}(\sin x-\cos x) \\$
$=I=\frac{e^{x}}{2}(\sin x-\cos x)$
Put in original equation
\begin{aligned} &y e^{x}=\frac{e^{x}}{2}(\sin x-\cos x)+C \\ & \end{aligned}
$y=c e^{-x}+\frac{1}{2}(\sin x-\cos x)$

Differential Equation Exercise 21.10 Question 13

Answer: $y=\frac{1}{2}(\cos x-\sin x)+C e^{-x}$
Hint: To solve this equation we use $e^{\int Pdx}$ formula.
Give:$\frac{d y}{d x}+y=\cos x$
Solution: $\frac{d y}{d x}+P(x) y=Q(x)$
\begin{aligned} &\text { If }=e^{\int P(x) d x} \\ &\end{aligned}
$y \text { If }=\int Q(x) \text { If }d x \\$
$\frac{d y}{d x}+y=\cos x \\$
$P(x)=1, Q(x)=\cos x \\$
$=e^{\int 1 d x} \\$
$=e^{x} \\$
$y e^{x}=\int \cos x e^{x} d x \ldots(i) \\$
$y\: \text{If}=\int Q(x)\text{If} \: d x$
Suppose \begin{aligned} & I=\int \cos x e^{x} d x\\ & \end{aligned}
$=\cos e^{x}-\int(-\sin x) e^{x} d x\\$
$=-\cos x e^{x}+-\int \sin x e^{x} d x\\$
$=I=\cos x e^{x}+\sin x e^{x}-\int \cos x e^{x} d x$
\begin{aligned} &=I=e^{x}(\cos x+\sin x)-I \\ \end{aligned}
$=2 I=e^{x}(\cos x+\sin x) \\$
$=I=\frac{e^{x}}{2}(\cos x+\sin x) \ldots(i i) \\$
$=y e^{x}=\frac{e^{x}}{2}(\cos x+\sin x)+C$

Differential Equation Exercise 21.10 Question 14

Answer: $y=C e^{-x}+\frac{1}{5}(2 \sin x-\cos x)$
Hint: To solve this equation we use $e\int P dx$ formula.
Give: \begin{aligned} &\frac{d y}{d x}+P y=Q \\ & \end{aligned}
Solution: $P=2, Q=\sin x$
\begin{aligned} &\text{If}=e^{\int f d x} \\ & \end{aligned}
$=e^{\int 1 d x} \\$
$=e^{x} \\$
$=y I f=\int Q\: \text{If} \: d x+C \\$
\begin{aligned} &=y e^{x}=\int e^{2 x} \sin x d x+C \\ & \end{aligned}
$=\int e^{a x} \sin b x=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x) \\$
$=y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+C \\$
$=y=\frac{2 \sin x-\cos x}{5}+C e^{-2 x}$

Differential Equation Exercise 21.10 Question 15

Answer: $2 y \cos x=\cos 2 x+C$
Hint: To solve this equation we use $e\int f dx$ formula.
Give: \begin{aligned} &\frac{d y}{d x}=y \tan x-2 \sin x \\ & \end{aligned}
Solution: $\frac{d y}{d x}-y \tan x=-2 \sin x$
$=\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=-\tan x, Q=-2 \sin x \\ & \end{aligned}
$\text{If}=e^{\int P d x} \\$
$=e^{\int 1 d x} \\$
$=e^{x} \\$
$=y I\! f=\int Q I\! f d x+C \\$
$=y \cos x=-\int 2 \sin x \cos x d x+C$
\begin{aligned} &=y \cos x=-\int \sin 2 x d x+C \\ & \end{aligned}
$=y \cos x=\frac{\cos 2 x}{2}+C \\$
$=2 y \cos x=\cos 2 x+C$

Differential Equation Exercise 21.10 question 16

Answer:$y=\tan ^{-1} x-1+C e^{-\tan ^{-1} x}$
Hint: To solve this equation we use $e\int f dx$ formula.
Give: \begin{aligned} &\left(1+x^{2}\right) \frac{d y}{d x}+y=\tan ^{-1} x\\ & \end{aligned}
Solution: $\left(1+x^{2}\right) \frac{d y}{d x}+y=\tan ^{-1} x\\$
$=\frac{d y}{d x}+\frac{y}{\left(1+x^{2}\right)}=\frac{\tan ^{-1} x}{1+x^{2}} \ldots(i)$
\begin{aligned} &P=\frac{1}{1+x^{2}}, \cos x=\frac{\tan ^{-1} x}{1+x^{2}} \\ & \end{aligned}
$I f=e^{\int P d x} \\$
$=e^{\int \frac{1}{1+x^{2}} d x} \\$
$=e^{\tan ^{-1} x} \\$
$=I f=e^{\tan ^{-1} x} \\$
$=4 e^{\tan ^{-1} x}=\int e^{\tan ^{-1} x} \frac{\tan ^{-1} x}{1+x^{2}} d x+C$
Put $\tan ^{-1} x=t$
\begin{aligned} &=\frac{1}{1+x^{2}} d x=d t \\ & \end{aligned}
$=y e^{\tan ^{-1} x}=\int e^{t} t d t \\$
$=y e^{\tan ^{-1} x}=t \int e^{t} d t-\int \frac{d}{d t}(t) \int e^{t} d t d x+C \\$
$=t e^{t}-\int e^{t} d t+C$
\begin{aligned} &=y e^{\tan ^{-1} x}=t e^{t}-e^{t}+C \\ & \end{aligned}
$=y e^{\tan ^{-1} x}=e^{t}(t-1)+C \\$
$=y e^{\tan ^{-1} x}=e^{t}(t-1)+C \\$
$=y e^{\tan ^{-1} x}=e^{\tan ^{-1} x}\left(\tan ^{-1} x-1\right)+C \\$
$=y=\tan ^{-1} x-1+\frac{C}{\tan ^{-1} x} \\$
$=y=\tan ^{-1} x-1+C e^{-\tan ^{-1} x}$

Differential Equation exercise 21.10 question 17

Answer: $y \sec x=x+C$
Hint: To solve this equation we use $e\int P dx$ formula.
Give:$\frac{d y}{d x}+y \tan x=\cos x$
Solution: $\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=\tan x, Q=\cos x \\ \end{aligned}
$\text { If }=e^{\int P d x} \\$
$=e^{\int \tan x d x} \\$
$=e^{\log \sec x} \\$
$=\sec x$
\begin{aligned} &=y I f=\int Q I f d x+C \\ &=y \sec x=\int \sec x \cos x d x \\ &=y \sec x=\int \frac{1}{\cos x} \cos x d x \\ &=y \sec x=\int 1 d x \\ &=y \sec x=x+C \end{aligned}

Differential Equation exercise 21.10 question 22

Answer: $2 x e^{\tan ^{-1} x}=e^{2 \tan ^{-1} y}+C$
Hint: To solve this equation we use $e\int f\left ( x \right )dx$ formula.
Give: $\left(1+y^{2}\right)+\left(x-e^{2 \tan ^{-1} y}\right) \frac{d y}{d x}=0$
Solution: \begin{aligned} &x-e^{\tan ^{-1} y} \frac{d y}{d x}=-\left(1+y^{2}\right) \\ & \end{aligned}
$=\left(e^{\tan ^{-1} y}-x\right) \frac{d y}{d x}=1+y^{2} \\$
$=\left(e^{\tan ^{-1} y}-x\right) d y=\left(1+y^{2}\right) d x$
Put $\tan^{-1}y= t$
\begin{aligned} &=\frac{1}{1+y^{2}} d y=d t \\ & \end{aligned}
$=\left(e^{t}-x\right) d t=d x \\$
$=\frac{d x}{d t}=e^{t}-x \\$
$=\frac{d x}{d t}+x=e^{t} \\$
$=\frac{d y}{d x}+P(x) y=Q x$
\begin{aligned} &=\frac{d x}{d t}+P(t) t=Q(x) \\ & \end{aligned}
$P(t)=1, Q t=e^{t} \\$
$I f=e^{\int P(t) d x} \\$
$=e^{\int 1 d t} \\$
$=e^{t} \\$
$=x e^{t}=\int e^{t} e^{t} d t+C \\$
$=x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{2}+C \\$
$=2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C$

Differential Equation Exercise 21.10 Question 18

Answer: $y \sin x=x^{2} \sin x+C$
Hint: To solve this we convert $\cot x$ to $\frac{\cos x}{\sin x}$ formula.
Give: $\frac{d y}{d x}+\cot x y=x^{2} \cot x+2 x$
Solution: \begin{aligned} & \frac{d y}{d x}+P y=Q\\ & \end{aligned}
$P=\cot x, Q=x^{2} \cot x+2 x\\$
$I f=e^{\int P d x}\\$
$=e^{\int \cot x d x}\\$
$=e^{\log \sin x}\\$
$=\sin x$
\begin{aligned} &=y I f=\int Q I f d x+C \\ & \end{aligned}
$=y \sin x=\int\left(x^{2} \cot x+2 x\right) \sin x d x+C \\$
$=y \sin x=\int x^{2} \frac{\cos x}{\sin x} \sin x+2 x \sin x d x+C \\$
$=y \sin x=\int x^{2} \cos x d x+\int 2 x \sin x d x+C$
\begin{aligned} &=x^{2} \int \cos x d x-\int \frac{d}{d x} x^{2} \int \cos x d x+2 \int x \sin x d x+C \\ & \end{aligned}
${\left[\int u v d x=u \int v d x-\int \frac{d y}{d x} \int v d x d x\right]} \\$
$=x^{2} \sin x-2 \int x \sin x d x+2 \int x \sin x d x+C \\$
$=y \sin x=x^{2} \sin x+C$

Differential Equation Exercise 21.10 Question 19

Answer: $y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C$
Hint: To solve this we convert $cotx$ to $\frac{\cos x}{\sin x}$ formula.
Give: $\frac{d y}{d x}+y \tan x=x^{2} \cos ^{2} x \\$
Solution: $\frac{d y}{d x}+P y=Q \\$
\begin{aligned} &P=\tan x, Q=x^{2} \cos ^{2} x \\ \end{aligned}
$I f=e^{\int P d x} \\$
$=e^{\int \tan x d x} \\$
$=e^{\log \sec x} \\$
$=\sec x$
\begin{aligned} &=y I f=\int Q I f d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(\sec x) d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(1 / \cos x) d x+C \\ &=y \sec x=\int x^{2} \cos x d x+C \\ &=y \sec x=x^{2} \int \cos x d x-\int(2 x \cos x d x) d x+C \end{aligned}
Using integration by parts
\begin{aligned} &=y \sec x=x^{2} \sin x-2 \int x^{2} \sin x d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C \end{aligned}

Differential Equation Exercise 21.10 Question 20

Answer: $2 y e^{\tan ^{-1} x}=e^{2 \tan ^{-1} y}+C$
Hint: To solve this equation we use $I\int f\left ( x \right )dx$ formula.
Give: $\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x} \$
Solution: $\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x} \\$
\begin{aligned} & \\ &\quad=\frac{d y}{d x}+\frac{y}{\left(1+x^{2}\right)}=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \ldots(i) \\\\ &\quad=\frac{d y}{d x}+P(x) y=Q x \end{aligned}
\begin{aligned} &P=\frac{1}{1+x^{2}}, Q=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \\\\ &I f=e^{\int P d x} \\\\ &=e^{\int \frac{1}{1+x^{2}} d x} \\\\ &=e^{\tan ^{-1} x} \ldots(i i) \end{aligned}
\begin{aligned} &=e^{\tan ^{-1} x} \frac{d y}{d x}+e^{\tan ^{-1} x} \frac{y}{1+x^{2}}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{1+x^{2}} \\\\ &=\frac{d}{d x}\left[y e^{\tan ^{-1} x}\right]=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{1+x^{2}} \\\\ & \end{aligned}

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

$=y e^{\tan ^{-1} x}=\int \frac{e^{\tan ^{-1} x^{2}}}{1+x^{2}} d x+C$

Differential Equation Exercise 21.10 Question 21

Answer:$y=x^{4}+x^{2} \log x+C x^{2}$
Hint: To solve this we convert formula.
Give: $x d y=\left(2 y+2 x^{4}+x^{2}\right) d x \\$
Solution:\begin{aligned} & & \frac{d y}{d x}=\frac{2 y+2 x^{4}+x^{2}}{x} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{2 y}{x}+2 x^{3}+x \\ &=\frac{d y}{d x}-\frac{2 y}{x}=2 x^{3}+x \\ &\frac{d y}{d x}+P y=Q \end{aligned}
\begin{aligned} &P=-\frac{2}{x}, Q=2 x^{3}+x \\ &I f=e^{\int P d x} \\ &=e^{-\int \frac{2}{x} d x} \\ &=e^{=2 \log x} \\ &=e^{\log x^{-2}} \\ &=x^{-2} \\ &=\frac{1}{x^{2}} \quad\left[e^{\log x}=x\right] \end{aligned}
\begin{aligned} &=y I f=\int Q I f d x+C \\ &=\frac{y}{x^{2}}=\int \frac{2 x^{3}+x}{x^{2}} d x+C \\ &=\frac{y}{x^{2}}=\int 2 x+\frac{1}{x} d x+C \\ &=\frac{y}{x^{2}}=x^{2}+\log x+C \\ &=y=x^{4}+x^{2} \log x+C x^{2} \end{aligned}

Differential Equation Exercise 21.10 Question 23

Answer: $y=\left(\frac{y+1}{y}\right)+C e^{\frac{1}{y}}$
Hint: To solve this equation we use formula.
Give: \begin{aligned} &y^{2} \frac{d x}{d y}+x-\frac{1}{y}=0 \\ & \end{aligned}
Solution: $\frac{1}{y^{2}}\left[y^{2}\left(\frac{d x}{d y}\right)+x-\frac{1}{y}\right]=0$
\begin{aligned} &=\frac{d y}{d x}+\frac{x^{2}}{y^{2}}-\frac{1}{y^{3}}=0 \\ & \end{aligned}
$=\frac{d y}{d x}+\frac{x^{2}}{y^{2}}=\frac{1}{y^{3}} \\$
$=\frac{d x}{d y}+P(x)=Q \\$
$P=\frac{1}{y^{\prime}} Q=\frac{1}{y^{3}} \\$
$I f=e^{\int P d y}$
\begin{aligned} &=x I f=\int I f Q d y \\ & \end{aligned}
$\frac{d x}{d y}+\frac{x}{y^{2}}=\frac{1}{y^{3}} \\$
$=I f=e^{\int \frac{1}{y^{2}} d y} \\$
$=e^{-\frac{1}{y}} \\$
$=x e^{-\frac{1}{y}}=e^{-\frac{1}{y}} \frac{1}{y^{3}} d y$
\begin{aligned} &=\int \frac{1}{y} e^{-\frac{1}{y}} \frac{1}{y^{2}} d y \\ & \end{aligned}
$=-\frac{1}{y}=t \\$
$=\frac{1}{y^{2}} d y=d t \\$
$=d y=y^{2} d t \\$
$-\int t e^{t} d t$
\begin{aligned} &=x e^{-\frac{1}{y}}=\left[t e^{t}-e^{t}\right]+C \\ & \end{aligned}
$=x e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C \\$
$=x e^{-\frac{1}{y}}=\frac{1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C \\$
$=x e^{-\frac{1}{y}}=e^{-\frac{1}{y}}\left(\frac{1}{y}+1+\frac{C}{e^{-\frac{1}{y}}}\right) \\$
$=x=\frac{1+y}{y}+C e^{\frac{1}{5}}$

Differential Equation Exercise 21.10 Question 24

Answer: $x=2 y^{3}+C y^{-2}$
Hint: To solve this equation we use formula.
Give: \begin{aligned} &\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0 \\ & \end{aligned}
Solution: $\frac{d y}{d x}=-\frac{y}{2 x-10 y^{3}}$
\begin{aligned} &=\frac{d y}{d x}=\frac{-\left(2 x-10 y^{3}\right)}{y} \\ & \end{aligned}
$=\frac{d y}{d x}=-\frac{2 x}{y}+\frac{10 y^{3}}{y} \\$
$=\frac{d x}{d y}+\frac{2 x}{y}=10 y^{2} \\$
$=\frac{d x}{d y}+R x=S \\$
$R=\frac{2 x}{y}, S=10 y^{2}$
\begin{aligned} &I f=e^{\int R d y} \\ & \end{aligned}
$=e^{2 \int \frac{1}{y} d y}$
\begin{aligned} &=e^{2 \log y} \\ & \end{aligned}
$=y^{2} \\$
$=x I f=\int S I f+C \\$
$=x y^{2}=\int S I f+C \\$
$=x y^{2}=\int 10 y^{2} y^{2}+C$
\begin{aligned} &=x y^{2}=\int 10 y^{4}+C \\ &=x y^{2}=\frac{10 y^{5}}{5}+C \\ & \end{aligned}
$=x y^{2}=2 y^{5}+C \\$
$=x=2 y^{3}+\frac{C}{y^{2}} \\$
$=x=2 y^{3}+C y^{-2}$

Differential Equation Exercise 21.10 Question 25

Answer: $x=\tan y+C \sqrt{\tan y}$
Hint: To solve this equation we use $\tan x$ and convert it to $\sin x$ and $\cos x$
Give: \begin{aligned} &(x+\tan y) d y=\sin 2 y d x \\ & \end{aligned}
Solution: $(x+\tan y) d y=\sin 2 y d x$
\begin{aligned} &=\frac{d x}{d y}=\frac{(x+\tan y)}{\sin 2 y} \\ &=\frac{d x}{d y}=\frac{x}{\sin 2 y}+\frac{\tan y}{\sin 2 y} \\ &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{\frac{\sin y}{\cos y}}{2 \sin y \cos y} \\ &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{\frac{\sin y}{\cos y}}{2 \sin y \cos y} \end{aligned}
\begin{aligned} &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{1}{2 \cos ^{2} y} \\ &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{1}{2} \sec ^{2} y \\ &=\frac{d x}{d y}+R x=S \end{aligned}
\begin{aligned} &R=\operatorname{cosec} 2 y, S=\frac{1}{2} \sec ^{2} y \\ & \end{aligned}
$\text { If }=e^{\int R d y} \\$
$=e^{\int(-\operatorname{cosec} 2 y) d y} \\$
$=e^{-\log |\operatorname{cosec} 2 y-\cot 2 y|} \\$
$=\operatorname{cosec} 2 y-\cot 2 y \\$
$=\frac{1}{\sin 2 y}-\frac{\cos 2 y}{\sin 2 y} \\$
$=\frac{1-\cos 2 y}{\sin 2 y}$
\begin{aligned} &=\frac{2 \sin ^{2} y}{2 \sin y \cos y} \\ & \end{aligned}
$=\frac{\sin y}{\cos y} \\$
$=\tan y \\$
$=e^{-\log |\tan y|} \\$
$=e^{\log |\cot y|} \\$
$=\cot y$
\begin{aligned} &=x I f=\int S I f+C \\ &=x \cot y=\int \frac{1}{2} \sec ^{2} y \cot y d y+C \\ &=x \cot y=\int \frac{1}{2 \cos ^{2} y} \frac{\cos y}{\sin y} d y+C \\ &=x \cot y=\int \frac{1}{2 \cos y} \frac{1}{\sin y} d y+C \\ &=x \cot y=\int \frac{1}{\sin 2 y} d y+C \\ &=x \cot y=\frac{1}{2} \log |\operatorname{cosec} 2 y-\cot 2 y|+C \end{aligned}

Differential Equation Excercise 21.10 Question 26

Answer: $x=(\tan y+C) e^{-y}$
Hint: To solve this equation we use where are constants.
Give: $d x+x d y=e^{-y} \sec ^{2} y d y$
Solution: \begin{aligned} & d x=e^{-y} \sec ^{2} y d y-x d y \\ & \end{aligned}
$=\frac{d x}{d y}+x=e^{-y} \sec ^{2} y$
First order linear differential equation form
\begin{aligned} &=\frac{d x}{d y}=P x-Q \\ &P=1 \text { and } e^{-y} \sec ^{2} y \end{aligned}
If of differential equation is

\begin{aligned} &I f=e^{\int R d y} \\ & \end{aligned}
$=e^{\int 1 d y} \\$
$\int d y=y+C \\$
$=I f=e^{y}$ $\quad\left[e^{\log x}=x\right]$
\begin{aligned} &=x(I f)=\int Q I F d y+C \\ & \end{aligned}
$=x e^{y}=\int e^{-y} \sec ^{2} y e^{y} d y+C \\$
$=x e^{y}=\int \sec ^{2} y d y+C \\$
$=x e^{y} \times \frac{1}{e^{y}}=(\tan y+C) \times \frac{1}{e^{y}}$ $\quad\left[\int \sec ^{2} y d y=\tan x+C\right] \\$
$=x=(\tan y+C) e^{-y}$

Differential Equation Excercise 21.10 Question 27

Answer: $2 y \cos x=\cos 2 x+C$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give: \begin{aligned} & \frac{d y}{d x}=y \tan x-2 \sin x \\ & \end{aligned}
Solution: $d x=\frac{d y}{d x}-y \tan x=-2 \sin x$
\begin{aligned} &=\frac{d x}{d y}+P y=Q \\ &P=-\tan x \text { and } Q=-2 \sin x \end{aligned}
$If$ of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}
$=e^{-\int \tan x d x} \\$
$=e^{\log \cos x}$
\begin{aligned} &=\cos x \quad\left[e^{\log x}=x\right] \\ & \end{aligned}
$y I f=\int \text { QIf } d x+C \\$
$=y \cos x=-\int 2 \sin x \cos x d x+C \\$
$=y \cos x=-\int \sin 2 x d x+C \\$
$=y \cos x=\frac{\cos 2 x}{2}+C \\$
$=2 y \cos x=\cos 2 x+C$

Differential Equation Excercise 21.10 Question 28

Answer:$y e^{\sin x}=(\sin x-1) e^{\sin x}+C$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give: \begin{aligned} &\frac{d y}{d x}+y \cos x=\sin x \cos x \\ & \end{aligned}
Solution:$\frac{d y}{d x}+y \cos x=\sin x \cos x$
\begin{aligned} &=\frac{d x}{d y}+P y=Q \\ &P=\cos x \text { and } Q=\sin x \cos x \end{aligned}
$If$ of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}
$=e^{\int \cos x d x} \\$
$=e^{\sin x} \\$
$=\sin x \quad\left[e^{\sin x}=\sin x\right] \\$
$y I f=\int \text { QIf } d x+C$
\begin{aligned} &y e^{\sin x}=\sin x \cos x e^{\sin x} d x \quad[\sin x=t, \cos x d x=d t] \\ & \end{aligned}
$=\int t e^{t} d t \\$
$=t e^{t}-\int e^{t}+C \\$
$=t e^{t}-e^{t}+C \\$
$=e^{t}(t-1)+C \\$
$=y e^{\sin x}=(\sin x-1) e^{\sin x}+C$

Differential Equation exercise 21.10 question 29

Answer: $y=\left(1+x^{2}\right)\left(x+\tan ^{-1} x+C\right)$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P, Q$ are constants.
Give: $\left(1+x^{2}\right) \frac{d y}{d x}-2 x y=\left(x^{2}+2\right)\left(x^{2}+1\right)$
Solution: $\frac{d y}{d x}+\left(\frac{-2 x y}{1+x^{2}}\right)=\frac{\left(x^{2}+2\right)\left(x^{2}+1\right)}{1+x^{2}}$
\begin{aligned} &\frac{d y}{d x}+\left(\frac{-2 x}{1+x^{2}}\right) y=x^{2}+2 \\ &=\frac{d x}{d y}+P y=Q \\ &P=-\frac{2 x}{1+x^{2}}, Q=x^{2}+2 \end{aligned}
$If$ of differential equation is

\begin{aligned} &\text { If }=e^{-\int \frac{2 x}{1+x^{2}} d x} \quad\left[1+x^{2}=t, 2 x d x=d t\right] \\ & \end{aligned}
$=e^{-\int \frac{d t}{t}} \\$
$=e^{-\ln t} \\$
$=e^{\ln \left(t^{-1}\right)} \\$
$=t^{-1} \\$
$=\frac{1}{t}$
\begin{aligned} &=\frac{1}{1+x^{2}} \\ & \end{aligned}
$y \text { If }=\int \text { QIf } d x+C \\$
$y \frac{1}{1+x^{2}}=\int\left(x^{2}+2\right) \frac{1}{x^{2}+1} d x+C \\$
$=\int \frac{x^{2}+1+1}{x^{2}+1} d x+C \\$
$=\int \frac{x^{2}+1}{x^{2}+1} d x+\int \frac{1}{x^{2}+1} d x+C \\$
$=\int d x+\tan ^{-1} x+C$
\begin{aligned} &=x+\tan ^{-1} x+C \\ & \end{aligned}
$=y=\left(1+x^{2}\right)\left(x+\tan ^{-1} x+C\right)$

Differential Equation exercise 21.10 question 30

Answer: $y \sin x=\frac{2}{3} \sin ^{3} x+C$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give: \begin{aligned} &(\sin x) \frac{d y}{d x}+y \cos x=2 \sin ^{2} x \cos x \\ & \end{aligned}
Solution: $\sin x \frac{d y}{d x}+y \cos x=2 \sin ^{2} x \cos x$
\begin{aligned} &=\frac{d y}{d x}+\frac{y \cos x}{\sin x}=\frac{2 \sin ^{2} x \cos x}{\sin x} \\ & \end{aligned}
$=\frac{d y}{d x}+y \cot x=2 \sin x \cos x \ldots(i) \\$
$=\frac{d y}{d x}+P y=Q \\$
$P=\cot x, Q=2 \sin x \cos x$
$If$ of differential equation is
\begin{aligned} &\text { If }=e^{\int P d x} \\ & \end{aligned}
$=e^{\int \cot x d x} \\$
$=e^{\log |\sin x|} \\$
$=\sin x \\$
$y \text { If }=\int \text { QIf } d x+C$
\begin{aligned} &=y \sin x=\int 2 \sin ^{2} x \cos x d x \\ & \end{aligned}
$=2 \int \sin ^{2} x \cos x d x$ $\quad[\sin x=t, \cos x d x=d t] \\$
$=2 \int t^{2} d t \\$
$=2\left[\frac{t^{3}}{3}\right]+C$
\begin{aligned} &=\frac{2}{3} t^{3}+C \\ & \end{aligned}
$=\frac{2}{3} \sin ^{3} x+C \\$
$y \sin x=\frac{2}{3} \sin ^{3} x+C$

Differential Equation exercise 21.10 question 31

Answer: $\frac{y(x-1)^{3}}{x+1}\left(x^{2}-6 x+8 \log |x+1|\right)+C$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give: $\left(x^{2}-1\right) \frac{d y}{d x}+2(x+2) y=2(x+1)$
Solution: \begin{aligned} &\frac{d y}{d x}+\frac{2(x+2)}{x^{2}-1} y=\frac{2}{x-1} \\ & \end{aligned}
$=\frac{d y}{d x}+P y=Q \\$
$P=\frac{2(x+2)}{x^{2}-1}, Q=\frac{2}{x-1}$
$If$ of differential equation is
\begin{aligned} &I f=e^{\int \frac{2(x+2)}{x^{2}-1} d x} \\ & \end{aligned}
$=e^{\int\left(\frac{2 x}{x^{2}-1}+\frac{4}{x^{2}-1}\right) d x} \\$
$=e^{\int \ln \left|x^{2}-1\right|+4 \times \frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|} \\$
$=e^{\int \ln \left|x^{2}-1\right|+2 \ln \left|\frac{x-1}{x+1}\right|} \\$
$=e^{\int \ln \left|x^{2}-1\right|+\ln \left|\frac{(x-1)^{2}}{(x+1)^{2}}\right|} \\$
$=e^{\ln \left(x^{2}-1\right) \frac{(x-1)^{2}}{(x+1)^{2}}}$
\begin{aligned} &=e^{\ln (x+1)(x-1) \frac{(x-1)^{2}}{(x+1)^{2}}} \\ &=e^{\ln (x-1) \frac{(x-1)^{2}}{(x+1)}} \\ &=e^{\ln \frac{(x-1)^{3}}{(x+1)}} \\ &=\frac{(x-1)^{3}}{x+1} \\ &y \text { If }=\int \text { QIf } d x+C \end{aligned}
\begin{aligned} &=y \frac{(x-1)^{3}}{x+1}=\int\left(\frac{2}{x-1}\right) \frac{(x-1)^{3}}{x+1} d x+C \\ & \end{aligned}
$=2 \int \frac{(x-1)^{3}}{x+1} d x+C \quad[x+1=t, d x=d t] \\$
$=2 \int \frac{\left(t^{2}+4-4 t\right)}{t} d t$
\begin{aligned} &=2\left[\frac{t^{2}}{2}\right]+8 \log |t|-8 t+C \\ & \end{aligned}
$=t^{2}+8 \log |t|-8 t+C \\$
$=(x+1)^{2}+8 \log |x+1|-8(x+1)+C \\$
$=x^{2}+2 x+1+8 \log |x+1|-8 x-8+C \\$
$=y \frac{(x-1)^{3}}{(x+1)}\left(x^{2}-6 x+8 \log |x+1|\right)+C$

Differential Equation exercise 21.10 question 32

Answer: $y=\sin x+\frac{2 \cos x}{x}-\frac{2 \sin x}{x}+\frac{c}{x^{2}}$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$+ are constants.
Give: \begin{aligned} &x \frac{d y}{d x}+2 y=x \cos x \\ & \end{aligned}
Solution: $\frac{d y}{d x}+\frac{2 y}{x}=\frac{x \cos x}{x}$
\begin{aligned} &=\frac{d y}{d x}+\frac{2}{x} y=\cos x \\ & \end{aligned}
$=\frac{d y}{d x}+P y=Q \\$
$P=\frac{2}{x^{\prime}} Q=\cos x$
$If$ of differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}
$=e^{\int \frac{2}{x} d x} \\$
$=e^{2 \log x} \\$
$=e^{\log x^{2}} \\$
$=x^{2} \\$
$y I f=\int \text { QIf } d x+C$
\begin{aligned} &=y x^{2}=\int \cos x x^{2} d x+C \\ & \end{aligned}
$=x^{2}(\sin x)-\int 2 \sin x d x+C \\$
$=x^{2}(\sin x)-2\left[x(\cos x)-\int 1(\cos x) d x+C\right] \\$
$=x^{2}(\sin x)-2 x(\cos x)+2 \int \cos x d x+C \\$
$=x^{2}(\sin x)-2 x(\cos x)-2 \sin x+C \\$
$=y x^{2}=x^{2}(\sin x)-2 x(\cos x)-2 \sin x+C$
\begin{aligned} &=y=\frac{x^{2}(\sin x)}{x^{2}}-\frac{2 x(\cos x)}{x^{2}}-\frac{2 \sin x}{x^{2}}+\frac{C}{x^{2}} \\ & \end{aligned}
$=y=(\sin x)-\frac{2(\cos x)}{x}-\frac{2 \sin x}{x^{2}}+\frac{C}{x^{2}}$

Differential Equation Exercise 21.10 Question 33

Answer: $y=\left(\frac{x^{2}}{2}+C\right) e^{x}$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give:\begin{aligned} &\frac{d y}{d x}-y=x e^{x} \\ & \end{aligned}
Solution: $\frac{d y}{d x}+(-1) y=x e^{x}$
\begin{aligned} &=\frac{d y}{d x}+P y=Q \\ &P=-1, Q=x e^{x} \end{aligned}
$If$ of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}
$=e^{\int-1 d x} \\$
$=e^{-\int d x} \\$
$=e^{-x} \\$
$y I f=\int Q I f d x+C \\$
$=y\left(e^{-x}\right)=\int x e^{x} e^{-x} d x+C$
\begin{aligned} &=y\left(e^{-x}\right)=\int x e^{x-x} d x+C \\ & \end{aligned}
$=y\left(e^{-x}\right)=\int x e^{0} d x+C \\$
$=y\left(e^{-x}\right)=\int x d x+C \\$
$=y\left(e^{-x}\right)=\frac{x^{1+1}}{1+1}+C \\$
$=y\left(e^{-x}\right)=\frac{x^{2}}{2}+C$
\begin{aligned} &=y\left(e^{-x}\right) e^{x}=e^{x}\left(\frac{x^{2}}{2}+C\right) \\ & \end{aligned}
$=y\left(e^{-x+x}\right)=e^{x}\left(\frac{x^{2}}{2}+C\right) \\$
$=y\left(e^{0}\right)=e^{x}\left(\frac{x^{2}}{2}+C\right) \\$
$=y=e^{x}\left(\frac{x^{2}}{2}+C\right)$

Differential Equation Exercise 21.10 Question 34

Answer: $y=\frac{x e^{4 x}}{6}-\frac{1}{36} e^{4 x}+C e^{-2 x}$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give: \begin{aligned} & \frac{d y}{d x}+2 y=x e^{4 x} \\ & \end{aligned}
Solution: $\frac{d y}{d x}+(2) y=x e^{4 x}$
\begin{aligned} &=\frac{d y}{d x}+P y=Q \\ &P=2, Q=x e^{4 x} \end{aligned}
$If$ of differential equation is

\begin{aligned} &\text { If }=e^{\int P d x} \\ & \end{aligned}
$=e^{\int 2 d x} \\$
$=e^{2 \int d x} \\$
$=e^{2 x} \\$
$y I f=\int \text { QIf } d x+C \\$
$=y\left(e^{2 x}\right)=\int x e^{4 x} e^{2 x} d x+C \\$
$=y e^{2 x}=\int x e^{6 x} d x+C$
\begin{aligned} &\quad=y e^{2 x}=\int x e^{6 x} d x+C\left[\int f(x) g(x) d c=f(x) \int[g(x) d x] \int f(x)\left[\int g(x) d x\right] d x\right] \\ & \end{aligned}
$=y e^{2 x}=x\left(\int e^{6 x} d x\right)-\int \frac{d}{d x}(x)\left(\int e^{6 x} d x\right) d x+C \\$
$=y e^{2 x}=x\left(\frac{e^{6 x}}{6}\right)-\int 1\left(\frac{e^{6 x}}{6}\right) d x+C \\$
$=y e^{2 x}=x\left(\frac{e^{6 x}}{6}\right)-\frac{1}{6}\left(\frac{e^{6 x}}{6}\right)+C \\$
$=y e^{2 x}=e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C$
\begin{aligned} &=y e^{2 x} e^{-2 x}=e^{-2 x}\left[e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C\right] \\ & \end{aligned}
$=y e^{2 x-2 x}=e^{-2 x} e^{6 x}\left(\frac{x}{6}\right)-e^{-2 x} \frac{1}{36}\left(e^{6 x}\right)+C e^{-2 x} \\$
$=y e^{0}=e^{6 x-2 x}\left(\frac{x}{6}\right)-e^{6 x-2 x} \frac{1}{36}+C e^{-2 x} \\$
$=y=e^{4 x}\left(\frac{x}{6}\right)-e^{4 x} \frac{1}{36}+C e^{-2 x}$

Differential Equation Exercise 21.10 Question 35

Answer: $x y^{-1}=2 y+C, C=0$
Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ are constants.
Give: \begin{aligned} &\left(x+2 y^{2}\right) \frac{d y}{d x}=y \text { when } x=2, y=1 \\ & \end{aligned}
Solution: $\frac{d x}{d y} y=x+2 y^{2}$
\begin{aligned} &=\frac{d x}{d y}=\frac{x+2 y^{2}}{y} \\ & \end{aligned}
$=\frac{d x}{d y}=\frac{x}{y}+\frac{2 y^{2}}{y} \\$
$=\frac{d x}{d y}-\frac{x}{y}=2 y \\$
$=\frac{d y}{d x}+P y=Q \\$
$P=-\frac{1}{y^{\prime}} Q=2 y$
$If$ of differential equation is
\begin{aligned} &I f=e^{\int P d y} \\ & \end{aligned}
$=e^{\int-\frac{1}{y} d y} \\$
$=e^{-\log y} \\$
$=\frac{1}{5} \\$
$x I f=\int \text { QIf } d x+C \\$
$=x\left(\frac{1}{y}\right)=\int 2 y\left(\frac{1}{y}\right) d y+C$
\begin{aligned} &=\frac{x}{y}=2 \int d y+C \\ & \end{aligned}
$=\frac{x}{y}=2 y+C \\$
$=x=y(2 y+C) \\$
$=x=2 y^{2}+C \\$
$\text { When } x =2, y=1 \\$
$=2=2+C \\$
$=C=2-2=0$

Differential Equation exercise 21 point 10 question 36 (i)

Answer:$y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.$ ,otherwise
Give:$\frac{d y}{d x}+3 y=e^{m x}, m$ is a given real number.
Hint: Use $\int e^{x} d x$
Explanation: \begin{aligned} & \frac{d y}{d x}+3 y=e^{m x} \\ & \end{aligned}
$=\frac{d y}{d x}+(3) y=e^{m x}$
This is a first order linear differential equation of the form
$=\frac{d y}{d x}+P y=Q$
Here $P=3 \text { and } Q=e^{m x}$
The integrating factor $If$ of the differential equation is
\begin{aligned} &I f=e^{\int P d x}\\ &=e^{\int 3 d x}\\ &=e^{3 \int d x}\\ &=e^{3 x} \quad\left[\int d c=x+C\right] \end{aligned}
Hence, the solution of differential equation is
\begin{aligned} &y(I f)=\int Q I f d x+C \\ &=y\left(e^{3 x}\right)=\int e^{m x} e^{3 x} d x+C \\ &=y\left(e^{3 x}\right)=\int e^{m x+3 x} d x+C \\ &=y\left(e^{3 x}\right)=\int e^{x(m+3)} d x+C \end{aligned}
Case 1: $m+3=0 \text { or } m=-3$
When $m+3=0$ , we have $e^{x\left ( m+3 \right )}$
\begin{aligned} &=e^{0}=1 \\ & \end{aligned}
$\Rightarrow y e^{3 x}=\int d x+C \\$
$\Rightarrow y e^{3 x}=x+C \\$
$\Rightarrow y e^{3 x} e^{-3 x}=(x+C) e^{-3 x} \\$
$\Rightarrow y e^{3 x-3 x}=(x+C) e^{-3 x}$
$\Rightarrow y=(x+C) e^{-3 x} \quad\left[e^{3 x-3 x}=e^{0}=1\right]$
Case 2: $m+3 \neq 0 \text { or } m \neq-3$
When $m+3 \neq 0$ we have
\begin{aligned} &y e^{3 x}=\int e^{x(m+3)} d x+C \\ & \end{aligned}
$\Rightarrow y e^{3 x}=\frac{e^{(m+3) x}}{m+3}+C \\$
$\Rightarrow y e^{3 x} e^{-3 x}=\left(\frac{e^{(m+3) x}}{m+3}+C\right) e^{-3 x} \\$
$\Rightarrow y e^{3 x} e^{-3 x}=\frac{\left(e^{m x} e^{3 x}\right) e^{-3 x}}{m+3}+C e^{-3 x} \\$
$\Rightarrow y=\frac{e^{m x}}{m+3}+C e^{-3 x}$
Thus the solution of the given differential equation is
$y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.$

Differential Equation exercise 21.10 question 36 subquetsion (ii)

Answer:$y=\frac{1}{5}(2 \sin 2 x-\cos 2 x)+C e^{x}$
Give: $\frac{d y}{d x}-y=\cos 2 x$
Hint: Using integration by parts
Explanation: \begin{aligned} & \frac{d y}{d x}-y=\cos 2 x \\ & \end{aligned}
$\frac{d y}{d x}+(-1) y=\cos 2 x$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=-1 \text { and } Q=\cos 2 x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}
$=e^{\int-1 d x} \\$
$=e^{-\int 1 d x} \quad\left[\int d c=x+C\right] \\$
$=e^{-x}$
Hence, the solution of differential equation is
\begin{aligned} &y(I f)=\int Q I f d x+C \\ & \end{aligned}
$\Rightarrow y\left(e^{-x}\right)=\int(\cos 2 x) e^{-x} d x+C \\$
$\Rightarrow y e^{-x}=\int e^{-x} \cos 2 x d x+C \\$
$\Rightarrow y e^{-x}=\int\left(e^{-x}\right)(\cos 2 x) d x+C$
\begin{aligned} \text { Let } I=& \int\left(e^{-x}\right)(\cos 2 x) d x+C \\ \end{aligned}
$\Rightarrow I=e^{-x} \int \cos 2 x d x-\int \frac{d}{d x}\left(e^{-x}\right)\left(\int \cos 2 x d x\right) d x+C \\$
$\Rightarrow I=e^{-x}\left(\frac{\sin 2 x}{2}\right)-\int-e^{-x}\left(\frac{\sin 2 x}{2}\right) d x+C \\$
$\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2} \int e^{-x} \sin 2 x d x+C$
\begin{aligned} &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[e^{-x}\left(\int \sin 2 x d x\right)\right]-\int \frac{d}{d x} e^{-x}\left(\int \sin 2 x d x\right) d x+C\right\} \\ & \end{aligned}
$\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[e^{-x}\left(\frac{-\cos 2 x}{2}\right)\right]-\int-e^{-x}\left(-\frac{\cos 2 x}{2}\right) d x+c\right\} \\$
$\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} \int-e^{-x}(-\cos 2 x) d x+C\right\} \\$
$\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} \int e^{-x} \cos 2 x d x+c\right\}$
\begin{aligned} &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} I\right\} \\ & \end{aligned}
$\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left(-\frac{1}{2} e^{-x} \cos 2 x\right)+\frac{1}{2}\left(-\frac{1}{2} I\right) \\$
$\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x-\frac{1}{4} I \\$
$\Rightarrow I+\frac{1}{4} I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x$
\begin{aligned} &\Rightarrow I+\frac{1}{4} I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x \\ & \end{aligned}
$\Rightarrow \frac{5}{4} I=\frac{1}{4} 2 e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x \\$
$\Rightarrow \frac{5}{4} I=\frac{1}{4} e^{-x}(2 \sin 2 x-\cos 2 x) \\$
$\Rightarrow 5 I=e^{-x}(2 \sin 2 x-\cos 2 x) \\$
$\therefore I=\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)$
By substituting the value of in the original integral we get
\begin{aligned} &\Rightarrow y e^{-x}=\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)+C \\ & \end{aligned}
$\Rightarrow y e^{-x} e^{x}=e^{x}\left[\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)+C\right] \\$
$\Rightarrow y e^{-x} e^{x}=\frac{e^{x-x}}{5}(2 \sin 2 x-\cos 2 x)+C e^{x} \\$
$\therefore y=\frac{1}{5}(2 \sin 2 x-\cos 2 x)+C e^{x}$

Differential Equation exercise 21.10 question 36 (iii)

Answer: \begin{aligned} & y=-e^{x}+C x \\ & \end{aligned}
Give: $x \frac{d y}{d x}-y=(x+1) e^{x}$
Hint: Using $\int \frac{1}{x}dx$
Explanation: $x \frac{d y}{d x}-y=(x+1) e^{x}$
Divide by $x$
\begin{aligned} &\frac{d y}{d x}-\frac{y}{x}=\frac{(x+1) e^{x}}{x} \\ &\frac{d y}{d x}+\left(-\frac{1}{x}\right) y=\frac{(x+1) e^{x}}{x} \end{aligned}
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=-\frac{1}{x} \text { and } Q=\frac{(x+1) e^{x}}{x} \end{aligned}
The integrating factor of this differential equation is
\begin{aligned} &\text { If }=e^{\int P d x} \\ &=e^{\int-\frac{1}{x} d x} \\ &=e^{-\int x^{-1} d x} \quad\left[\int d c=x+C\right] \\ &=e^{-x} \\ &=x^{-1}=\frac{1}{x} \end{aligned}
Hence, the solution of differential equation is
\begin{aligned} &y(I f)=\int Q I f d x+C \\ &\Rightarrow y\left(e^{-x}\right)=\int \frac{(x+1) e^{x}}{x}\left(\frac{1}{x}\right) d x+C \end{aligned}
\begin{aligned} &\Rightarrow y\left(e^{-x}\right)=\int \frac{(x+1) e^{x}}{x^{2}} d x+C \\ &=y\left(e^{-x}\right)=I_{1}+C \ldots(i) \\ &=I_{1}=\int \frac{(x+1) e^{-x}}{x^{2}} d x \\ &=I_{1}=\int \frac{(x) e^{-x}}{x^{2}} d x+\int \frac{(1) e^{-x}}{x^{2}} d x \\ &=I_{1}=\int \frac{e^{-x}}{x} d x+\int \frac{e^{-x}}{x^{2}} d x \end{aligned}
\begin{aligned} &=I_{1}=\frac{1}{x} \frac{e^{-x}}{(-1)}-\int\left(\frac{1}{x^{2}}\right) \frac{e^{-x}}{1} d x+\int \frac{1}{x^{2}} e^{x} d x \\ &=I_{1}=\frac{-e^{-x}}{x}-\int \frac{e^{-x}}{x^{2}} d x+\int \frac{e^{x}}{x^{2}} d x \\ &=I_{1}=\frac{-e^{-x}}{x} \end{aligned}
By $\frac{y}{x}=-\frac{e^{-x}}{x}+C$
Multiplying by x we get
$y=-e^{-x}+C x$

Differential Equation exercise 21.10 question 36 (iv)

Answer: \begin{aligned} &\frac{1}{5} x^{4}+\frac{c}{x} \\ & \end{aligned}
Give: $x \frac{d y}{d x}+y=x^{4} \\$
Hint: Using $\int x^{n} d x$
Explanation: $x \frac{d y}{d x}+y=x^{4}$
Divide by x
\begin{aligned} &\frac{d y}{d x}+\frac{y}{x}=\frac{x^{4}}{x} \\ &\frac{d y}{d x}+\left(\frac{1}{x}\right) y=x^{3} \end{aligned}

This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{x} \text { and } Q=x^{3} \end{aligned}

The integrating factor $If$ of this differential equation is

$I f=e^{\int P d x}$
\begin{aligned} &=e^{\int \frac{1}{x} d x} \\ &=e^{\log x} \quad\left[\int \frac{1}{x} d c=\log |x|+C\right] \\ &=x \end{aligned}
Hence, the solution of differential equation is
\begin{aligned} &y(I f)=\int Q I f d x+C \\ &=y x=\int x^{3} x d x+C \\ &=y x=\int x^{4} d x+C \\ &=y x=\frac{x^{5}}{5}+C \end{aligned}
Divide by x, we get
$=y=\frac{x^{4}}{5}+\frac{C}{x}$

Differential Equation Exercise 21.10 Question 36 (v)

Answer: \begin{aligned} &y=\frac{1}{2} \log x+\frac{c}{\log x}\\ & \end{aligned}
Give:$x \log x) \frac{d y}{d x}+y=\log x\\$
Hint: Using $\int \frac{1}{x} d x\\$
Explanation:$(x \log x) \frac{d y}{d x}+y=\log x\\$
Divide by $x \log x$
$\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{x \log x} \text { and } Q=\frac{1}{x} \end{aligned}

The integrating factor $If$ of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{1}{x \log x} d x} \quad\left[\log x=t, \frac{1}{x} d x=d t\right] \\ &=e^{\int \frac{1}{t} d t} \\ &=e^{\log t} \end{aligned}
\begin{aligned} &=e^{\log (\log x)} \quad\left[e^{\log x}=x\right] \\ &=\log x \end{aligned}
Hence, the solution of the differential equation is
\begin{aligned} &y(I f)=\int \text { QIfd } x+C \\ &=y \log x=\int \frac{1}{x} \log x d x+C \\ &=y \log x=\frac{\log ^{2} x}{x}+C \quad\left[\int x d x=\frac{x^{2}}{2}\right] \\ &=y=\frac{1}{\log x}\left(\frac{\log ^{2} x}{2}+C\right) \\ &=y=\frac{\log ^{2} x}{2 \log x}+\frac{C}{\log x} \\ &=y=\frac{\log x}{2}+\frac{C}{\log x} \end{aligned}

Differential Equation Exercise 21.10 Question 36 (vi)

Answer: \begin{aligned} &\left(1+x^{2}\right)\left(x+\tan ^{-1} x+C\right) \\ & \end{aligned}
Give: $\frac{d y}{d x}-\frac{2 x y}{1+x^{2}}=x^{2}+2 \\$
Hint: Using $\int \frac{1}{1+x^{2}} d x$
Explanation:\begin{aligned} & \frac{d y}{d x}-\frac{2 x y}{1+x^{2}}=x^{2}+2 \\ & \end{aligned}
$=\frac{d y}{d x}-\left(\frac{2 x}{1+x^{2}}\right) y=x^{2}+2$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{-2 x}{1+x^{2}} \text { and } Q=x^{2}+2 \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-\frac{2 x}{1+x^{2}} d x} \\ &=e^{-2 \int \frac{x}{1+x^{2}} d x} \\ &=e^{-\log \left|1+x^{2}\right|} \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \\ &=e^{\log \left|1+x^{2}\right|^{-1}} \\ &=\left(1+x^{2}\right)^{-1} \end{aligned}
$=\frac{1}{1+x^{2}}$
Hence, the solution of different equation is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y\left(\frac{1}{1+x^{2}}\right)=\int\left(x^{2}+2\right) \frac{1}{x^{2}+1} d x+C \\ &=\frac{y}{1+x^{2}}=\int \frac{x^{2}+1+1}{x^{2}+1} d x+C \\ &=\frac{y}{1+x^{2}}=\int \frac{x^{2}+1}{x^{2}+1}+\frac{1}{x^{2}+1} d x+C \end{aligned}
\begin{aligned} &=\frac{y}{1+x^{2}}=\int 1+\frac{1}{x^{2}+1} d x+C \\ &=\frac{y}{1+x^{2}}=\int 1 d x+\int \frac{1}{x^{2}+1} d x+C \\ &=\frac{y}{1+x^{2}}=x+\tan ^{-1} x+C \quad\left[\int \frac{1}{x^{2}+1} d x=\tan ^{-1} x\right] \\ &=y=\left(1+x^{2}\right)\left(x+\tan ^{-1} x+C\right) \end{aligned}

Differential Equation Exercise 21.10 Question 36 (vii)

Answer: \begin{aligned} &y=\frac{1}{2} e^{\sin x}+\frac{c}{e^{\sin x}} \\ & \end{aligned}
Give: $\frac{d y}{d x}+y \cos x=e^{\sin x} \cos x \\$
Hint: Using $\int \cos x d x \\$
Explanation: $\frac{d y}{d x}+\cos x y=e^{\sin x} \cos x$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\cos x \text { and } Q=e^{\sin x} \cos x \end{aligned}

The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int \cos x d x} \\ &=e^{\sin x} \qquad\left[\int \sin x d x=\cos x+C\right] \end{aligned}

Hence, the solution of different equation is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y e^{\sin x}=\int e^{\sin x} \cos x e^{\sin x} d x+C \end{aligned}
\begin{aligned} &=y e^{\sin x}=\int e^{2 \sin x} \cos x d x+C \\ &=y e^{\sin x}=\int e^{2 t} d t+C \|[\sin x=t, \cos x d x=d t] \\ &=y e^{\sin x}=\left[\frac{e^{2 t}}{2}\right]+C \end{aligned}
\begin{aligned} &=y e^{\sin x}=\frac{e^{2 \sin x}}{2}+C \\ &=y=\frac{e^{2 \sin x}}{2 e^{\sin x}}+\frac{C}{e^{\sin x}} \\ &=y=\frac{e^{\sin x}}{2}+\frac{C}{e^{\sin x}} \end{aligned}

Differential Equation Exercise 21.10 Question 36 (viii)

Answer: \begin{aligned} & x+y-1=C e^{-y} \\ & \end{aligned}
Give: $(x+y) \frac{d y}{d x}=1$
Hint: Using integration by parts.
Explanation: \begin{aligned} &(x+y) \frac{d y}{d x}=1 \\ & \end{aligned}
$=\frac{d x}{d y}=x+y \\$
$=\frac{d x}{d y}-x=y$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=-1 \text { and } Q=y \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d y}\\ &=e^{\int-1 d y}\\ &=e^{-\int d y}\\ &=e^{-y} \quad\left[\int d y=y+C\right] \end{aligned}
Hence, the solution of different equation is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y e^{-y}=\int y e^{-y} d y+C \ldots(i) \end{aligned}
Using integration by parts we have
\begin{aligned} &=\int y e^{-y} d y=\frac{y e^{-y}}{-1}-\int(1) e^{-y} d y+C \\ &=-y e^{-y}-\frac{e^{-y}}{-1}+C \\ &=-y e^{-y}+e^{-y}+C \end{aligned}
From i

$=x e^{-y}=-y e^{-y}+e^{-y}+C$
Divide by $e^{-y}$
\begin{aligned} &=x=-y+1+\frac{C}{e^{-y}} \\ &=x=-y+1+C e^{-y} \end{aligned}

Differential Equation Exercise 21.10 Question 36 (ix)

Answer: \begin{aligned} &y=\tan x-1+\frac{c}{e^{\tan x}}\\ & \end{aligned}
Give: $\frac{d y}{d x} \cos ^{2} x=\tan x-y$
Hint: Using integration by parts.
Explanation: \begin{aligned} &\frac{d y}{d x} \cos ^{2} x=\tan x-y \\ & \end{aligned}
$=\frac{d x}{d y} \cos ^{2} x+y=\tan x$
Divide by $\cos ^{2}x$
\begin{aligned} &=\frac{d y}{d x}+\frac{y}{\cos ^{2} x}=\frac{\tan x}{\cos ^{2} x} \\ \\&=\frac{d y}{d x}+\left(\frac{1}{\cos ^{2} x}\right) y=\frac{\tan x}{\cos ^{2} x} \end{aligned}
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{\cos ^{2} x} \operatorname{and} Q=\frac{\tan x}{\cos ^{2} x} \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{1}{\cos ^{2} x} d x} \\ &=e^{\int \sec ^{2} x d x} \\ &=e^{\tan x} \quad\left[\int \sec ^{2} x d x=\tan x+C\right] \end{aligned}
Hence, the solution of different equation is
\begin{aligned} &y I f=\int \text { QIfd } x+C \\ &=y e^{\tan x}=\int \frac{\tan x}{\cos ^{2} x} e^{\tan x} d x+C \ldots(i) \end{aligned}
Using integration by parts,

$=\int \tan x \sec ^{2} x e^{\tan x} d x+C$
Put \begin{aligned} &\tan x=t, \sec ^{2} x d x=d t \\ & \end{aligned}
\begin{aligned} & \\ &\quad=\int t e^{t} d t+C \\ &\quad=t e^{t}-\int(1) e^{t} d t \\ &\quad=t e^{t}-e^{t}+C \\ &=\tan x e^{\tan x}-e^{\tan x}+C \end{aligned}
Put in $\left ( i \right )$
$=y e^{\tan x}=\tan x e^{\tan x}-e^{\tan x}+C$
Divide by $e^{\tan x}$
$=y=\tan x-1+\frac{C}{e^{\tan x}}$

Differential Equation Exercise 21.10 Question 36 (x)

Answer: \begin{aligned} &x e^{y}=\tan y+C \\ & \end{aligned}
Give: $e^{-y} \sec ^{2} y d y=d x+x d y \\$
Hint: Using $\int \sec ^{2} x d x$
Explanation: \begin{aligned} & e^{-y} \sec ^{2} y d y=d x+x d y \\ & \end{aligned}
\begin{aligned} &=e^{-y} \sec ^{2} y d y-x d y=d x \\ &=\left(e^{-y} \sec ^{2} y-x\right) d y=d x \\ &=\frac{d x}{d y}=e^{-y} \sec ^{2} y-x \\ &=\frac{d y}{d x}+x=e^{-y} \sec ^{2} y \end{aligned}
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=1 \text { and } Q=e^{-y} \sec ^{2} y \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int 1 d y} \\ &=e^{\int d y} \\ &=e^{y} \quad\left[\int d y=y+C\right] \end{aligned}
Hence, the solution of different equation is
\begin{aligned} &x I f=\int Q I f d y+C \\ &=x e^{y}=\int e^{-y} \sec ^{2} y e^{y} d y+C \\ &=x e^{y}=\int \sec ^{2} y d y+C \quad\left[e^{-y} e^{y}=e^{-y+y}=e^{0}=1\right] \\ &=x e^{y}=\tan y+C \quad\left[\int \sec ^{2} x d x=\tan x+C\right] \end{aligned}

Differential Equation Exercise 21.10 Question 36 (xi)

Answer: $y=\log x+\frac{c}{\log x} \\$
Give: $x \log x \frac{d y}{d x}+y=2 \log x \\$
Hint: Using $\int \frac{1}{x} d x \\$
Explanation: \begin{aligned} & & x \log x \frac{d y}{d x}+y=2 \log x \end{aligned}
Divide by $x \log x \\$
\begin{aligned} & &=\frac{d y}{d x}+\left(\frac{1}{x \log x}\right) y=\frac{2}{x} \end{aligned}
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=\frac{1}{x \log x} \text { and } Q=\frac{2}{x} \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &\text { If } f=e^{\int \frac{1}{x \log x} d x} \\ &=e^{\log |\log x|} \\ &=\log x \quad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution of different equation is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \log x=\int\left(\frac{2}{x} \log x\right) d x+C \end{aligned}
\begin{aligned} &=y \log x=2 \int\left(\frac{1}{x} \log x\right) d x+C \\ &=y \log x=\frac{2(\log x)^{2}}{2}+C \quad\left[\int x d x=\frac{x^{2}}{2}+C\right] \\ &=y \log x=(\log x)^{2}+C \end{aligned}
Divide by $\log x$ , we get
\begin{aligned} &=\frac{y \log x}{\log x}=\frac{(\log x)^{2}}{\log x}+\frac{C}{\log x} \\ &=y=\log x+\frac{C}{\log x} \end{aligned}

Differential Equation Exercise 21.10 Question 36 (xii)

Answer: $y=\frac{x^{2}}{4}+\frac{x^{2}}{16}+C$
Give: $x \frac{d y}{d x}+2 y=x^{2} \log x$
Hint: Using integration by parts and $\int \frac{1}{x} d x$
Explanation: $x \frac{d y}{d x}+2 y=x^{2} \log x$
Divide by x
$=\frac{d y}{d x}+\left(\frac{2}{x}\right) y=x \log x$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=\frac{2}{x} \text { and } Q=x \log x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{2}{x}^{2} d x} \\ &=e^{2 \int \frac{1}{x} d x} \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \\ &=e^{2 \log |x|} \\ &=e^{\log \left|x^{2}\right|} \\ &=x^{2} \quad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution of different equation is
$y I f=\int Q I f d x+C$
\begin{aligned} &=y x^{2}=\int x \log x x^{2} d x+C \\ &=y x^{2}=\int x^{3} \log x d x+C \ldots(i) \end{aligned}
We have \begin{aligned} & \int x^{3} \log x d x=\log x \frac{x^{4}}{4}-\int \frac{1}{x} \frac{x^{4}}{4} d x+C\\ & \end{aligned}
\begin{aligned} & =\log x \frac{x^{4}}{4}-\frac{1}{4} \int x^{3} d x+C\\ &=\log x \frac{x^{4}}{4}-\frac{1}{4} \frac{x^{4}}{4}+C\\ &=\log x \frac{x^{4}}{4}-\frac{x^{4}}{16}+C \end{aligned}
From i
$=y x^{2}=\log x \frac{x^{4}}{4}-\frac{x^{4}}{16}+C$
Divide by $x^{2}$
\begin{aligned} &=y=\frac{1}{x^{2}} \log x \frac{x^{4}}{4}-\frac{1}{x^{2}} \frac{x^{4}}{16}+C \\\\ &=y=\frac{x^{2}}{4} \log x-\frac{x^{2}}{16}+C \end{aligned}

Differential Equation Exercise 21.10 Question 37 (i)

Answer: $y=\frac{e^{x}}{2} \\$
Give: \begin{aligned} & & y^{\prime}+y=e^{x} \end{aligned}
Hint: Using integration
Explanation: $\frac{d y}{d x}+y=e^{x}$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=1 \text { and } Q=e^{x} \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int 1 d x} \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \\ &=e^{\int d x} \\ &=e^{x} \end{aligned}
Hence, the solution of different equation is
\begin{aligned} &y I f=\int \text { QIfd } x+C \\ &=y e^{x}=\int e^{x} e^{x} d x+C \\ &=y e^{x}=\int e^{2 x} d x+C \\ &=y e^{x}=\frac{e^{2 x}}{2}+C \quad\left[\int e^{2 x}=\frac{e^{2 x}}{2}+C\right] \end{aligned}
Divide by
\begin{aligned} &=y=\frac{1}{e^{x}} \frac{e^{2 x}}{2}+\frac{1}{e^{x}} C \\ &=y=\frac{e^{x}}{2}+\frac{C}{e^{x}} \end{aligned}
Given when $y(0)=\frac{1}{2}, \text { when } x=0, y=\frac{1}{2}$
\begin{aligned} &=\frac{1}{2}=\frac{e^{0}}{2}+\frac{C}{e^{0}} \\ &=\frac{1}{2}=\frac{1}{2}+C \\ &=C=0 \end{aligned}
By \begin{aligned} y=& \frac{e^{x}}{2}+\frac{0}{e^{x}} \\ \end{aligned}
$=y=\frac{e^{x}}{2}$

Differential Equation Exercise 21.10 Question 37 (ii)

Answer: $x+y+\log x=1$
Give: $x \frac{d y}{d x}-y=\log x, y(1)=0$
Hint: Using \begin{aligned} &\int \frac{1}{x} d x\\ & \end{aligned}
Explanation: $x \frac{d y}{d x}-y=\log x$
Divide by x
$=\frac{d y}{d x}+\left(\frac{-1}{x}\right) y=\frac{\log x}{x}$
This is a first order linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ & \end{aligned}
$P=\frac{-1}{x} \text { and } Q=\frac{\log x}{x}$
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{-1}{x} d x} \\ &=e^{-\int \frac{1}{x} d x} \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \\ &=e^{-\log x} \\ &=e^{\log x^{-1}} \\ &=x^{-1} \quad\left[e^{\log e^{x}}=x\right] \\ &=\frac{1}{x} \end{aligned}
Hence, the solution of different equation is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \frac{1}{x}=\int \frac{\log x}{x} \frac{1}{x} d x+C \\ &=\frac{y}{x}=\int \frac{\log x}{x} \frac{1}{x} d x+C \ldots(i) \end{aligned}
We have \begin{aligned} &\int \frac{\log x}{x} \frac{1}{x} d x \\ & \end{aligned}
Put $x=t \Rightarrow x=e^{t}$
Using integration by parts
\begin{aligned} &=t \frac{e^{-t}}{-1}-\int(1) \frac{e^{-t}}{-1} d t \\ &=-t e^{-t}+e^{-t}+C \\ &=-\frac{t}{e^{t}}+\frac{1}{e^{t}}+C \\ &=-\frac{\log x}{x}+\frac{1}{x}+C \end{aligned}
Substituting in i
$=\frac{y}{x}=-\frac{\log x}{x}+\frac{1}{x}+C$
Multiplying by x
$=y=1-\log x+C \ldots(i i)$
Given $y\left ( 1 \right )= 0$
\begin{aligned} \text { When } x &=1, y=0 \\ &=0=1-0+C \quad[\log 0=1] \\ &=C=-1 \end{aligned}
Substituting in ii
\begin{aligned} &=y=1-\log x-1 x \\ &=y=1-\log x-x \\ &=y++\log x=1 \end{aligned}

Differential Equation Exercise 21.10 Question 37 (iii)

Answer: \begin{aligned} & \cos x+y e^{2 x}=1 \\ & \end{aligned}
Give: $\frac{d y}{d x}+2 y=e^{-2 x} \sin x, y(0)=0$
Hint: Using integrating factor
Explanation: $\frac{d y}{d x}+2 y=e^{-2 x} \sin x$
This is a linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=2 \text { and } Q=e^{-2 x} \sin x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int 2 d x} \\ &=e^{2 \int d x} \\ &=e^{2 x} \quad\left[\int d x=x+C\right] \end{aligned}
Hence, the solution is

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y\left(e^{2 x}\right)=\int e^{-2 x} \sin x e^{2 x} d x+C \\ &=y e^{2 x}=\int \sin x d x+C \quad\left[e^{-2 x} e^{2 x}=e^{-2 x+2 x}=e^{0}=1\right] \\ &=y e^{2 x}=-\cos x+C \ldots(i) \end{aligned}
Now $\begin{gathered} y(0)=0, \text { When } x=0, y=0 \\ \end{gathered}$
$\quad=0 e^{2(0)}=-\cos 0+C$
\begin{aligned} &=0=-1+C \\ &=C=1 \\ & \end{aligned}
By i
$=y e^{2 x}=-\cos x+1 \\$
$=\cos x+y e^{2 x}=1$

Differential Equation exercise 21.10 question 37 (iv)

Answer: \begin{aligned} & y=-e^{-x}+\frac{x}{e} \\ & \end{aligned}
Give: $x \frac{d y}{d x}-y=(x+1) e^{-2 x}, y(1)=0$
Hint: Using integration by parts and $\int \frac{1}{x}dx$
Explanation: $x \frac{d y}{d x}-y=(x+1) e^{-x}$
Divide by x , we get
\begin{aligned} &=\frac{d y}{d x}-y\left(\frac{1}{x}\right)=\frac{(x+1)}{x} e^{-x} \\ &=\frac{d y}{d x}+y\left(\frac{-1}{x}\right)=\frac{(x+1)}{x} e^{-x} \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=-\frac{1}{x} \text { and } Q=\frac{(x+1)}{x} e^{-x} \end{aligned}
The integrating factor of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-\frac{1}{x} d x} \\ &=e^{-\int \frac{1}{x} d x} \\ &=e^{-\log |x|} \quad\left[\int \frac{1}{x} d x=\log x+C\right] \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y\left(\frac{1}{x}\right)=\int \frac{(x+1)}{x} e^{-x} \frac{1}{x} d x+C \\ &=\frac{y}{x}=\int \frac{(x+1)}{x} e^{-x} \frac{1}{x} d x+C \end{aligned}
\begin{aligned} &=\frac{y}{x}=I_{\prime}+C \ldots(i) \\ &=I_{1}=\int \frac{(x+1)}{x} e^{-x} \frac{1}{x} d x+C \\ &=\frac{1}{x} \frac{e^{-x}}{x}-\int\left(\frac{-1}{x^{2}}\right) \frac{e^{-x}}{-1} d x+\int \frac{e^{-x}}{x^{2}} d x \\ &=-\frac{e^{-x}}{x}-\int \frac{e^{-x}}{x^{2}} d x+\int \frac{e^{-x}}{x^{2}} d x \\ &=-\frac{e^{-x}}{x} \end{aligned}
By $\frac{y}{x}=-\frac{e^{-x}}{x}+C$
Multiply by x
$=y=-e^{-x}+C x \ldots(i i)$
Now $y(1)=0 \text { when } x=1, y=0$
\begin{aligned} &=0=-e^{-1}+C(1) \\ &=0=-\frac{1}{e}+C \\ &=C=\frac{1}{e} \end{aligned}
Put in (ii)
\begin{aligned} &=y=-e^{-x}+\frac{1}{e} x \\ &=y=-e^{-}+\frac{x}{e} \end{aligned}

Differential Equation exercise 21.10 question 37 (v)

Answer: $x e^{\tan ^{-1} y}=\tan ^{-1} y$
Give: \begin{aligned} &\left(1+y^{2}\right) d x+\left(x-e^{\tan ^{-1} y}\right) d y=0 \\ & \end{aligned}
Hint: Using $\int \frac{1}{1+x^{2}} d x$
Explanation: \begin{aligned} &\left(1+y^{2}\right) d x+\left(x-e^{\tan ^{-1} y}\right) d y=0 \\ & \end{aligned}
$=\left(x-e^{\tan ^{-1} y}\right) d y=-\left(1+y^{2}\right) d x \\$
$=x-e^{\tan ^{-1} y}=-\left(1+y^{2}\right) \frac{d x}{d y} \\$

$=\frac{x-e^{\tan ^{-1} y}}{1+y^{2}}=-\frac{d x}{d y}$
\begin{aligned} &=\frac{x}{1+y^{2}}-\frac{e^{\tan ^{-1} y}}{1+y^{2}}+\frac{d x}{d y}=0 \\ &\end{aligned}
$=\frac{d x}{d y}+\left(\frac{1}{1+y^{2}}\right) x=\frac{e^{\tan ^{-1} y}}{1+y^{2}}$
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\frac{1}{1+y^{2}} \text { and } Q=\frac{e^{\tan ^{1} y}}{1+y^{2}} \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &\text { If }=e^{\int \operatorname{Pdx}} \\ &=e^{\int \frac{1}{1+y^{2}} d x} \\ &=e^{\tan ^{-1} y} \quad\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+C\right] \end{aligned}
Hence, the solution is
\begin{aligned} &x I f=\int Q I f d y+C \\ &=x\left(e^{\tan ^{-1} y}\right)=\int \frac{e^{-\tan ^{-1} y}}{1+y^{2}} e^{\tan ^{-1} y} d y+C \\ &=x e^{\tan ^{-1} y}=\int \frac{1}{1+y^{2}} d y+C \end{aligned} $\quad\left[e^{-\tan ^{-1} y+\tan ^{-1} y}=e^{0}=1\right]$
$\\ =x e^{\tan ^{-1} y}=\tan ^{-1} y+C \ldots \text { (i) }$
$\quad\left[\int \frac{1}{1+y^{2}} d y=\tan ^{-1} y\right]$
Now \begin{aligned} &y(0)=0 \\ & \end{aligned}
$\quad=0 e^{\tan ^{-1}(0)}=\tan ^{-1}(0)+C \\$
$\quad=C=0$
Substituting in (i)
\begin{aligned} &=x e^{\tan ^{-1} y}=\tan ^{-1} y+0 \\ &=x e^{\tan ^{-1} y}=\tan ^{-1} y \end{aligned}

Differential Equation exercise 21.10 question 37 (vi)

Answer: \begin{aligned} & y=x^{2}+\cos x \\ & \end{aligned}
Give:$\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x \\$
Hint: Using $\int \frac{1}{1+x^{2}} d x \\$
Explanation: $\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x$
$=\frac{d x}{d y}+(\tan x) y=2 x+x^{2} \tan x$
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\tan x \text { and } Q=2 x+x^{2} \tan x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \tan x d x} \\ &=e^{\log |\sec x|} \end{aligned} $\quad\left[\int \tan x d x=\sec x+C\right] \\$
$=\sec x \quad\left[e^{\log e^{x}}=x\right]$
Hence, the solution is
\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y \sec x=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C \\ &=y \sec x=I+C \ldots(i) \\ &=I=\int\left(2 x+x^{2} \tan x\right) \sec x d x \\ &=\int 2 x \sec x d x+\int x^{2} \tan x \sec x d x \end{aligned}
\begin{aligned} &=2\left(\frac{x^{2}}{2} \sec x-\int \frac{x^{2}}{2}(\tan x \sec x) d x\right)+\int x^{2} \tan x \sec x d x+C \\ &=\frac{2 x^{2}}{2} \sec x-\int \frac{2 x^{2}}{2}(\tan x \sec x) d x+\int x^{2} \tan x \sec x d x+C \\ &=x^{2} \sec x-\int x^{2} \tan x \sec x d x+\int x^{2} \tan x \sec x d x+c \\ &=x^{2} \sec x+C \end{aligned}
Substituting in (i)

$=y \sec x=x^{2} \sec x+C$
Dividing by $\sec x$
\begin{aligned} &=y=x^{2}+\frac{C}{\sec x} \\ &=y=x^{2}+C \cos x \ldots(i i) \end{aligned}
Now \begin{aligned} &y(0)=1 \text { when } x=0, y=1 \\ & \end{aligned}

$\quad=1=0^{2}+C \cos (0) \\$
$\quad=1=+C(1) \quad[\cos 0=1] \\$
$\quad=C=1$
Substituting in (ii)
\begin{aligned} &=y=x^{2}+(1) \cos x \\ &=y=x^{2}+\cos x \end{aligned}

Differential Equation exercise 21.10 question 37 (vii)

Answer: \begin{aligned} & y=\sin x\\ & \end{aligned}
Give: $x \frac{d y}{d x}+y=x \cos x+\sin x, y\left(\frac{\pi}{2}\right)=1\\$
Hint: Using integration by parts $\int \frac{1}{1+x^{2}} d x\\$
Explanation: $x \frac{d y}{d x}+y=x \cos x+\sin x$
Divide by x
\begin{aligned} &=\frac{d x}{d y}+\frac{y}{x}=\cos x+\frac{\sin x}{x} \\ &=\frac{d x}{d y}+\left(\frac{1}{x}\right) y=\cos x+\frac{\sin x}{x} \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\frac{1}{x} \text { and } Q=\cos x+\frac{\sin x}{x} \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} & \\ & \end{aligned}
$=e^{\int \frac{1}{x}} d x \\$
$=e^{\log |x|}$ ${\left[\int \frac{1}{x} d x=\log |x|+C\right]} \\$
$=x$ ${\left[e^{\log e^{x}}=x\right]}$
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y(x)=\int\left(\cos x+\frac{\sin x}{x}\right) x d x+C \end{aligned}
\begin{aligned} &=y x=\int(x \cos x+\sin x) d x+C \\ &=y x=\int x \cos x d x+\int \sin x d x+C \ldots(i) \end{aligned}
Using integration by parts
\begin{aligned} &\int x \cos x d x \\ &=x \sin x-\int \sin x d x \end{aligned}
Substituting in (i)
\begin{aligned} &=y x=x \sin x-\int \sin x d x+\int \sin x d x+C \\ &=y x=x \sin x+C \end{aligned}
Divide by x

$=y=\sin x+\frac{C}{x} \ldots(i i)$
Now \begin{aligned} &y\left(\frac{\pi}{2}\right)=1 \text { when } x=\frac{\pi}{2}, y=1 \\ & \end{aligned}
$\quad=1=\sin \frac{\pi}{2}+\frac{C}{\frac{\pi}{2}} \\$
$\quad=1=1+\frac{2 C}{\pi} \quad\left[\sin \frac{\pi}{2}=1\right]$
\begin{aligned} &=\frac{2 C}{\pi}=0\\ & \end{aligned}
$=C=0\\$
Substituting in (ii)
$=y=\sin x+(0) \frac{1}{x}\\$
$=y=\sin x+0\\$
$=y=\sin x$

Differential Equation Excercise 21.10 Question 37 (viii)

Answer:\begin{aligned} & y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\\ & \end{aligned}
Give:$\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x, y\left(\frac{\pi}{2}\right)=0\\$
Hint: Using $\int \cot x d x \text { and } \int x d x\\$
Explanation: $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x$
$\frac{d y}{d x}+(\cot x) y=4 x \operatorname{cosec} x$
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\cot x \text { and } Q=4 x \operatorname{cosec} x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}
$=e^{\int \cot x d x} \\$
$x=e^{\log |\sin x|}$ $\quad\left[\int \cot x d x=\log |\sin x|+C\right] \\$
$=\sin x$ $\quad\left[e^{\log e^{x}}=x\right]$
Hence, the solution is
\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y(\sin x)=\int(4 x \operatorname{cosec} x) \sin x d x+C \\ &=y \sin x=4 \int x \frac{1}{\sin x} \sin x d x+C \quad\left[\operatorname{cosec} x=\frac{1}{\sin x}\right] \end{aligned}
\begin{aligned} &=y \sin x=4 \int x d x+C \\ &=y \sin x=4\left(\frac{x^{2}}{2}\right)+C \\ &=y \sin x=2 x^{2}+C \ldots(i) \end{aligned}
Now \begin{aligned} &y\left(\frac{\pi}{2}\right)=0 \text { when } x=\frac{\pi}{2}, y=0 \\ & \end{aligned}
$\quad=0 \sin x=2\left(\frac{\pi}{2}\right)^{2}+C \\$
$=0=2 \frac{\pi^{2}}{4}+C \\$
$=C=-\frac{\pi^{2}}{2}$
Substituting in (i)
$=y \sin x=2 x^{2}-\frac{\pi^{2}}{2}$

Differential Equation Excercise 21.10 Question 37 (ix)

Answer: $y=\cos x-2 \cos ^{2} x$
Give: \begin{aligned} & \frac{d y}{d x}+2 y \tan x=\sin x, y=0 \text { when } x=\frac{\pi}{3} \\ & \end{aligned}
Hint: Using $\int \tan x d x$
Explanation: $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x \\$
\begin{aligned} &\frac{d y}{d x}+2 y \tan x=\sin x \\ &=\frac{d y}{d x}+(2 \tan x) y=\sin x \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=2 \tan x \text { and } Q=\sin x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int 2 \tan x d x} \\ &=e^{2 \int \tan x d x} \end{aligned}
\begin{aligned} &=e^{2 \log |\sec x|} \qquad\left[\int \tan x d x=\log |\sec x|+C\right] \\ &=e^{\log \sec ^{2} x} \\ &=\sec ^{2} x \qquad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \sec ^{2} x=\int\left(\sin x \sec ^{2} x\right) d x+C \\ &=y \sec ^{2} x=\int \sin x \frac{1}{\cos ^{2} x} d x+C \end{aligned}
\begin{aligned} &=y \sec ^{2} x=\int \frac{\sin x}{\cos x} \frac{1}{\cos x} d x+C \\ &=y \sec ^{2} x=\int \tan x \sec x d x+C \qquad\left[\tan x=\frac{\sin x}{\cos x}\right] \\ &=y \sec ^{2} x=\sec x+C \qquad \quad\left[\int \tan x \sec x d x=\sec x+C\right] \end{aligned}
Divide by $\sec x$
\begin{aligned} &=y \sec x=1+\frac{C}{\sec x} \\ &=y \sec x=1+C \cos x \ldots(i) \end{aligned}
Now \begin{aligned} y &=0 \text { when } x=\frac{\pi}{3} \\ & \end{aligned}
$=0 \sec x=1+C \cos \frac{\pi}{3}$
\begin{aligned} &=0=1+C\left(\frac{1}{2}\right) \qquad\left[\cos \frac{\pi}{3}=\frac{1}{2}\right] \\ &=\frac{C}{2}=-1 \\ &=C=-2 \end{aligned}
Substituting in (i)
$=y \sec x=1-2 \cos x$
Divide by $\sec x$
\begin{aligned} &=y=\frac{1}{\sec x}-\frac{2 \cos x}{\sec x} \\ &=y=\cos x-2 \cos x \cos x \qquad\left[\frac{1}{\sec x}=\cos x\right] \\ &=y=\cos x-2 \cos ^{2} x \end{aligned}

Differential Equation Excercise 21.10 Question 37 (x)

Answer: $y \operatorname{cosec}^{2} x=4 \sin x-2\\$
Give: $\frac{d y}{d x}-3 y \cot x=\sin 2 x, y=2 \text { when } x=\frac{\pi}{2}\\$
Hint: Using $\int \operatorname{cosec} x \cot x d x$
Explanation: $\frac{d y}{d x}-3 y \cot x=\sin 2 x \\$
\begin{aligned} & &\frac{d y}{d x}+(-3 \cot x) y=\sin 2 x \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=-3 \cot x \text { and } Q=\sin 2 x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-3 \cot x d x} \\ &=e^{-3 \int \cot x d x} \\ &=e^{-3 \log |\sin x|} \qquad\left[\int \cot x d x=\log |\sin x|+C\right] \\ & \end{aligned}
$=e^{\log \sin ^{-3} x} \\$
$=\sin ^{-3} x \qquad\left[e^{\log e^{x}}=x\right]$
\begin{aligned} &=\frac{1}{\sin ^{3} x} \\ &=\operatorname{cosec}^{3} x \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \operatorname{cosec}^{3} x=\int\left(\sin 2 x \operatorname{cosec}^{3} x\right) d x+C \\ &=y \operatorname{cosec}^{3} x=\int\left(2 \sin x \cos x \frac{1}{\sin ^{3} x}\right) d x+C \end{aligned}
\begin{aligned} &=y \operatorname{cosec}^{3} x=2 \int\left(\cos x \frac{1}{\sin ^{2} x}\right) d x+C \\ &=y \operatorname{cosec}^{3} x=2 \int\left(\frac{\cos x}{\sin x} \frac{1}{\sin x}\right) d x+C \\ &=y \operatorname{cosec}^{3} x=2 \int \cot x \operatorname{cosec} x d x+C \qquad\left[\frac{\cos x}{\sin x}=\cot x, \frac{1}{\sin x}=\operatorname{cosec} x\right] \\ &=y \operatorname{cosec}^{3} x=2(-\operatorname{cosec} x)+C \qquad\left[\int \cot x \operatorname{cosec} x d x=-\operatorname{cosec} x\right] \end{aligned}
Divide by cosecx
\begin{aligned} &=y \operatorname{cosec}^{2} x=-2+\frac{C}{\operatorname{cosec} x} \\ &=y \operatorname{cosec}^{2} x=-2+C \sin x \ldots(i) \quad\left[\frac{1}{\operatorname{cosec} x}=\sin x\right] \end{aligned}
Now \begin{aligned} y &=2 \text { when } x=\frac{\pi}{2} \\ & \end{aligned}

$=2 \operatorname{cosec}^{2}\left(\frac{\pi}{2}\right)=-2+C \sin \frac{\pi}{2}$
\begin{aligned} &=2(1)^{2}=-2+C(1) \quad\left[\operatorname{cosec} \frac{\pi}{2}=\frac{1}{\sin \frac{\pi}{2}}=\frac{1}{1}=1\right] \\ &=2=-2+C \\ &=C=4 \end{aligned}
Substituting in (i)
$=y \operatorname{cosec}^{2} x=+4 \sin x-2$

Differential Equation Excercise 21.10 Question 37 (xi)

Answer: $y \sin x+\cos ^{2} x=0 \\$
Give: $\frac{d y}{d x}+y \cot x=2 \cos x, y\left(\frac{\pi}{2}\right)=0 \\$
Hint: Using $\int \cot x d x$
Explanation: $\frac{d y}{d x}+(\cot x) y=2 \cos x$
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\cot x \text { and } Q=2 \cos x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \cot x d x} \\ &=e^{\log |\sin x|} \qquad\qquad\left[\int \cot x d x=\log |\sin x|+C\right] \\ &=\sin x \qquad\qquad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y(\sin x)=\int 2 \cos x \sin x d x+C \\ &=y \sin x=\int \sin 2 x d x+C \qquad \qquad \qquad[2 \cos x \sin x=\sin 2 x] \\ &=y \sin x=-\frac{\cos 2 x}{2}+C \ldots(i) \end{aligned}
We have $y\left(\frac{\pi}{2}\right)=0 \text { when } x=\frac{\pi}{2}, y=0$
\begin{aligned} &\\ &=(0) \sin \frac{\pi}{2}=-\frac{\cos 2\left(\frac{\pi}{2}\right)}{2}+C \end{aligned}
\begin{aligned} &=C=\frac{\cos \pi}{2} \\\\ &=C=\frac{(-1)}{2} \qquad\qquad[\cos \pi=-1] \\\\ &=C=-\frac{1}{2} \end{aligned}
Substituting in (i)

\begin{aligned} &=y \sin x=-\frac{\cos 2 x}{2}-\frac{1}{2} \\\\ &\end{aligned}
$=y \sin x=-\frac{1}{2}(\cos 2 x+1)$
\begin{aligned} &=2 y \sin x=-(\cos 2 x+1) \\\\ &=2 y \sin x=-\cos 2 x-1 \\ \\&=2 y \sin x+\cos 2 x+1=0 \\ \\&=2 y \sin x+\cos ^{2} x=0 \quad\left[1+\cos 2 x=2 \cos ^{2} x\right] \end{aligned}

Differential Equation exercise 21.10 question 37 (xii)

Answer: $y \sin x=-\frac{\cos 2 x}{2}+C \\$
Give: $d y=\cos x(2-y \operatorname{cosec} x) d x \\$
Hint: Using $\int \cot x d x \\$
Explanation: $d y=\cos x(2-y \operatorname{cosec} x) d x \\$
\begin{aligned} & &=\frac{d y}{d x}=\cos x(2-y \operatorname{cosec} x) \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=2 \cos x-y \cos x \frac{1}{\sin x} \qquad\left[\operatorname{cosec} x=\frac{1}{\sin x}\right] \\ &=\frac{d y}{d x}=2 \cos x-y \cot x \qquad\left[\frac{\cos x}{\sin x}=\cot x\right] \\ &=\frac{d y}{d x}+(\cot x) y=2 \cos x \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\cot x \operatorname{and} Q=2 \cos x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &\text { If } f=e^{\int \operatorname{Pdx}} \\\\ &=e^{\int \cot x d x} \\\\ &\begin{array}{ll} =e^{\log |\sin x|} & \qquad\left[\int \cot x d x=\log |\sin x|+C\right] \\ \\=\sin x & {\left[e^{\log e^{x}}=x\right]} \end{array} \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y(\sin x)=\int 2 \cos x \sin x d x+C \\ &=y \sin x=\int \sin 2 x d x+C \qquad \qquad \qquad[2 \cos x \sin x=\sin 2 x] \\ &=y \sin x=-\frac{\cos 2 x}{2}+C \end{aligned}

Differential Equation exercise 21.10 question 37 (xiii)

Answer: \begin{aligned} &y=x^{2}-\frac{\pi}{4} \operatorname{cosec} x+C \\ & \end{aligned}
Give: $\tan x \frac{d y}{d x}=2 x \tan x+x^{2}-y, \tan x \neq 0, y=0 \text { when } x=\frac{\pi}{2}$
Hint: Using integration by parts and $\int cot x dx$
Explanation: $\tan x \frac{d y}{d x}=2 x \tan x+x^{2}-y\\$
Divide by $\tan x\\$
$=\frac{d y}{d x}=2 x+\frac{x^{2}}{\tan x}-\frac{y}{\tan x}$
\begin{aligned} &=\frac{d y}{d x}=2 x+\cot x x^{2}-y \cot x \quad\left[\frac{1}{\tan x}=\cot x\right] \\ &=\frac{d y}{d x}+y \cot x=2 x+\cot x x^{2} \\ &=\frac{d y}{d x}+(\cot x) y=2 x+\cot x x^{2} \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\cot x \text { and } Q=2 x+\cot x x^{2} \end{aligned}

The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \cot x d x} \\ &=e^{\log |\sin x|} \quad\left[\int \cot x d x=\log |\sin x|+C\right] \\ &=\sin x \quad\left[e^{\log e^{x}}=x\right] \end{aligned}

Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \sin x=\int\left(2 x+x^{2} \cot x\right) \sin x d x+C \\ &=y \sin x=\int 2 x \sin x+x^{2} \cot x \sin x d x+C \end{aligned}
\begin{aligned} &=y \sin x=2 \int x \sin x d x+\int x^{2} \cot x \sin x d x+C \\ &=y \sin x=2\left[\frac{x^{2}}{2} \sin x-\int \cos x \frac{x^{2}}{2} d x\right]+\int \frac{\cos x}{\sin x} \sin x x^{2} d x+C \qquad\left[\cot x=\frac{\cos x}{\sin x}\right] \\ &=y \sin x=x^{2} \sin x-\int \cos x x^{2} d x+\int \cos x x^{2} d x+C \\ &=y \sin x=x^{2} \sin x+C \end{aligned}

Divide by sinx
\begin{aligned} &=y=x^{2}+\frac{C}{\sin x} \\ &=y=x^{2}+C \operatorname{cosec} x \ldots(i) \end{aligned}
Now $y =0 \text { when } x=\frac{\pi}{2} \\$
\begin{aligned} &=0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2} \\ &=0=\frac{\pi^{2}}{4}+C(1) \quad\left[\operatorname{cosec} \frac{\pi}{2}=1\right] \\ &=C=-\frac{\pi^{2}}{4} \end{aligned}
Substituting in (i)
$=y=x^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} x$

Differential Equation Exercise 21.10 Question 38

Answer: $y=\frac{x^{2}}{4}+\frac{c}{x^{2}}$
Give: $\tan x \frac{d y}{d x}+2 y=x^{2}$
Hint: Using $\int \frac{1}{x} d x \text { and } e^{\log e^{x}}=x$
Explanation: $\tan x \frac{d y}{d x}+2 y=x^{2}$
Divide by x
\begin{aligned} &=\frac{d y}{d x}+\frac{2 y}{x}=x \\ &=\frac{d y}{d x}=\left(\frac{2}{x}\right) y=x \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\frac{2}{x} \text { and } Q=x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &\text { If }=e^{\int \operatorname{Pd} x} \\ &=e^{\int \frac{2}{x} d x} \\ &=e^{2 \int \frac{1}{x} d x} \\ &=e^{2 \log x} \quad\left[\int \frac{1}{x} d x=\log x+C\right] \\ &=e^{\log x^{2}} \\ &=x^{2} \quad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y x^{2}=\int x x^{2} d x+C \\ &=y x^{2}=\int x^{3} d x+C \\ &=y x^{2}=\frac{x^{4}}{4}+C \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}
Divide by $x^{2}$

$=y=\frac{x^{2}}{4}+\frac{c}{x^{2}}$

Differential Equation Exercise 21.10 Question 39

Answer: $y=\frac{1}{2}(\sin x-\cos x)+\frac{c}{e^{-x}} \\$
Give: \begin{aligned} & &\frac{d y}{d x}-y=\cos x \end{aligned}
Hint: Using integrating factor and integration by parts
Explanation: $\frac{d y}{d x}-y=\cos x$
\begin{aligned} &\\ &=\frac{d y}{d x}+(-1) y=\cos x \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=-1 \text { and } Q=\cos x \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-1 d x} \\ &=e^{-\int d x} \\ &=e^{-x} \quad\left[\int d x=x+C\right] \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y\left(e^{-x}\right)=\int \cos x e^{-x} d x+C \ldots(i) \end{aligned}
Using integration by parts
$I=\int \cos x e^{-x} d x=\cos x \frac{e^{-x}}{-1}-\int \sin x \frac{e^{-x}}{-1} d x+C$
\begin{aligned} &=I=-\cos x e^{-x}-\left[\sin x \frac{e^{-x}}{-1}-\int \cos x \frac{e^{-x}}{-1} d x\right] \\ &=I=-\cos x e^{-x}+\sin x e^{-x}-\int \cos x e^{-x} d x \end{aligned}
\begin{aligned} &=I=-\cos x e^{-x}+\sin x e^{-x}-I \\ &=2 I=-\cos x e^{-x}+\sin x e^{-x} \\ &=2 I=e^{-x}(\sin x-\cos x) \\ &=I=\frac{1}{2} e^{-x}(\sin x-\cos x) \end{aligned}
Substituting in (i)
$=y e^{-x}=\frac{1}{2} e^{-x}(\sin x-\cos x)+C$
Divide by $e^{-x}$
$=y=\frac{1}{2}(\sin x-\cos x)+\frac{C}{e^{-x}}$

Differential Equation Exercise 21.10 Question 40

Answer: \begin{aligned} &y=3 x^{2}+C x \\ \end{aligned}
Give:\begin{aligned} & &\left(y+3 x^{2}\right) \frac{d x}{d y}=x \\ \end{aligned}
Hint: Using \begin{aligned} & & \int \frac{1}{x} d x \end{aligned}
Explanation: $\left(y+3 x^{2}\right) \frac{d x}{d y}=x$
\begin{aligned} & \\ &=x \frac{d y}{d x}=y+3 x^{2} \end{aligned}
Divide by x
\begin{aligned} &=\frac{d y}{d x}=\frac{y+3 x^{2}}{x} \\ &=\frac{d y}{d x}=\frac{y}{x}+\frac{3 x^{2}}{x} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{y}{x}+3 x \\ &=\frac{d y}{d x}-\frac{y}{x}=3 x \\ &=\frac{d y}{d x}+\left(-\frac{1}{x}\right) y=3 x \end{aligned}
This is a linear differential equation of the form
$\frac{d x}{d y}+P x=Q$
$P=-\frac{1}{x} \text { and } Q=3 x$
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-\frac{1}{x} d x} \\ &=e^{-\int \frac{1}{x} d x} \end{aligned}
\begin{aligned} &=e^{-\log x} \quad\left[\int \frac{1}{x} d x=\log x+C\right] \\ &=e^{\log x^{-1}} \\ &=x^{-1} \quad\left[e^{\log e^{x}}=x\right] \\ &=\frac{1}{x} \end{aligned}
Hence, the solution is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y\left(\frac{1}{x}\right)=\int 3 x\left(\frac{1}{x}\right) d x+C \end{aligned}
\begin{aligned} &=\frac{y}{x}=\int 3 d x+C \\ &=\frac{y}{x}=3 \int d x+C \\ &=\frac{y}{x}=3 x+C \quad\left[\int d x=x+C\right] \end{aligned}
Multiply by x

$=y=3 x^{2}+C x$

Differential Equation Exercise 21.10 Question 41

Answer: $x=y^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} y \\$
Give: \begin{aligned} & & \frac{d x}{d y}+x \cot y=2 y+y^{2} \cot y, y \neq 0, x=0 \text { when } y=\frac{\pi}{2} \end{aligned}
Hint: Using integration by parts and $\int x^{n}dx$
Explanation: $\frac{d x}{d y}+x \cot y=2 y+y^{2} \cot y \\$
\begin{aligned} & &=\frac{d x}{d y}+(\cot y) x=2 y+y^{2} \cot y \end{aligned}
This is a linear differential equation of the form
$\frac{d x}{d y}+P x=Q$
$P=\cot x \text { and } Q=2 y+y^{2} \cot y$
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \cot y d x} \\ &=e^{\log |\sin y|} \qquad\qquad\qquad\left[\int \cot x d x=\log |\sin x|+C\right] \\ &=\sin y \quad\qquad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution is
\begin{aligned} &x I f=\int Q I f d y+C \\\\ &=x(\sin y)=\int\left(2 y+y^{2} \cot y\right) \sin y d y+C \\\\ &=x \sin y=\int 2 y \sin y+y^{2} \cot y \sin y d y+C \end{aligned}
\begin{aligned} &=x \sin y=\int 2 y \sin y+y^{2} \frac{\cos y}{\sin y} \sin y d y+C \\\\ &=x \sin y=\int 2 y \sin y d y+\int y^{2} \cos y d y+C \\\\ &=x \sin y=2 \int y \sin y d y+\int y^{2} \cos y d y+C \ldots(i) \end{aligned}
Using integration by parts
\begin{aligned} &=2 \int y \sin y d y \\ &=2\left[\sin y \frac{y^{2}}{2}-\int \cos y \frac{y^{2}}{2} d y\right] \\ &=\sin y y^{2}-\int \cos y y^{2} d y \end{aligned}
Substituting on (i)
\begin{aligned} &=x \sin y=\sin y y^{2}-\int \cos y y^{2} d y+\int y^{2} \cos y d y+C \\ &=x \sin y=\sin y y^{2}+C \end{aligned}
Divide by $\sin y$

\begin{aligned} &=x=y^{2}+\frac{C}{\sin y} \\ &=x=y^{2}+C \operatorname{cosec} y \ldots(i i) \qquad\left[\frac{1}{\sin y}=\operatorname{cosec} y\right] \end{aligned}
Now $x =0 \text { when } y=\frac{\pi}{2} \\$
\begin{aligned} &=0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2} \end{aligned}
\begin{aligned} &=0=\frac{\pi^{2}}{4}+C(1) \qquad\left[\operatorname{cosec} \frac{\pi}{2}=1\right] \\ &=C=-\frac{\pi^{2}}{4} \end{aligned}
Substituting in (ii)
$=x=y^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} y$

Differential Equation Exercise 21.10 Question 42

Answer: \begin{aligned} & x+\cot ^{-1} y=1+C e^{\tan ^{-1} y}\\ & \end{aligned}
Give: \begin{aligned} & \left(\cot ^{-1} y+x\right) d y=\left(1+y^{2}\right) d x\\ \end{aligned}
Hint: Using integration by parts and $\int \frac{1}{1+x^{2}} d x$
Explanation: $\left(\cot ^{-1} y+x\right) d y=\left(1+y^{2}\right) d x$
\begin{aligned} &=\frac{d x}{d y}=\frac{\cot ^{-1} y+x}{1+y^{2}} \\\\ &=\frac{d x}{d y}=\frac{\cot ^{-1} y}{1+y^{2}}+\frac{x}{1+y^{2}} \end{aligned}
\begin{aligned} &=\frac{d x}{d y}-\frac{x}{1+y^{2}}=\frac{\cot ^{-1} y}{1+y^{2}} \\\\ &=\frac{d x}{d y}+\left(\frac{-1}{1+y^{2}}\right) x=\frac{\cot ^{-1} y}{1+y^{2}} \end{aligned}
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\frac{-1}{1+y^{2}} \text { and } Q=\frac{\cot ^{-1} y}{1+y^{2}} \end{aligned}
The integrating factor $If$ of this differential equation is
\begin{aligned} &I f=e^{\int P d y} \\\\ &=e^{\int \frac{-1}{1+y^{2}} d y} \\\\ &=e^{-\int \frac{1}{1+y^{2}} d y} \end{aligned}
$\begin{array}{ll} =e^{-\tan ^{-1} y} &\qquad\qquad {\left[\int \frac{1}{1+\mathrm{y}^{2}} d y=\tan ^{-1} y+C\right]} \\\\ =e^{\cot ^{-1} y} &\qquad {\left[\tan ^{-1} y=\cot ^{-1}\left(\frac{1}{y}\right)\right]} \end{array}$
Hence, the solution is
$x\: \text{If}=\int Q \: \text{If} \: d y+C$
$=x\left(e^{\cot ^{-1} y}\right)=\int \frac{\cot ^{-1} y}{1+y^{2}} e^{\cot ^{-1} y} d y$
Put $\cot ^{-1} y=t \\$
\begin{aligned} & &\qquad \begin{aligned} =&-\frac{d y}{1+y^{2}}=d t \\ =& \frac{d y}{1+y^{2}}=-d t \end{aligned} \end{aligned}
So,

$=-\int t e^{t} d t$
Using integration by parts
\begin{aligned} &=-\left[t e^{t}-\int e^{t} d t\right] \\\\ &=-\left[t e^{t}-e^{t}-C\right] \\\\ &=-t e^{t}+e^{t}+C \\ \\&=x e^{\cot ^{-1} y}=-\cot ^{-1} y e^{\cot ^{-1} y}+e^{\cot ^{-1} y}+C \end{aligned}
Divide by $e^{cot^{-1}}y$

\begin{aligned} &=x=-\cot ^{-1} y+1+\frac{c}{e^{\cot ^{-1} y}} \\ &=x+\cot ^{-1} y=1+C e^{\tan ^{-1} y} \quad\left[\cot ^{-1} y=\tan ^{-1}\left(\frac{1}{y}\right)\right] \end{aligned}

In the 21st chapter of mathematics, class 12, there are eleven exercises. The tenth exercise in this Differential Equation chapter, ex 21.10, has 65 questions in the textbook. The concept in this exercise is to solve the differential equations, initial value problem, general solution, and particular solution of differential equations. Few questions have subparts, while most of the questions do not. Only Level 1 questions are present in this tenth exercise. However, even though there is no Level 2 part, the students face challenges even in the Level 1 questions as the chapter moves towards its end. Hence, the usage of RD Sharma Class 12 Chapter 21 Exercise 21.10 solution book is vital.

Most of the CBSE schools recommend the RD Sharma books to their students because it follows the NCERT syllabus. And to add another point, the RD Sharma Class 12th Exercise 21.10 book contains many practice questions for the students to work out on an extra basis. This makes the students understand the concept in-depth and prevent making mistakes in the exams.

The Class 12 RD Sharma Chapter 21 Exercise 21.10 Solution contains accurate answers from experts in the mathematical field. Begin your practice today in the presence of the RD Sharma solution materials to observe the rise in your marks. RD Sharma solutions Students who face challenges in solving the differential equation sums will soon start feeling it easy. The RD Sharma Class 12th Exercise 21.10 book has rescued many students who faced the same difficulties.

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## RD Sharma Chapter wise Solutions

1. Which RD Sharma book is prescribed for the students who want to be clear about the concepts and sums in the mathematics chapter 21?

The RD Sharma Class 12th Exercise 21.10 is the most prescribed reference book for the students who wish to learn the concepts in this chapter.

2. Which is the best website to refer to the RD Sharma solution books?

All the RD Sharma solution books are available at the top educational website, Career 360. Students can access the books from this site.

3. What concept does the class 12 mathematics chapter 21 focus on?

The central concept that the class 12 mathematics chapter 21 focuses on is Differential Integration. Therefore, most of the sums are based on this topic.

4. What are the benefits of the practice sums present in the RD Sharma solution books?

The additional sums given in the RD Sharma Class 12th Exercise 21.10 make these students well-versed in the concept. It also makes them exam-ready effortlessly.

5. How many exercises are there in the class 12 mathematics in chapter 21?

There are eleven exercises in the class 12 mathematics chapter 21 syllabus.

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