RD Sharma Class 12 Exercise 21.10 Differential Equation Solutions Maths - Download PDF Free Online
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.
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- RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
- 21.10
- RD Sharma Chapter wise Solutions
RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
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:$y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.$ ,otherwiseGive:$\frac{d y}{d x}+3 y=e^{m x}, m$ 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
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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