The most preferred set of solution books by the students of CBSE board schools are the RD Sharma books. It helps the students prepare well for all their public exams in every subject and chapter. For the students who encounter lots of doubts and confusion while solving Differential Equations, the RD Sharma Class 12th Exercise 21.7 is the rescuer. A good guide is essential for the students who are preparing for their public exams.
Differential Equations Excercise:21.7
Differential Equations exercise 21.7 question 1
Answer: $\log y+c=2 x+2 \log |x-1|$Hint:Separate the terms of x and y and then integrate them.
Given: $(x-1) \frac{d y}{d x}=2 x y$Solution:$\begin{aligned} &(x-1) \frac{d y}{d x}=2 x y \\\\ &\frac{d y}{y}=\left(\frac{2 x}{x-1}\right) d x \\\\ &\frac{d y}{y}=2\left(1+\frac{1}{x-1}\right) d x \end{aligned}$Integrating both sides
$\log y+c=2 x+2 \log |x-1|$Differential Equations exercise 21.7 question 2
Answer: $C \sqrt{1+x^{2}}$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(1+x^{2}\right) d y=x y d x$Solution: $\left(1+x^{2}\right) d y=x y d x$$\begin{aligned} &\frac{d y}{y}=\frac{x d x}{1+x^{2}} \\\\ &\text { Put } 1+x^{2}=t \\\\ &2 x d x=d t \end{aligned}$$\begin{aligned} &x d x=\frac{d t}{2} \\\\ &\frac{d y}{y}=\frac{d t}{2 t} \end{aligned}$Integrating both sides
$\begin{aligned} &\log |y|=\frac{1}{2} \log |t|+\log C \\\\ &\log |y|=\frac{1}{2} \log \left|1+x^{2}\right|+\log C \end{aligned}$$\begin{aligned} &\log |y|=\left(\log \left|1+x^{2}\right|\right)^{\frac{1}{2}}+\log C \\\\ &\log |y|=\log \left[C \sqrt{1+x^{2}}\right] \\\\ &y=C \sqrt{1+x^{2}} \end{aligned}$Differential Equations exercise 21.7 question 3
Answer: $\log |y|=e^{x}+x+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=\left(e^{x}+1\right) y$Solution: $\frac{d y}{d x}=\left(e^{x}+1\right) y$$\frac{d y}{y}=\left(e^{x}+1\right) d x$Integrating on both sides
$\log |y|=e^{x}+x+c$Differential Equations exercise 21.7 question 4 maths
Answer: $\log |y|=\frac{2}{3} x^{3}+x^{2}+2 x+2 \log |x-1|+c$Hint: Separate the terms of x and y and then integrate them.
Given: $(x-1) \frac{d y}{d x}=2 x^{3} y$Solution: $(x-1) \frac{d y}{d x}=2 x^{3} y$$\begin{gathered} \frac{d y}{y}=\frac{2 x^{8}}{x-1} d x \\\\ \frac{d y}{y}=\frac{2\left((x-1)\left(x^{2}+x+1\right)+1\right)}{(x-1)} d x \\\\ \frac{d y}{y}=2\left(x^{2}+x+1+\frac{1}{(x-1)}\right) d x \end{gathered}$Integrating both sides
$\begin{aligned} &\int \frac{d y}{y}=\int 2\left(x^{2}+x+1+\frac{1}{(x-1)}\right) d x \\\\ &\log |y|=\frac{2 x^{3}}{3}+\frac{2 x^{2}}{2}+2 x+2 \log |x-1|+c \\\\ &\log |y|=\frac{2 x^{3}}{3}+x^{2}+2 x+2 \log |x-1|+c \end{aligned}$Differential Equations exercise 21.7 question 5
Answer: $\frac{y^{3}}{3}+\frac{y^{2}}{2}=\frac{x^{2}}{2}+\log |x|+c$Hint: Separate the terms of x and y and then integrate them.
Given: $x y(y+1) d y=\left(x^{2}+1\right) d x$Solution:$x y(y+1) d y=\left(x^{2}+1\right) d x$$\begin{aligned} &y(y+1) d y=\frac{\left(x^{2}+1\right) d x}{x} \\\\ &\left(y^{2}+y\right) d y=\left(\frac{x^{2}}{x}+\frac{1}{x}\right) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int\left(y^{2}+y\right) d y=\int\left(x+\frac{1}{x}\right) d x \\\\ &\frac{y^{3}}{3}+\frac{y^{2}}{2}=\frac{x^{2}}{2}+\log |x|+c \end{aligned}$Differential Equations exercise 21.7 question 6
Answer: $e^{y}=e^{x}+c$Hint: Separate the terms of x and y and then integrate them.
Given: $e^{y-x} \frac{d y}{d x}=1$Solution: $e^{y-x} \frac{d y}{d x}=1$$\begin{aligned} &e^{y} \cdot e^{-x} \frac{d y}{d x}=1 \\\\ &\frac{e^{y}}{e^{x}} d y=d x \\\\ &e^{y} d y=e^{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int e^{y} d y=\int e^{x} d x \\\\ &e^{y}=e^{x}+c \end{aligned}$Differential Equations exercise 21.7 question 7
Answer: $\sin y=e^{x} \log x+c$Hint:Separate the terms of x and y and then integrate them.
Given: $x \cos y d y=\left(x e^{x} \log x+e^{x}\right) d x$Solution: $x \cos y d y=\left(x e^{x} \log x+e^{x}\right) d x$$\cos y d y=\left[\frac{\left(x e^{x} \log x+e^{x}\right)}{x}\right] d x$Integrating both sides,
$\begin{aligned} &\int \cos y d y=\int \frac{\left(x e^{x} \log x+e^{x}\right)}{x} d x \\\\ &\sin y=\int\left(e^{x} \log x+\frac{e^{x}}{x}\right) d x \\\\ &\sin y=\int e^{x} \log x d x+\int \frac{e^{x}}{x} d x \end{aligned}$ [Integrating by parts]
$\begin{aligned} &\sin y=\log x \int e^{x} d x-\int \frac{1}{x}\left(\int e^{x} d x\right) d x+\int \frac{e^{x}}{x} d x \\\\ &\sin y=\log x e^{x}-\int \frac{e^{x}}{x} d x+\int \frac{e^{x}}{x} d x+c \\\\ &\sin y=e^{x} \log x+c \end{aligned}$Differential Equations exercise 21.7 question 8 maths
Answer: $-e^{-y}=e^{x}+\frac{x^{3}}{3}+3+c$Hint:Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=e^{x+y}+x^{2} e^{y}$Solution: $\frac{d y}{d x}=e^{x+y}+x^{2} e^{y}$$\begin{aligned} &\frac{d y}{d x}=e^{x} \cdot e^{y}+x^{2} e^{y} \\\\ &d y=e^{y}\left[e^{x}+x^{2}\right] d x \\\\ &\frac{d y}{e^{y}}=\left[e^{x}+x^{2}\right] d x \end{aligned}$ Integrating both sides
$\begin{array}{r} \int e^{-y} d y=\int e^{x} d x+\int x^{2} d x \\\\ -e^{-y}=e^{x}+\frac{x^{3}}{3}+3+c \end{array}$Differential Equations exercise 21.7 question 9
Answer: $y-1=C x y$Hint:Separate the terms of x and y and then integrate them.
Given: $x \frac{d y}{d x}+y=y^{2}$Solution: $x \frac{d y}{d x}+y=y^{2}$$\begin{aligned} &x \frac{d y}{d x}+y=y^{2} \\\\ &x \frac{d y}{d x}=y^{2}-y \\\\ &x d y=\left(y^{2}-y\right) d x \\\\ &\frac{d y}{y(y-1)}=\frac{d x}{x} \end{aligned}$Integrating on both sides
$\int \frac{d y}{y(y-1)}=\int \frac{d x}{x} \\\\$$\begin{aligned} &\int \frac{y-(y-1)}{y(y-1)} d y=\int \frac{d x}{x} \\\\ &\int \frac{1}{(y-1)} d y-\int \frac{1}{y} d y=\int \frac{d x}{x} \end{aligned}$We get,
$\begin{aligned} &-\log y+\log (y-1)=\log x+\log c \\\\ &\log \frac{(y-1)}{y}=\log x C \\\\ &\frac{y-1}{y}=C x \\\\ &y-1=C x y \end{aligned}$Differential Equations exercise 21.7 question 10
Answer: $\left(e^{y}+1\right) \sin x=c$Hint:Separate the terms of x and y and then integrate them.
Given: $\left(e^{y}+1\right) \cos x d x+e^{y} \sin x d y=0$Solution: $\left(e^{y}+1\right) \cos x d x+e^{y} \sin x d y=0$$e^{y} \sin x d y=-\left(e^{y}+1\right) \cos x d x$$\frac{-e^{y}}{\left(e^{y}+1\right)} d y=\frac{\cos x}{\sin x} d x$ Integrating both sides
$\begin{aligned} &\int \frac{-e^{y}}{\left(e^{y}+1\right)} d y=\int \frac{\cos x}{\sin x} d x \\\\ &\text { Put } e^{y}+1=t \Rightarrow e^{y} d y=d t \\\\ &\sin x=u \Rightarrow d x \cos x=d u \end{aligned}$$\begin{aligned} &-\int \frac{d t}{t}=\int \frac{d u}{u} \\\\ &-\log |t|=\log |u|+\log c \\\\ &-\log \left|e^{y}+1\right|=\log |\sin x|+\log c \end{aligned}$$\begin{aligned} &-\log \left|e^{y}+1\right|-\log |\sin x|=\log c \\\\ &-\left[\log \left(e^{y}+1\right)(\sin x)\right]=\log c \\\\ &{\left[\log \left(e^{y}+1\right)(\sin x)\right]=-\log c} \\\\ &\left(e^{y}+1\right) \sin x=c \end{aligned}$Differential Equations exercise 21.7 question 11
Answer: $x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+c$Hint: Separate the terms of x and y and then integrate them.
Given: $x \cos ^{2} y d x=y \cos ^{2} x d y$Solution: $x \cos ^{2} y d x=y \cos ^{2} x d y$$\begin{aligned} &\frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y \\\\ &x \sec ^{2} x d x=y \sec ^{2} y d y \end{aligned}$Integrating both sides
$\int x \sec ^{2} x d x=\int y \sec ^{2} y d y$Using integration by parts
$\begin{aligned} &x \int \sec ^{2} x d x=y \int \sec ^{2} y d y \\\\ &x \tan x-\int \tan x d x=y \tan y-\int \tan y d y \end{aligned}$Using identity,
$\int \tan x d x=\log |\sec x|$$\begin{aligned} &x \tan (x)-\log |\sec x|=y \tan y-\log |\sec y|+c \\\\ &x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+c \end{aligned}$Differential Equations exercise 21.7 question 12 maths
Answer: $y-x=\log |x|-\log |y-1|+c$Hint: Separate the terms of x and y and then integrate them.
Given: $x y d y=(y-1)(x+1) d x$Solution: $x y d y=(y-1)(x+1) d x$$\begin{aligned} &\frac{y d y}{y-1}=\frac{(x+1)}{x} d x \\\\ &\left(\frac{1+y-1}{y-1}\right) d y=\left(1+\frac{1}{x}\right) d x \\\\ &\left(\frac{1}{y-1}+\frac{(y-1)}{(y-1)}\right) d y=\left(1+\frac{1}{x}\right) d x \end{aligned}$ Integrating both sides
$\begin{aligned} &\int \frac{1}{y-1} d y+\int 1 d y=\int 1 d x+\int \frac{1}{x} d x \\\\ &\log |y-1|+y=x+\log |x|+c \\\\ &y-x=\log |x|-\log |y-1|+c \end{aligned}$Differential Equations exercise 21.7 question 13
Answer: $x c=\cos y$Hint: Separate the terms of x and y and then integrate them.
Given: $x \frac{d y}{d x}+\cot y=0$Solution: $x \frac{d y}{d x}+\cot y=0$$\begin{aligned} &x \frac{d y}{d x}=-\cot y \\\\ &d y=-\cot y \frac{d x}{x} \\\\ &\frac{d y}{-\cot y}=\frac{d x}{x} \end{aligned}$Integrating both sides
$\begin{aligned} &-\tan y d y=\frac{d x}{x} \\\\ &-\int \tan y d y=\int \frac{d x}{x} \end{aligned}$$\begin{aligned} &-[-\log |\cos y|]=\log |x|+\log C \\\\ &\log |\cos y|=\log |x|+\log C \\\\ &\cos y=x C \end{aligned}$Differential Equations exercise 21.7 question 14
Answer: $\sin y=e^{x} \log x+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=\frac{x e^{x} \log x+e^{x}}{x \cos y}$Solution: $\frac{d y}{d x}=\frac{x e^{x} \log x+e^{x}}{x \cos y}$$\cos y d y=\frac{x \cdot e^{x} \log x+e^{x}}{x} d x$Integrating both sides
$\begin{aligned} &\int \cos y d y=\int e^{x} \log x d x+\int \frac{e^{x}}{x} d x \\\\ &\sin y=\log x \int e^{x} d x-\int \frac{1}{x}\left(\int e^{x} d x\right) d x+\int \frac{e^{x}}{x} d x \end{aligned}$$\begin{aligned} &\sin y=e^{x} \log x-\int \frac{e^{x}}{x} d x+\int \frac{e^{x}}{x} d x+c \\\\ &\sin y=e^{x} \log x+c \end{aligned}$Differential Equations exercise 21.7 question 15
Answer: $c=e^{x}+e^{-y}+\frac{x^{4}}{4}$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=e^{x+y}+e^{y} x^{3}$Solution: $\frac{d y}{d x}=e^{x+y}+e^{y} x^{3}$$\frac{d y}{d x}=e^{y}\left(e^{x}+x^{3}\right) \\$$\begin{aligned} &\frac{d y}{e^{y}}=\left(e^{x}+x^{3}\right) d x \\\\ &e^{-y} d y=\left(e^{x}+x^{3}\right) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int e^{-y} d y=\int e^{x} d x+\int x^{3} d x \\\\ &-e^{-y}+c=e^{x}+\frac{x^{4}}{4} \\\\ &c=e^{x}+e^{-y}+\frac{x^{4}}{4} \end{aligned}$Differential Equations exercise 21.7 question 16 maths
Answer: $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} \log \left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+\frac{1}{2} \log \left|\frac{\sqrt{1+y^{2}}-1}{\sqrt{1+y^{2}}+1}\right|=c$Hint: Separate the terms of x and y and then integrate them.
Given: $y \sqrt{1+x^{2}}+x \sqrt{1+y^{2}} \frac{d y}{d x}=0$Solution: $y \sqrt{1+x^{2}}+x \sqrt{1+y^{2}} \frac{d y}{d x}=0$$\begin{aligned} &x \sqrt{1+y^{2}} d y=-y \sqrt{1+x^{2}} d x \\\\ &\frac{\sqrt{1+y^{2}}}{-y} d y=\frac{\sqrt{1+x^{2}}}{x} d x \end{aligned}$$\begin{aligned} &1+y^{2}=m^{2} \Rightarrow y=\sqrt{m^{2}-1} \\\\ &2 y d y=2 m d m \Rightarrow d y=\frac{m}{y} d m \\\\ &\text { And Put } 1+x^{2}=n^{2} \Rightarrow x=\sqrt{n^{2}-1} \\\\ &2 x d x=2 n d n \Rightarrow x d x=n d n \Rightarrow d x=\frac{n}{x} d n \end{aligned}$Integrating both sides
$\begin{aligned} &=>-\int \frac{\sqrt{1+y^{2}}}{y} d y=\int \frac{\sqrt{1+x^{2}}}{x} d x \\\\ &=>-\int \frac{m}{m^{2}-1} d m . m=\int \frac{n}{n^{2}-1} d n \cdot n \end{aligned}$$=>-\int \frac{m^{2}}{m^{2}-1} d m=\int \frac{n^{2}}{n^{2}-1} d n$$=>-\int \frac{m^{2}-1+1}{m^{2}-1} d m=\int \frac{n^{2}-1+1}{n^{2}-1} d n$$\begin{aligned} &=-\int 1 d m-\int \frac{1}{\left(m^{2}-1\right)} d m=\int 1 d n+\int \frac{1}{n^{2}-1} d n \\\\ &=>-m-\frac{1}{2} \log \left(\frac{m-1}{m+1}\right)=n+\frac{1}{2} \log \left(\frac{n-1}{n+1}\right)+c \end{aligned}$$\begin{aligned} &m=\sqrt{1+y^{2}} \& n=\sqrt{1+x^{2}} \\\\ &\Rightarrow-\sqrt{1+y^{2}}-\frac{1}{2} \log \left(\frac{\sqrt{1+y^{2}}-1}{\sqrt{1+y^{2}}+1}\right)=\sqrt{1+x^{2}}+\frac{1}{2} \log \left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+c \end{aligned}$$c=\sqrt{1+x^{2}}+\sqrt{1+y^{2}}+\frac{1}{2} \log \left(\frac{\sqrt{1+y^{2}}-1}{\sqrt{1+y^{2}}+1}\right)+\frac{1}{2} \log \left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)$Differential Equations exercise 21.7 question 17
Answer: $\log \left(y+\sqrt{1+y^{2}}\right)\left(x+\sqrt{1+x^{2}}\right)=C$Hint: Separate the terms of x and y and then integrate them.
Given: $\sqrt{1+x^{2}} d y+\sqrt{1+y^{2}} d x=0$Solution: $\sqrt{1+x^{2}} d y+\sqrt{1+y^{2}} d x=0$$\begin{aligned} &-\sqrt{1+x^{2}} d y=\sqrt{1+y^{2}} d x \\\\ &\frac{d y}{\sqrt{1+y^{2}}}=-\frac{d x}{\sqrt{1+x^{2}}} \end{aligned}$Integrating both sides
$\begin{aligned} & \int \frac{d y}{\sqrt{1+y^{2}}}=-\int \frac{d x}{\sqrt{1+x^{2}}} \end{aligned}$$\text { Let } I=\int \frac{d x}{\sqrt{1+x^{2}}}$Put
$x=\tan \theta$$d x=\sec ^{2} \theta d \theta$$\begin{aligned} &I=\int \frac{1}{\sqrt{1+\tan ^{2} \theta}} \sec ^{2} \theta d \theta \\\\ &=\int \frac{1}{\sqrt{\sec ^{2} \theta}} \sec ^{2} \theta d \theta \\\\ &\left(\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}\right) \end{aligned}$$\begin{aligned} &I=\int \sec \theta d \theta \\\\ &=\log |\sec \theta+\tan \theta| \\\\ &\tan \theta=x \& \sec \theta=\sqrt{1+\tan ^{2} \theta}=\sqrt{1+x^{2}} \\\\ &I=\log \left|x+\sqrt{1+x^{2}}\right|+c \end{aligned}$Similarly,
$\int \frac{1}{\sqrt{1+y^{2}}}=\log \left|y+\sqrt{1+y^{2}}\right|+c$Hence,
$\begin{aligned} &\log \left|y+\sqrt{1+y^{2}}\right|=-\log \left|x+\sqrt{1+x^{2}}\right|+c \\\\ &\log \left|y+\sqrt{1+y^{2}}\right|+\log \left|x+\sqrt{1+x^{2}}\right|=c \\\\ &\log \left(y+\sqrt{1+y^{2}}\right)\left(x+\sqrt{1+x^{2}}\right)=c \end{aligned}$Differential Equations exercise 21.7 question 18
Answer: $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} l \operatorname{og}\left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+c=0$Hint: Separate the terms of x and y and then integrate them.
Given: $\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$Solution: $\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$$\begin{aligned} &\Rightarrow x y \frac{d y}{d x}=-\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \\\\ &\Rightarrow x y \frac{d y}{d x}=-\sqrt{\left(1+x^{2}\right)\left(1+y^{2}\right)} \\\\ &\Rightarrow \frac{y d y}{\sqrt{1+y^{2}}}=\frac{-\sqrt{1+x^{2}}}{x} \end{aligned}$Integrating both sides
$\Rightarrow \int \frac{y d y}{\sqrt{1+y^{2}}}=\int \frac{-\sqrt{1+x^{2}}}{x} d x$$\text { Let } 1+y^{2}=t \Rightarrow 2 y d y=d t \Rightarrow y d y=\frac{d t}{2}$$1+x^{2}=m^{2} \Rightarrow x=\sqrt{m^{2}-1} \Rightarrow 2 x d x=2 m d m \Rightarrow d x=\frac{m}{x} d m$$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t=-\int \frac{m}{m^{2}-1} d m \cdot m$$\Rightarrow \frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\int \frac{m^{2}}{m^{2}-1} d m=0$$\begin{aligned} &\Rightarrow \sqrt{t}+\int \frac{m^{2}-1+1}{m^{2}-1} d m=0 \\\\ &\Rightarrow \sqrt{t}+\int \frac{m^{2}-1}{m^{2}-1} d m+\int \frac{1}{m^{2}-1} d m=0 \\\\ &\Rightarrow \sqrt{t}+m+\frac{1}{2} \log \frac{(m-1)}{m+1}=0 \end{aligned}$ $\cdots \cdot \int \frac{1}{m^{2}-1} d m=\frac{1}{2} \log \frac{(m-1)}{m+1}$$\begin{aligned} &\text { Where } m=\sqrt{1+x^{2}}, 1+y^{2}=t \\\\ &\sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} l \operatorname{og}\left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+c=0 \end{aligned}$Differential Equations exercise 21.7 question 19
Answer: $y^{2} \log y=e^{x} \sin ^{2} x+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=\frac{e^{x}\left(\sin ^{2} x+\sin 2 x\right)}{y(2 \log y+1)}$Solution: $\frac{d y}{d x}=\frac{e^{x}\left(\sin ^{2} x+\sin 2 x\right)}{y(2 \log y+1)}$$\begin{aligned} &\Rightarrow y(2 \log y+1) d y=e^{x}\left(\sin ^{2} x+\sin 2 x\right) d x \\\\ &\Rightarrow(2 y \log y+y) d y=\left(e^{x} \sin ^{2} x+e^{x} \sin 2 x\right) d x \\\\ &\Rightarrow 2 y \log y d y+y d y=e^{x} \sin ^{2} x d x+e^{x} \sin 2 x d x \end{aligned}$Integrating both sides and using integrating by parts
$\Rightarrow 2\left[\log y \int y d y-\int\left\{\frac{d}{d y}(\log y) \int y d y\right\}\right] d y+\int y d y$ $=\sin ^{2} x \int e^{x} d x-\int\left[\frac{d}{d x} \sin ^{2} x \int\left(e^{x} d x\right)\right] d x+\int e^{x} \sin 2 x d x$$\Rightarrow 2\left[\log y\left(\frac{y^{2}}{2}\right)-\int\left(\frac{1}{y}\right)\left(\frac{y^{2}}{2}\right) d y\right]+y d y$ $=\sin ^{2} x e^{x}-\int\left[2 \sin x \cos x e^{x}\right] d x+\int e^{x} \sin 2 x+c$$\begin{aligned} &\Rightarrow y^{2} \log y-\int y d y+\int y d y=-\int e^{x} \sin 2 x d x+\int e^{x} \sin 2 x d x+e^{x} \sin ^{2} x \\\\ &y^{2} \log y=e^{x} \sin ^{2} x+c \end{aligned}$Differential Equations exercise 21.7 question 20 maths
Answer: $y \sin y=x^{2} \log x+c$Hint: You must know about the rules of solving differential equation and integration
Given: $\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}$Solution: $\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}$$(\sin y+y \cos y) d y=x(2 \log x+1) d x$Integrating both sides
$\int \sin y d y+\int y \cos y d y=2 \int x \log x d x+\int x d x$$\begin{aligned} &-\cos y+y \int \cos y d y-\int\left[\frac{d}{d y} y \cdot \int \cos y d y\right] d y=2\left\{\log x \int x-\int\left[\frac{d}{d x} \log x \cdot \int x d x\right] d x\right\} \\\\ &-\cos y+[y \sin y+\cos y]=2\left[\frac{x^{2}}{2} \log x-\frac{1}{2} \int x d x\right]+\frac{x^{2}}{2} \\\\ &y \sin y=x^{2} \log x+c \end{aligned}$Differential Equations exercise 21.7 question 21
Answer: $-\log |y|+\log |y-1|=-\frac{1}{2} \log \left|1-x^{2}\right|+\log c$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(1-x^{2}\right) d y+x y d x=x y^{2} d x$Solution: $\left(1-x^{2}\right) d y+x y d x=x y^{2} d x$$\begin{aligned} &\left(1-x^{2}\right) d y=\left(x y^{2}-x y\right) d x \\\\ &\left(1-x^{2}\right) d y=x y(y-1) d x \\\\ &\frac{d y}{y(y-1)}=\frac{x d x}{\left(1-x^{2}\right)} \end{aligned}$Integrating both sides
$\int \frac{d y}{y(y-1)}=\int \frac{x d x}{\left(1-x^{2}\right)}$$\int\left(\frac{1}{y-1}-\frac{1}{y}\right) d y=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)} d x$Put
$\begin{aligned} &1-x^{2}=t \\\\ &-2 x d x=d t \\\\ &-\log |y|+\log |y-1|=-\frac{1}{2} \int \frac{d t}{t} \end{aligned}$$\begin{aligned} &-\log |y|+\log |y-1|=-\frac{1}{2} \log |t|+\log c \\\\ &-\log |y|+\log |y-1|=-\frac{1}{2} \log \left|1-x^{2}\right|+\log c \end{aligned}$Differential Equations exercise 21.7 question 22
Answer: $\tan x \times \sin x=c$Hint: Separate the terms of x and y and then integrate them.
Given: $\tan y d x+\sec ^{2} y \tan x d y=0$Solution: $\tan y d x+\sec ^{2} y \tan x d y=0$$\begin{aligned} &\Rightarrow \sec ^{2} y \tan x d y=-\tan y d x \\\\ &\Rightarrow \frac{\sec ^{2} y}{\tan y} d y=-\frac{1}{\tan x} d x \\\\ &\Rightarrow \frac{1}{\cos ^{2} y} \times \frac{\cos y}{\sin y} d y=-\cot x d x \end{aligned}$$\begin{aligned} &\Rightarrow \frac{1}{\sin y \cos y} d y=-\cot x d x \\\\ &\Rightarrow \frac{2}{\sin 2 y}=-\cot x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\Rightarrow 2 \int \operatorname{cosec} 2 y d y=-\int \cot x d x \\\\ &\Rightarrow \log \tan x=-\log \sin x+\log c \\\\ &\Rightarrow \log \tan x+\log \sin x=\log c \\\\ &\Rightarrow \log (\tan x \times \sin x)=\log c \\\\ &\Rightarrow \tan x \times \sin x=c \end{aligned}$Differential Equations exercise 21.7 question 23
Answer: $\tan ^{-1} x+\tan ^{-1} y+\frac{1}{2} \log \left|1+x^{2}\right|+\frac{1}{2} \log \left|1+y^{2}\right|=c$Hint: Separate the terms of x and y and then integrate them.
Given: $(1+x)\left(1+y^{2}\right) d x+(1+y)\left(1+x^{2}\right) d y=0$Solution: $\begin{aligned} &(1+x)\left(1+y^{2}\right) d x+(1+y)\left(1+x^{2}\right) d y=0 \\\\ &(1+x)\left(1+y^{2}\right) d x=-(1+y)\left(1+x^{2}\right) d y \\\\ &\frac{(1+x)}{1+x^{2}} d x=-\frac{(1+y)}{1+y^{2}} d y \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{(1+x)}{1+x^{2}} d x=-\int \frac{(1+y)}{1+y^{2}} d y \\\\ &\Rightarrow \int \frac{1}{1+x^{2}} d x+\int \frac{x}{1+x^{2}} d x=-\int \frac{1}{1+y^{2}} d y-\int \frac{y}{1+y^{2}} d y \end{aligned}$$\begin{aligned} &1+x^{2}=t \Rightarrow 2 x d x=d t \\\\ &1+y^{2}=u \Rightarrow 2 y d y=d u \\\\ &\int \frac{1}{1+x^{2}} d x+\frac{1}{2} \int \frac{1}{t} d t=-\int \frac{1}{1+y^{2}} d y-\frac{1}{2} \int \frac{1}{u} d u \end{aligned}$$\begin{aligned} &\tan ^{-1} x+\frac{1}{2} \log |t|=-\tan ^{-1} y-\frac{1}{2} \log |u|+c \\\\ &\tan ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|=-\tan ^{-1} y-\frac{1}{2} \log \left|1+y^{2}\right|+c \\\\ &\tan ^{-1} x+\tan ^{-1} y+\frac{1}{2} \log \left|1+x^{2}\right|+\frac{1}{2} \log \left|1+y^{2}\right|=c \end{aligned}$Differential Equations exercise 21.7 question 24 maths
Answer: $\sec y=-2 \cos x+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)$Solution: $\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)$$\frac{d y}{d x}=\frac{\sin (x+y)+\sin (x-y)}{\tan y}$$=\frac{2 \sin x \cos y}{\tan y}$$\begin{aligned} &\Rightarrow \frac{\tan y}{\cos y} d y=2 \sin x d x \\\\ &\Rightarrow \frac{\sin y}{\cos ^{2} y} d y=2 \sin x d x \end{aligned}$Integrating both sides
$\Rightarrow \int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin x d x$$\begin{aligned} &\Rightarrow \sec y=-2 \cos x+c \\\\ &\Rightarrow \sec y=-2 \cos x+c \end{aligned}$Differential Equations exercise 21.7 question 25
Answer: $\sin y=C \cos x$Hint: Separate the terms of x and y and then integrate them.
Given: $\cos x \cos y \frac{d y}{d x}=-\sin x \sin y$Solution: $\cos x \cos y \frac{d y}{d x}=-\sin x \sin y$$\begin{aligned} &\frac{\cos y}{\sin y} d y=\frac{-\sin x}{\cos x} d x \\\\ &\cot y d y=-\tan x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \cot y d y=-\int \tan x d x \\\\ &\log |\sin y|=-[-\log |\cos x|]+\log C \\\\ &\log \sin y=\log \cos x+\log C \end{aligned}$$\begin{aligned} &\log \sin y=\log C(\cos x) \\\\ &\sin y=C \cos x \end{aligned}$Differential Equations exercise 21.7 question 26
Answer: $\log |\sin y|=-\sin x+C$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}+\frac{\cos x \sin y}{\cos y}=0$Solution: $\frac{d y}{d x}+\frac{\cos x \sin y}{\cos y}=0$$\begin{aligned} &\frac{d y}{d x}=-\frac{\cos x \sin y}{\cos y} \\\\ &\frac{\cos y}{\sin y} d y=-\cos x d x \\\\ &\cot y d y=-\cos x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \cot y d y=-\int \cos x d x \\\\ &\log |\sin y|=-\sin x+c \end{aligned}$Differential Equations exercise 21.7 question 27
Answer: $-\sqrt{1-y^{2}}=\sqrt{1-x^{2}}+c \text { or } \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=c$Hint: You must know about the rules of solving differential equation and integration
Given: $x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}} d y=0$Solution: $y \sqrt{1-x^{2}} d y=-x \sqrt{1-y^{2}} d x$$\int \frac{y}{\sqrt{1-y^{2}}} d y=\int \frac{-x}{\sqrt{1-x^{2}}} d x$We know
$\frac{d}{d x}\left(\sqrt{1-x^{2}}\right)=\frac{-x}{\sqrt{1-x^{2}}}$Integration of
$\frac{-x}{\sqrt{1-x^{2}}} d x=\sqrt{1-x^{2}}$Similarly Integration of
$\frac{y}{\sqrt{1-y^{2}}} d y=-\sqrt{1-y^{2}}$$\begin{aligned} &\Rightarrow-\sqrt{1-y^{2}}=\sqrt{1-x^{2}}+c \\\\ &\Rightarrow-\sqrt{1-y^{2}}-\sqrt{1-x^{2}}=c \\\\ &\Rightarrow-\left[\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\right]=c \end{aligned}$$\begin{aligned} &\Rightarrow \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=-c \\\\ &\Rightarrow \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=c \end{aligned}$Differential Equations exercise 21.7 question 28 maths
Answer: $y=\log \left(c(1+y)\left(e^{x}+1\right)\right)$Hint: Separate the terms of x and y and then integrate them.
Given: $y\left(1+e^{x}\right) d y=(y+1) e^{x} d x$Solution: $y\left(1+e^{x}\right) d y=(y+1) e^{x} d x$$\frac{y d y}{(y+1)}=\frac{e^{x}}{\left(1+e^{x}\right)} d x$Integrating both sides
$\begin{aligned} &\Rightarrow y-\log |y+1|=\log \left|e^{x}+1\right|+\log c \\\\ &\Rightarrow y=\log |y+1|+\log \left|e^{x}+1\right|+\log c \\\\ &y=\log \left(c(1+y)\left(e^{x}+1\right)\right) \end{aligned}$Differential Equations exercise 21.7 question 29
Answer: $\log (x y)+x-\frac{y^{2}}{2}=C$Hint: Separate the terms of x and y and then integrate them.
Given: $(y+x y) d x+\left(x-x y^{2}\right) d y=0$Solution: $\begin{aligned} &(y+x y) d x+\left(x-x y^{2}\right) d y=0 \\\\ &y(1+x) d x+x\left(1-y^{2}\right) d y=0 \\\\ &\frac{1+x}{x} d x+\frac{\left(1-y^{2}\right)}{y} d y=0 \end{aligned}$Integrating both sides and also separate the terms
$\begin{aligned} &\int \frac{1}{x} d x+\int 1 d x+\int \frac{1}{y} d y-\int y d y=0 \\\\ &\log |x|+x+\log |y|-\frac{y^{2}}{2}=C \\\\ &\log (x y)+x-\frac{y^{2}}{2}=C \end{aligned}$Differential Equations exercise 21.7 question 30
Answer: $\log (1+y)=x-\frac{x^{2}}{2}+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=1-x+y-x y$Solution: $\frac{d y}{d x}=1-x+y-x y$$\begin{aligned} &\frac{d y}{d x}=(1-x)(1+y) \\\\ &\frac{d y}{(1+y)}=(1-x) d x \\\\ &\int \frac{d y}{(1+y)}=\int 1 d x-\int x d x \\ \end{aligned}$$\log (1+y)=x-\frac{x^{2}}{2}+c$Differential Equations exercise 21.7 question 31
Answer: $\tan ^{-1} y=\tan ^{-1} x+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(y^{2}+1\right) d x-\left(x^{2}+1\right) d y=0$Solution: $\left(y^{2}+1\right) d x-\left(x^{2}+1\right) d y=0$$\left(y^{2}+1\right) d x=\left(x^{2}+1\right) d y$$\begin{aligned} &\frac{d y}{\left(y^{2}+1\right)}=\frac{d x}{\left(x^{2}+1\right)} \\\\ &\tan ^{-1} y=\tan ^{-1} x+c \end{aligned}$Differential Equations exercise 21.7 question 32 maths
Answer: $\log (y+1)+x+\frac{x^{2}}{2}=c$Hint: Separate the terms of x and y and then integrate them.
Given: $d y+(x+1)(y+1) d x=0$Solution: $d y+(x+1)(y+1) d x=0$$\begin{aligned} &d y=-(x+1)(y+1) d x \\\\ &\frac{d y}{(y+1)}=-(x+1) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{y+1} d y=\int-x d x-\int 1 d x \\\\ &\Rightarrow \log (y+1)=-\frac{x^{2}}{2}-x+c \\\\ &\Rightarrow \log (y+1)+x+\frac{x^{2}}{2}=c \end{aligned}$Differential Equations exercise 21.7 question 33
Answer: $\tan ^{-1} y=x+\frac{x^{3}}{3}+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$Solution: $\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$$\frac{d y}{\left(1+y^{2}\right)}=\left(1+x^{2}\right) d x$Integrating both sides
$\begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int 1 d x+\int x^{2} d x \\\\ &\tan ^{-1} y=x+\frac{x^{3}}{3}+c \end{aligned}$Differential Equations exercise 21.7 question 34
Answer: $\log |y|=\frac{2 x^{3}}{3}+x^{2}+2 x+2 \log (x-1)+c$Hint: Separate the terms of x and y and then integrate them.
Given: $(x-1) \frac{d y}{d x}=2 x^{3} y$Solution: $(x-1) \frac{d y}{d x}=2 x^{3} y$$\begin{aligned} &\Rightarrow \frac{d y}{y}=\frac{2 x^{3}}{(x-1)} d x \\\\ &\Rightarrow \frac{d y}{y}=\frac{2\left((x-1)\left(x^{2}+x+1\right)+1\right)}{(x-1)} d x \\\\ &\Rightarrow \frac{d y}{y}=2\left(x^{2}+x+1+\frac{1}{x-1}\right) d x \end{aligned}$Integrating both sides
$\int \frac{d y}{y}=2\left[\int x^{2} d x+\int x d x+\int 1 d x+\int \frac{1}{x-1}\right] d x$$\begin{aligned} &\log |y|=2\left[\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+\log |x-1|\right]+c \\\\ &\log |y|=\frac{2 x^{3}}{3}+x^{2}+2 x+2 \log (x-1)+c \end{aligned}$Differential Equations exercise 21.7 question 35
Answer: $e^{-x}-e^{-y}-e^{x}=c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=e^{x+y}+e^{-x+y}$Solution: $\frac{d y}{d x}=e^{x+y}+e^{-x+y}$$\begin{aligned} &\frac{d y}{d x}=e^{x} e^{y}+e^{-x} e^{y} \\\\ &\frac{d y}{d x}=e^{y}\left[e^{x}+e^{-x}\right] \\\\ &\frac{d y}{e^{y}}=\left[e^{x}+e^{-x}\right] d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int e^{-y} d y=\int e^{x} d x+\int e^{-x} d x \\\\ &-e^{-y}=e^{x}+\left[-e^{-x}\right]+c \\\\ &-e^{-y}=e^{x}-e^{-x}+c \\\\ &e^{-x}-e^{-y}-e^{x}=c \end{aligned}$Differential Equations exercise 21.7 question 36 maths
Answer: $\tan y=\frac{\sin 2 x}{2}+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=\left(\cos ^{2} x-\sin ^{2} x\right) \cos ^{2} y$Solution: $\frac{d y}{d x}=\left(\cos ^{2} x-\sin ^{2} x\right) \cos ^{2} y$$\begin{aligned} &\frac{d y}{\cos ^{2} y}=\left(\cos ^{2} x-\sin ^{2} x\right) d x \\\\ &\frac{d y}{\cos ^{2} y}=\cos 2 x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \sec ^{2} y d y=\int \cos 2 x d x \\\\ &\tan y=\frac{\sin 2 x}{2}+c \end{aligned}$Differential Equations exercise 21.7 question 37 (i)
Answer: $y^{2}+2=\frac{c^{1}}{x^{2}+2}$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(x y^{2}+2 x\right) d x+\left(x y^{2}+2 y\right) d y=0$Solution: $\left(x y^{2}+2 x\right) d x+\left(x y^{2}+2 y\right) d y=0$$\begin{aligned} &x\left(y^{2}+2\right) d x+y\left(x^{2}+2\right) d y=0 \\\\ &y\left(x^{2}+2\right) d y=-x\left(y^{2}+2\right) d x \\\\ &\frac{y}{\left(y^{2}+2\right)} d y=\frac{-x}{\left(x^{2}+2\right)} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\Rightarrow \int \frac{y}{\left(y^{2}+2\right)} d y=-\int \frac{x}{x^{2}+2} d x \\\\ &\Rightarrow \frac{1}{2} \int \frac{2 y}{y^{2}+2} d y=-\frac{1}{2} \int \frac{2 x}{x^{2}+2} d x \end{aligned}$$\begin{aligned} &\Rightarrow \frac{1}{2} \log \left|y^{2}+2\right|=-\frac{1}{2} \log \left|x^{2}+2\right|+\log c \\\\ &\Rightarrow \frac{1}{2}\left[\log \left|y^{2}+2\right|+\log \left|x^{2}+2\right|\right]=\log c \end{aligned}$$\begin{aligned} &\Rightarrow \log \left(y^{2}+2\right)\left(x^{2}+2\right)=2 \log c \\\\ &\Rightarrow \log \left(y^{2}+2\right)\left(x^{2}+2\right)=\log c^{2} \\\\ &\Rightarrow\left(y^{2}+2\right)\left(x^{2}+2\right)=c^{1} \\\\ &\Rightarrow\left(y^{2}+2\right)=\frac{c^{1}}{\left(x^{2}+2\right)} \end{aligned}$Differential Equations exercise 21.7 question 37 (ii)
Answer: $-\frac{\log y}{y}-\frac{1}{y}=-\left[-x^{2} \cos x+2 x \sin x+2 \cos x\right]+c$Hint: Separate the terms of x and y and then integrate them.
Given: $\operatorname{cosec} x \log y \frac{d y}{d x}+x^{2} y^{2}=0$Solution: $\operatorname{cosec} x \log y \frac{d y}{d x}+x^{2} y^{2}=0$$\operatorname{cosec} x \log y \frac{d y}{d x}=-x^{2} y^{2}$$\frac{\log y}{y^{2}} d y=-x^{2} \frac{1}{\operatorname{cosec} x} d x$Integrating both sides
$\int \frac{\log y}{y^{2}} d y=\int-x^{2} \sin x d x$$\int \frac{\log y}{y^{2}} d y=-\left[-x^{2} \cos x+2 \int x \cos x d x\right]$ { using integration by parts}
$\begin{aligned} &-\frac{\log y}{y}-\frac{1}{y}=-\left[-x^{2} \cos x+2 x \sin x-2 \int \sin x d x\right] \\\\ &-\frac{\log y}{y}-\frac{1}{y}=-\left[-x^{2} \cos x+2 x \sin x+2 \cos x\right]+c \end{aligned}$Differential Equations exercise 21.7 question 38 (i)
Answer: $(y-x)-(\log (x(1+y)))=c$Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x}=1+x+y+x y$Solution: $x y \frac{d y}{d x}=1+x+y+x y$$\begin{aligned} &x y \frac{d y}{d x}=(1+x)+y(1+x) \\\\ &x y \frac{d y}{d x}=(1+x)(1+y) \\\\ &\frac{y}{1+y} d y=\frac{1+x}{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int 1 d y-\int \frac{1}{1+y} d y=\int \frac{1}{x} d x+\int 1 d x \\\\ &y-\log |1+y|=\log |x|+x+c \end{aligned}$$\begin{aligned} &y-x-\log |1+y|-\log |x|=c \\\\ &(y-x)-\log [x(1+y)]=c \end{aligned}$Differential Equations exercise 21.7 question 38 (ii) maths
Answer: $\left(1+y^{2}\right)=\frac{c^{1}}{\left(1-x^{2}\right)}$Hint: Separate the terms of x and y and then integrate them.
Given: $y\left(1-x^{2}\right) \frac{d y}{d x}=x\left(1+y^{2}\right)$Solution: $y\left(1-x^{2}\right) \frac{d y}{d x}=x\left(1+y^{2}\right)$$\frac{y d y}{\left(1+y^{2}\right)}=\frac{x}{\left(1-x^{2}\right)} d x$Integrating both sides
$\int \frac{y d y}{\left(1+y^{2}\right)}=\int \frac{x}{\left(1-x^{2}\right)} d x$$\frac{1}{2} \int \frac{2 y}{\left(1+y^{2}\right)} d y=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)} d x$$\frac{1}{2} \log \left|1+y^{2}\right|=-\frac{1}{2} \log \left|1-x^{2}\right|+\log c$$\begin{aligned} &\frac{1}{2}\left[\log \left|1+y^{2}\right|+\log \left|1-x^{2}\right|\right]=\log c \\\\ &\log \left(1+y^{2}\right)\left(1-x^{2}\right)=\log c^{2} \\\\ &\left(1+y^{2}\right)\left(1-x^{2}\right)=c^{2} \\\\ &\left(1+y^{2}\right)=\frac{c^{1}}{1-x^{2}} \end{aligned}$Differential Equations exercise 21.7 question 38 (iii)
Answer: $y=e^{\frac{x}{y}}+c$Hint: Separate the terms of x and y and then integrate them.
Given: $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y$Solution: $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y$$\frac{d x}{d y}=\frac{x e^{\frac{x}{y}}+y^{2}}{y_{e}^{\frac{x}{y}}}$$x=v y=>\frac{d x}{d y}=v+y \cdot \frac{d v}{d y}$$v+y \cdot \frac{d v}{d y}=\frac{v y \cdot e^{v}+y^{2}}{y e^{v}}$$\begin{aligned} &y \cdot \frac{d v}{d y}=\frac{v y \cdot e^{v}+y^{2}}{y e^{v}}-v \\\\ &y \cdot \frac{d v}{d y}=\frac{v y \cdot e^{v}+y^{2}-v y e^{v}}{y e^{v}} \end{aligned}$$\begin{aligned} &e^{v} d v=\frac{y d y}{y} \\\\ &e^{v} d v=d y \end{aligned}$Integrating both sides
$\begin{aligned} &\int e^{y} d v=\int 1 d y \\\\ &y=e^{v}+c \end{aligned}$Put
$\begin{aligned} &v=\frac{x}{y} \\\\ &y=e^{\frac{x}{y}}+c \end{aligned}$Differential Equations exercise 21.7 question 38 (iv)
Answer: $\frac{1}{2}\left(\tan ^{-1} x\right)^{2}+\log \left|1+y^{2}\right|=c$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(1+y^{2}\right) \tan ^{-1} x d x+2 y\left(1+x^{2}\right) d y=0$Solution:$\left(1+y^{2}\right) \tan ^{-1} x d x+2 y\left(1+x^{2}\right) d y=0$$\begin{aligned} &\left(1+y^{2}\right) \tan ^{-1} x d x=-2 y\left(1+x^{2}\right) d y \\\\ &\frac{\tan ^{-1} x d x}{\left(1+x^{2}\right)}=-\frac{2 y}{\left(1+y^{2}\right)} d y \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{\tan ^{-1} x d x}{\left(1+x^{2}\right)}=-\int \frac{2 y}{\left(1+y^{2}\right)} d y \\\\ &\frac{1}{2}\left(\tan ^{-1} x\right)^{2}=-\log \left(1+y^{2}\right)+c \\\\ &\frac{1}{2}\left(\tan ^{-1} x\right)^{2}+\log \left|1+y^{2}\right|=c \end{aligned}$Differential Equations exercise 21.7 question 39
Answer: $y^{2}=4|\sec 2 x|$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=y \tan 2 x, y(0)=2$Solution:$\frac{d y}{d x}=y \tan 2 x, y(0)=2$$\frac{d y}{y}=\tan 2 x d x$Integrating both sides
$\begin{aligned} &\int\left(\frac{d y}{y}\right)=\int \tan 2 x d x \\\\ &\log y=\frac{1}{2} \log |\sec 2 x|+c \\\\ &y(0)=2 \text { where, } y=2 \; \& \; x=0 \end{aligned}$$\begin{aligned} &\log (2)=\frac{1}{2} \log |\sec 2(0)|+c \\\\ &c=\log 2 \\\\ &\log y=\frac{1}{2} \log |\sec 2 x|+\log 2 \end{aligned}$$\begin{aligned} &\log y=\log \sqrt{\sec 2 x}+\log 2 \\\\ &y=2 \sqrt{\sec 2 x} \\\\ &y^{2}=4|\sec 2 x| \end{aligned}$Differential Equations exercise 21.7 question 40 maths
Answer: $3 \sqrt{\frac{y}{2}}=\sqrt{x}$Hint: Separate the terms of x and y and then integrate them.
Given: $2 x \frac{d y}{d x}=3 y, y(1)=2$Solution:$2 x \frac{d y}{d x}=3 y$$\frac{d x}{3 y}=\frac{d x}{2 x}$Integrating both sides
$\begin{aligned} &\int \frac{d x}{3 y}=\int \frac{d x}{2 x} \\\\ &\frac{1}{3} \log |y|=\frac{1}{2} \log |x|+c \end{aligned}$$\begin{aligned} &\text { Put } y(1)=2 \text { where, } y=2 \& x=1 \\\\ &\frac{1}{3} \log |2|=\frac{1}{2} \log |1|+c \\\\ &\frac{1}{3} \log |2|=c \end{aligned}$$\begin{aligned} &{[\log 1=0]} \\\\ &\frac{1}{3} \log |y|=\frac{1}{2} \log |x|+\frac{1}{3} \log |2| \\\\ &\frac{1}{3} \log |y|-\frac{1}{3} \log |2|=\frac{1}{2} \log |x| \end{aligned}$$\begin{aligned} &\frac{1}{3} \log \frac{y}{2}=\frac{1}{2} \log |x| \\\\ &\log \sqrt[3]{\frac{y}{2}}=\log \sqrt{x} \\\\ &\sqrt[3]{\frac{y}{2}}=\sqrt{x} \end{aligned}$Differential Equations exercise 21.7 question 41
Answer: $y=\log \left[\frac{(y+2)^{2} x}{8}\right]$Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x}=y+2, y(2)=0$Solution:$\begin{aligned} &x y \frac{d y}{d x}=y+2 \\\\ &\Rightarrow \frac{y d y}{y+2}=\frac{d x}{x} \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{y}{y+2} d y=\int \frac{d x}{x} \\\\\ &\Rightarrow \int \frac{(y+2)-2}{(y+2)} d y=\int \frac{d x}{x} \end{aligned}$$\begin{aligned} &\Rightarrow \int\left(1-\frac{2}{y+2}\right) d y=\int \frac{d x}{x} \\\\ &\Rightarrow y-2 \log (y+2)=\log |x|+c \end{aligned}$ ....................(1)
$y(2)=0 \text { at } x=2, y=0$Put in (1)
$\begin{aligned} &\Rightarrow 0-2 \log |0+2|=\log |2|+c \\\\ &\Rightarrow-2 \log 2=\log 2+c \\\\ &\Rightarrow c=-3 \log 2 \end{aligned}$Put in (1), we get
$\begin{aligned} &y-2 \log (y+2)=\log |x|-3 \log 2 \\\\ &\Rightarrow y=\log |y+2|^{2}+\log |x|-\log 2^{3} \\\\ &\Rightarrow y=\log |y+2|^{2}+\log |x|-\log 8 \\\\ &\Rightarrow y=\log \left[\frac{(y+2)^{2} x}{8}\right] \end{aligned}$Differential Equations exercise 21.7 question 42
Answer: $4 \cdot y^{2}\left(2-e^{x}\right)=1$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=2 e^{x} y^{3}, y(0)=\frac{1}{2}$Solution:$\begin{aligned} &\frac{d y}{d x}=2 e^{x} y^{3} \\\\ &\frac{d y}{y^{3}}=2 e^{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{d y}{y^{3}}=2 \int e^{x} d x \\\\ &\Rightarrow \int y^{-3} d y=2 \int e^{x} d x \Rightarrow \frac{y^{-3+1}}{-3+1}=2 e^{x}+c \\\\ &{\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]} \end{aligned}$$\begin{aligned} &\Rightarrow \frac{y^{-2}}{-2}=2 e^{x}+c \\\\ &\Rightarrow \frac{-1}{2 y^{2}}=2 e^{x}+c \\\\ &\Rightarrow-1=2 y^{2}\left(2 e^{x}+c\right) \end{aligned}$ ..............(1)
Now given that
$y(0)=\frac{1}{2} \text { i.e. at } x=0\: \&\; y=\frac{1}{2}$Put in (1)
$\begin{aligned} &-1=2\left(\frac{1}{2}\right)^{2}\left(2 e^{0}+c\right) \\\\ &\Rightarrow-1=\left(\frac{1}{2}\right)(2(1)+c) \Rightarrow-2=2+c \Rightarrow c=-4 \end{aligned}$Put in (1) we get
$\begin{aligned} &\Rightarrow-1=2 y^{2}\left(2 e^{x}-4\right) \Rightarrow-1=-2\left(2 y^{2}\right)\left(2-e^{x}\right) \\\\ &\Rightarrow 4 y^{2}\left(2-e^{x}\right)=1 \end{aligned}$Differential Equations exercise 21.7 question 43
Answer: $r=r_{0} e^{\frac{-t^{2}}{2}}$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d r}{d t}=-r t, r(0)=r_{0}$Solution:$\begin{aligned} &\frac{d r}{d t}=-r t \\\\ &\frac{d r}{r}=-t d t \end{aligned}$Integrating both sides
$\begin{aligned} &\Rightarrow \int \frac{d r}{r}=-\int t d t \\\\ &\Rightarrow \log |r|=-\frac{t^{2}}{2}+c \end{aligned}$ ...............(1)
Given that
$r(0)=r_{0} \text { i.e. at } t=0, r=r_{0}$Put in (1)
$\log r_{0}=0+c \Rightarrow c=\log r_{0}$Put in (1) we get
$\begin{aligned} &\log |r|=-\frac{t^{2}}{2}+\log r_{0} \Rightarrow \log |r|-\log r_{0}=-\frac{t^{2}}{2} \\\\ &\Rightarrow \log \frac{r}{r_{0}}=-\frac{t^{2}}{2} \end{aligned}$$\begin{aligned} &{\left[\begin{array}{l} \log _{e} a=x \\ \Rightarrow a=e^{x} \end{array}\right]} \\\\ &\Rightarrow \frac{r}{r_{0}}=e^{\frac{-t^{2}}{2}} \\\\ &\Rightarrow r=r_{0} e^{\frac{-t^{2}}{2}} \end{aligned}$Differential Equations exercise 21.7 question 44 maths
Answer: $y=e^{\sin ^{2} x}$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=y \sin 2 x ; y(0)=1$Solution:$\begin{aligned} &\frac{d y}{d x}=y \sin 2 x \\\\ &\Rightarrow \frac{d y}{y}=\sin 2 x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{d y}{y}=\int \sin 2 x d x \\\\ &\Rightarrow \log |y|=\frac{-\cos 2 x}{2}+c \end{aligned}$ ...............(1)
Now Given that
$\text { at } x=0 ; y=1[y(0)=1]$$\begin{aligned} &\log |1|=\frac{-\cos 2(0)}{2}+c \\\\ &\Rightarrow 0=\frac{-1}{2}+c \Rightarrow c=\frac{1}{2} \end{aligned}$Put in (1) we get
$\begin{aligned} &\log |y|=\frac{-\cos 2 x}{2}+\frac{1}{2} \\\\ &\Rightarrow 2 \log |y|+\cos 2 x=1 \Rightarrow 2 \log y=1-\cos 2 x \\\\ &\Rightarrow \log y=\frac{2 \sin ^{2} x}{2} \end{aligned}$$\begin{aligned} &{\left[1-\cos 2 x=2 \sin ^{2} x\right]} \\\\ &\Rightarrow \log y=\sin ^{2} x \\\\ &{\left[\log _{\varepsilon} a=x \Rightarrow x=e^{x}\right]} \\\\ &\Rightarrow y=e^{\sin ^{2} x} \end{aligned}$Differential Equations exercise 21.7 question 45 (i)
Answer: $z=\sec x$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=y \tan x, y(0)=1$Solution: $\frac{d y}{d x}=y \tan x, y(0)=1$$\Rightarrow \frac{d y}{y}=\tan x \; d x$Integrating both sides
$\begin{aligned} &\Rightarrow \int \frac{1}{y} d y=\int \tan x d x \\\\ &\Rightarrow \log |y|=\log |\sec x|+\log c \\\\ &\Rightarrow \log |y|=\log |c \sec x| \\\\ &\Rightarrow y=c \sec x \end{aligned}$ ...............(1)
Now Given that
$y=1 \text { when } x=0$$\begin{aligned} &1=c \sec (0) \Rightarrow c=1 \quad[\sec 0=1] \\\\ &y=(1) \sec x \Rightarrow y=\sec x \end{aligned}$Differential Equations exercise 21.7 question 45 (ii)
Answer: $\log |y|=\frac{5}{2} \log |x|$Hint: Separate the terms of x and y and then integrate them.
Given: $2 x \frac{d y}{d x}=5 y, y(1)=1$Solution:$\begin{aligned} &2 x \frac{d y}{d x}=5 y \\\\ &\Rightarrow \frac{d y}{5 y}=\frac{d x}{2 x} \end{aligned}$Integrating both sides
$\Rightarrow \int \frac{d y}{5 y}=\int \frac{d x}{2 x} \Rightarrow \frac{1}{5} \log |y|=\frac{1}{2} \log |x|+c$ ...................(1)
Now given that
$y(1)=1 \quad \therefore y=1 \text { at } x=1$$\therefore \frac{1}{5} \log |y|=\frac{1}{2} \log |x|+c \Rightarrow c=0[\therefore \log 1=0]$Put in (1)
$\frac{1}{5} \log |y|=\frac{1}{2} \log |x|+0 \Rightarrow \log |y|=\frac{5}{2} \log |x|$Differential Equations exercise 21.7 question 45 (iii)
Answer: $y e^{2 x}+1=0$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=2 e^{2 x} y^{2}, y(0)=-1$Solution:$\begin{aligned} &\frac{d y}{d x}=2 e^{2 x} y^{2} \\\\ &\Rightarrow \frac{d y}{y^{2}}=2 e^{2 x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{y^{2}} d y=\int 2 e^{2 x} d x \\\\ &\Rightarrow \int y^{-2} d y=\int 2 e^{2 x} d x \end{aligned}$$\Rightarrow \frac{y^{-2+1}}{-2+1}=\frac{2 e^{2 x}}{2}+c \Rightarrow \frac{-1}{y}=e^{2 x}+c$ ..............(1)
Given that
$y(0)=-1 \text { i.e at } x=0 ; y=-1$$\begin{aligned} &\Rightarrow \frac{-1}{-1}=e^{2(0)}+c \Rightarrow 1=e^{0}+c \Rightarrow 1=1+c \Rightarrow c=0 \\\\ &{\left[\therefore e^{0}=1\right]} \end{aligned}$Put in (1)
$\Rightarrow \frac{-1}{y}=e^{2 x}+0 \Rightarrow y e^{2 x}+1=0$Differential Equations exercise 21.7 question 45 (iv) maths
Answer: $\sin y=e^{x}$Hint: Separate the terms of x and y and then integrate them.
Given: $\cos y \frac{d y}{d x}=e^{x}, y(0)=\frac{\pi}{2}$Solution:$\begin{aligned} &\cos y \frac{d y}{d x}=e^{x} \\\\ &\cos y d y=e^{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \cos y d y=\int e^{x} d x \\\\ &\sin y=e^{x}+c \end{aligned}$ ..............(1)
Given that
$y(0)=\frac{\pi}{2} \text { i.e. at } x=0, y=\frac{\pi}{2}$$\begin{aligned} &\therefore \sin \frac{\pi}{2}=e^{0}+c \quad[\therefore o f(1)] \\\\ &\qquad \Rightarrow 1=1+c \Rightarrow c=0\left[\therefore \sin \frac{\pi}{2}=1, e^{0}=1\right] \end{aligned}$Put in (1) we get
$\sin y=e^{x}+0 \Rightarrow \sin y=e^{x}$Differential Equations exercise 21.7 question 45 (v)
Answer: $x^{2}=\log y$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=2 x y, y(0)=1$Solution:$\begin{aligned} &\frac{d y}{d x}=2 x y \\\\ &\Rightarrow \frac{d y}{y}=2 x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{y} d y=\int 2 x d x \\\\ &\Rightarrow \log y=\frac{2 x^{2}}{2}+c \Rightarrow \log y=x^{2}+c \end{aligned}$ .............(1)
Given that at
$x=0, y=1$$\therefore \log 1=0+c \Rightarrow 0=0+c \Rightarrow c=0 \quad[\therefore \log 1=0]$Put in (1) we get
$\log y=x^{2}+0 \Rightarrow x^{2}=\log y$Differential Equations exercise 21.7 question 45 (vi)
Answer: $\tan ^{-1} y=\frac{x^{3}}{3}+x+\frac{\pi}{4}$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=1+x^{2}+y^{2}+x^{2} y^{2}, y(0)=1$Solution:$\begin{aligned} &\frac{d y}{d x}=1+x^{2}+y^{2}+x^{2} y^{2} \\\\ &\Rightarrow \frac{d y}{d x}=\left(1+x^{2}\right)+y^{2}\left(1+x^{2}\right) \\\\ &\Rightarrow \frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right) \\\\ &\Rightarrow \frac{d y}{1+y^{2}}=\left(1+x^{2}\right) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int\left(1+x^{2}\right) d x \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+c \end{aligned}$ ...............(1)
Given that
$y(0)=1 \text { i.e. at } x=0, y=1$$\begin{aligned} &\tan ^{-1}(1)=0+\frac{0}{3}+c \Rightarrow c=\tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \\\\ &{\left[\therefore \tan \frac{\pi}{4}=1\right]} \\\\ &\Rightarrow c=\frac{\pi}{4} \\\\ &\tan ^{-1} y=x+\frac{x^{3}}{3}+\frac{\pi}{4} \end{aligned}$Differential Equations exercise 21.7 question 45 (vii)
Answer: $x+2+\log \left|\frac{(x y+2 x)^{2}}{9}\right|$Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x}=(x+2)(y+2), y(1)=-1$Solution:$\begin{aligned} &x y \frac{d y}{d x}=(x+2)(y+2) \\\\ &\Rightarrow \frac{y}{y+2} d y=\frac{x+2}{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{y+2-2}{y+2} d y=\int 1 d x+\int \frac{2}{x} d x \\\\ &\int\left(1-\frac{2}{y+2}\right) d y=\int 1 d x+\int \frac{2}{x} d x \end{aligned}$$\begin{aligned} &\Rightarrow \int 1 d y-2 \int \frac{1}{y+2} d y=\int 1 d x+2 \int \frac{1}{x} d x \\\\ &\Rightarrow y-2 \log |y+2|=x+2 \log |x|+c \\\\ &\Rightarrow y-x=2 \log |x|+2 \log |y+2|+c \\\\ &y=x+2 \log |(x)(y+2)|+c \end{aligned}$.................(1)
Given that
$y(1)=-1$ i.e.
$y=-1$ when
$x=1$$\begin{aligned} &\therefore-1=1+2 \log |(1)(1+2)|+c \\\\ &\Rightarrow 2 \log 3+c=2 \Rightarrow c=2-2 \log 3=2-\log 3^{2} \\\\ &\Rightarrow c=2-\log 9 \end{aligned}$Put in (1)
$\begin{aligned} &y=x+2 \log |(x)(y+2)|+2-\log 9 \\\\ &{\left[\begin{array}{l} \therefore \log m+\log n=\log m n \\\\ \log m-\log n=\log \frac{m}{n} \end{array}\right]} \\\\ &\Rightarrow y=x+2+\log \left|\frac{(x y+2 x)^{2}}{9}\right| \end{aligned}$Differential Equations exercise 21.7 question 45 (viii) maths
Answer: $\tan ^{-1} y=\frac{x^{2}}{2}+x$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=1+x+y^{2}+x y^{2}, \text { when } y=0, x=0$Solution:$\begin{aligned} &\frac{d y}{d x}=1+x+y^{2}+x y^{2} \\\\ &\frac{d y}{d x}=(1+x)+y^{2}(1+x)=(1+x)\left(1+y^{2}\right) \\\\ &\Rightarrow \frac{d y}{1+y^{2}}=(1+x) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int(1+x) d x \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{2}}{2}+c \end{aligned}$ ..............(1)
Given that when
$y=0, x=0$$\therefore \tan ^{-1}(0)=0+\frac{0}{2}+c \Rightarrow c=0$Put in (1) we get
$\begin{aligned} &\tan ^{-1} y=x+\frac{x^{2}}{2}+0 \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{2}}{2} \end{aligned}$Differential Equations exercise 21.7 question 45 (ix)
Answer: $y=\log \left|(y+3)^{3} x^{2}\right|-2$Hint: Separate the terms of x and y and then integrate them.
Given: $2(y+3)-x y \frac{d y}{d x}=0, y(1)=-2$Solution:$\begin{aligned} &2(y+3)-x y \frac{d y}{d x}=0 \\\\ &\Rightarrow x y \frac{d y}{d x}=2(y+3) \Rightarrow \frac{y d y}{y+3}=\frac{2}{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{y d y}{y+3}=\int \frac{2}{x} d x \\\\ &\Rightarrow \int \frac{y+3-3}{y+3} d y=2 \int \frac{1}{x} d x \\\\ &\Rightarrow \int\left(1-\frac{3}{y+3}\right) d y=2 \int \frac{1}{x} d x \end{aligned}$$\begin{aligned} &\Rightarrow y-3 \log |y+3|=2 \log |x|+c \\\\ &\Rightarrow y=3 \log |y+3|+2 \log |x|+c \\\\ &\Rightarrow y=\log \left|(y+3)^{3}\right|+\log x^{2}+c \\\\ &\Rightarrow y=\log \left|(y+3)^{3} x^{2}\right|+c \end{aligned}$ ..............(1)
Given that
$y(1)=-2 \text { i.e. at } x=1, y=2$$\begin{aligned} &\therefore-2=\log \left|(-2+3)^{3}(1)^{2}\right|+c \\\\ &-2=\log 1+c \Rightarrow-2=0+c \Rightarrow c=-2 \\\\ &{[\therefore \log 1=0]} \end{aligned}$Put in (1) we get
$y=\log \left|(y+3)^{3} x^{2}\right|-2$Differential Equations exercise 21.7 question 45 (x)
Answer: $e^{x}-\tan y=2$Hint: Separate the terms of x and y and then integrate them.
Given: $: e^{x} \tan y d x+\left(2-e^{x}\right) \sec ^{2} y d y=0 ; y(0)=\frac{\pi}{4}$Solution:$\begin{aligned} &\quad e^{x} \tan y d x+\left(2-e^{x}\right) \sec ^{2} y d y=0 \\\\ &\Rightarrow e^{x} \tan y d x=-\left(2-e^{x}\right) \sec ^{2} y d y \\\\ &\Rightarrow e^{x} \tan y d x=\left(e^{x}-2\right) \sec ^{2} y d y \\\\ &\Rightarrow \frac{e^{x}}{e^{x}-2} d x=\frac{\sec ^{2} y}{\tan y} d y \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{e^{x}}{e^{x}-2} d x=\int \frac{\sec ^{2} y}{\tan y} d y \\\\ &\Rightarrow \log \left|e^{x}-2\right|=\log |\tan y|+\log |c| \\\\ &{\left[\therefore \int \frac{1}{x} d x=\log |x|+c\right]} \end{aligned}$$\begin{aligned} &\Rightarrow \log \left|e^{x}-2\right|-\log |\tan y|=\log |c| \\\\ &\Rightarrow \log \left|\frac{e^{x}-2}{\tan y}\right|=\log |c| \\\\ &{\left[\therefore \log m-\log n=\log \frac{m}{n}\right]} \end{aligned}$Given that
$y(0)=\frac{\pi}{4} \text { i.e when } x=0 ; y=\frac{\pi}{4}$$\begin{aligned} &\Rightarrow \log \left|\frac{e^{0}-2}{\tan \frac{\pi}{4}}\right|=\log |c| \Rightarrow \log \left|\frac{1-2}{1}\right|=\log c \\\\ &{\left[\therefore e^{0}=1, \tan \frac{\pi}{4}=1\right]} \\\\ &\Rightarrow \log 1=\log c \Rightarrow \log c=0 \end{aligned}$Put in (1)
$\begin{aligned} &\Rightarrow \log \left|\frac{e^{x}-2}{\tan y}\right|=0 \Rightarrow \frac{e^{x}-2}{\tan y}=e^{0} \\\\ &\Rightarrow e^{x}-2=\tan y(1) \\\\ &\begin{array}{l} {\left[\therefore e^{0}=1\right]\left[\begin{array}{l} \therefore \log _{e} a=x \\ a=e^{x} \end{array}\right]} \\\\ \Rightarrow e^{x}-\tan y=2 \end{array} \end{aligned}$Differential Equations exercise 21.7 question 46
Answer: $x=2 \cos y$Hint: Separate the terms of x and y and then integrate them.
Given: $x \frac{d y}{d x}+\cot y=0, y=\frac{\pi}{4} \text { when } x=\sqrt{2}$Solution: $\begin{aligned} &x \frac{d y}{d x}+\cot y=0 \\\\ &\Rightarrow x \frac{d y}{d x}=-\cot y \\\\ &\Rightarrow \frac{d y}{\cot y}=-\frac{d x}{x} \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{\cot y} d y=-\int \frac{1}{x} d x \\\\ &-\int \frac{\sin y}{\cos y} d y=\int \frac{1}{x} d x \\\\ &\Rightarrow \log |\cos y|=\log |x|+c \end{aligned}$$\begin{aligned} &{\left[\therefore \int \frac{1}{x} d x=\log |x|+c\right]} \\\\ &\Rightarrow \log |\cos y|-\log |x|=c \end{aligned}$ ...............(1)
Now, when
$x=\sqrt{2}, y=\frac{\pi}{4}$$\begin{aligned} &\therefore \log \left|\cos \frac{\pi}{4}\right|-\log |\sqrt{2}|=c \\\\ &\Rightarrow \log \frac{1}{\sqrt{2}}-\log \sqrt{2}=c \Rightarrow \log (2)^{-\frac{1}{2}}-\log (2)^{\frac{1}{2}}=c \end{aligned}$$\begin{aligned} &\Rightarrow-\frac{1}{2} \log 2-\frac{1}{2} \log 2=c \Rightarrow-2\left(\frac{1}{2} \log 2\right)=c \\\\ &\Rightarrow-\log 2=c \end{aligned}$Put in (1)
$\begin{aligned} &\Rightarrow \log |\cos y|-\log |x|=-\log 2 \\\\\ &\Rightarrow \log \left|\frac{\cos y}{x}\right|+\log 2=0 \end{aligned}$$\left[\begin{array}{l} \therefore \log (m)+\log (n)=\log m n \\\\ \log (m)-\log (n)=\log \frac{m}{n} \end{array}\right]$$\begin{aligned} &\Rightarrow \log \left|\frac{2 \cos y}{x}\right|=0 \\\\ &{\left[\begin{array}{l} \therefore \log _{\varepsilon} a=x \\\\ x=a^{x} \end{array}\right]} \\\\ &\Rightarrow \frac{2 \cos y}{x}=e^{0} \end{aligned}$$\begin{aligned} &{\left[e^{0}=1\right]} \\\\ &\Rightarrow 2 \cos y=x(1) \\\\ &\Rightarrow 2 \cos y=x \end{aligned}$Differential Equations exercise 21.7 question 47
Answer: $x+y+x y=1$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0, y=1 \text { when } x=0$Solution: $\begin{aligned} &\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0 \\\\ &\left(1+x^{2}\right) \frac{d y}{d x}=-\left(1+y^{2}\right) \\\\ &\Rightarrow \frac{d y}{1+y^{2}}=-\frac{1}{1+x^{2}} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\Rightarrow \int \frac{1}{\left(1+y^{2}\right)} d y=-\int \frac{1}{\left(1+x^{2}\right)} \mathrm{dx} \\\\ &\tan ^{-1} y=-\tan ^{-1} x+c \end{aligned}$ ................(1)
Given that
$y=1 \text { when } x=0$$\therefore \tan ^{-1}(1)=-\tan ^{-1}(0)+c \Rightarrow \frac{\pi}{4}=0+c \Rightarrow c=\frac{\pi}{4}$Put in (1) we get
$\begin{aligned} &\Rightarrow \tan ^{-1} y=-\tan ^{-1} x+\frac{\pi}{4} \\\\ &{\left[\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]} \\\\ &\Rightarrow \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4} \end{aligned}$$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4} \\\\ &\Rightarrow\left(\frac{x+y}{1-x y}\right)=\tan \frac{\pi}{4} \\\\ \end{aligned}$$\begin{aligned} &\Rightarrow\left(\frac{x+y}{1-x y}\right)=1 \\\\ &\Rightarrow x+y=1-x y \\\\ &\Rightarrow x+y+x y=1 \end{aligned}$Differential Equations exercise 21.7 question 48 maths
Answer: $2 y \sin y=2 x^{2} \log x+x^{2}-1$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=\frac{2 x(\log x+1)}{\sin y+y \cos y}, y=0 \text { when } x=1$Solution: $\begin{aligned} &\frac{d y}{d x}=\frac{2 x(\log x+1)}{\sin y+y \cos y} \\\\ &\Rightarrow(\sin y+y \cos y) d y=2 x(\log x+1) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\Rightarrow \int(\sin y+y \cos y) d y=\int 2 x(\log x+1) d x \\\\ &\Rightarrow \int \sin y d y+\int y \cos y d y=\int 2 x \log x d x+\int 2 x d x \end{aligned}$ [?Integration by parts]
$\begin{aligned} &\Rightarrow-\cos y+\left[y \sin y-\int \sin y d y\right]=2\left[\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \frac{x^{2}}{2} d x\right] \\\\ &\Rightarrow-\cos y+y \sin y+\cos y=x^{2} \log x-\frac{x^{2}}{2}+c \\\\ &\Rightarrow y \sin y=x^{2} \log x+\frac{x^{2}}{2}+c \end{aligned}$ ...........(1)
Given that
$y=0 \text { when } x=1$$\begin{aligned} &\Rightarrow 0 \sin 0=1 \cdot \log 1+\frac{1}{2}+c \\\\ &\Rightarrow 0=0+\frac{1}{2}+c \\\\ &\Rightarrow c=-\frac{1}{2} \end{aligned}$Put in (1)
$\begin{aligned} &\Rightarrow y \sin y=x^{2} \log x+\frac{x^{2}}{2}-\frac{1}{2} \\\\ &\Rightarrow 2 y \sin y=2 x^{2} \log x+x^{2}-1 \end{aligned}$Differential Equations exercise 21.7 question 49
Answer: $x \log (x+1)-x+\log (x+1)+3$Hint: Separate the terms of x and y and then integrate them.
Given: $e^{\frac{d y}{d x}}=x+1 \text { given that } y=3 \text { when } x=0$Solution: $e^{\frac{d y}{d x}}=x+1$ Integrating both sides
$\begin{aligned} &\log e^{\frac{dy}{dx}}=\log (x+1) \\\\ &\Rightarrow \frac{d y}{d x}=\log (x+1) d x\left[\because \log _{e} e=1\right] \\\\ &\Rightarrow d y=\log (x+1) d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int d y=\int \log (x+1) \cdot 1 d x \\ &\Rightarrow y=\log (x+1) x-\int \frac{1}{x+1} x d x \\\\ &\Rightarrow y=\log (x+1) x-\int \frac{x+1-1}{x+1} d x \\\\ &\Rightarrow y=x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) d x \end{aligned}$ [Integration by parts]
$\begin{aligned} &\Rightarrow y=x \log (x+1)-[x-\log |x+1|]+c \\\\ &\Rightarrow y=x \log (x+1)-x+\log |x+1|+c \end{aligned}$ ..............(1)
Now
$y=3$ when
$x = 0$$\begin{aligned} &3=0 \cdot \log (0+1)-0+\log (0+1)+c \\\\ &\Rightarrow 3=0-0+0+c \Rightarrow c=3[\because \log 1=0] \end{aligned}$Put in (1)
$\Rightarrow y=x \log (x+1)-x+\log |x+1|+3$Differential Equations exercise 21.7 question 50
Answer: $\sin x+\log |\sin y|=1$Hint: Separate the terms of x and y and then integrate them.
Given: $\cos y d y+\cos x \sin y d x=0, y=\frac{\pi}{2} \text { when } x=\frac{\pi}{2}$Solution: $\begin{aligned} &\cos y d y+\cos x \sin y d x=0 \\\\ &\cos y d y=-\cos x \sin y d y \\\\ &\quad \Rightarrow \frac{-\cos y}{\sin y} d y=\cos x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{-\cos y}{\sin y} d y=\int \cos x d x \\\\ &\Rightarrow-\log |\sin y|=\sin x+c \end{aligned}$ ..............(1)
When $x=\frac{\pi}{2}, y=\frac{\pi}{2}$
$\begin{aligned} &\Rightarrow-\log \left|\sin \frac{\pi}{2}\right|=\sin \frac{\pi}{2}+c \\\\ &-\log |1|=1+c \Rightarrow 0-1=c \Rightarrow c=-1 \\\\ &{\left[\therefore \log 1=0, \sin \frac{\pi}{2}=1\right]} \end{aligned}$
Put in (1) we get
$\begin{gathered} \Rightarrow-\log |\sin y|=\sin x-1 \\\\ \quad \Rightarrow \sin x+\log |\sin y|=1 \end{gathered}$
Differential Equations exercise 21.7 question 51
Answer: $2 x^{2} y+y=1$Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x}=-4 x y^{2}, y=1 \text { when } x=0$Solution: $\begin{aligned} &\frac{d y}{d x}=-4 x y^{2} \\ &\Rightarrow \frac{d y}{y^{2}}=-4 x d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1}{y^{2}}=-4 \int x d x \\\\ &\Rightarrow \frac{y^{-2+1}}{-2+1}=\frac{-4 x^{2}}{2}+c \\\\ &\Rightarrow \frac{-1}{y}=-2 x^{2}+c \\ &\Rightarrow-\frac{1}{y}=-2 x^{2}+c \end{aligned}$ ...............(1)
Now
$y=1$ when
$x=0$$\Rightarrow-\frac{1}{1}=-2.0+c=>c=-1$Put in (1)
$\begin{aligned} \Rightarrow \frac{-1}{y} &=-2 x^{2}-1 \end{aligned}$$\Rightarrow 2 x^{2} y+y=1$Differential Equations exercise 21.7 question 52 maths
Answer: $2 y=e^{x}(\sin x-\cos x)+1$Hint: Separate the terms of x and y and then integrate them.
Given: Curve passing through (0,0) with differential equation
$\frac{d y}{d x}=e^{x} \sin x$
Solution:
$\begin{aligned} &\frac{d y}{d x}=e^{x} \sin x \\\\ &\Rightarrow d y=e^{x} \sin x d x \end{aligned}$
Integrating both sides
$\int d y=\int e^{x} \sin x d x \Rightarrow y=\int e^{x} \sin x d x$ .................(*)
$\begin{aligned} &y=\sin x e^{x}-\int \cos x e^{x} d x \\\\ &\Rightarrow y=\sin x e^{x}-\left[\cos x e^{x}+\int \sin x e^{x} d x\right]+c \\\\ &\Rightarrow y=\sin x e^{x}-\cos x e^{x}-y+c \\\\ &\Rightarrow 2 y=e^{x}(\sin x-\cos x)+c \end{aligned}$..............(1)
The curve passes through (0,0) [∴ of given]
$\begin{aligned} &0=e^{0}(\sin 0-\cos 0)+c\\\\ &\Rightarrow 0=1(0-1)+c \Rightarrow c=1\\\\ &\operatorname{Put\; in}(1)\\\\ &2 y=e^{x}(\sin x-\cos x)+1 \end{aligned}$
Differential Equations exercise 21.7 question 53
Answer: $y=x+\log \left(\frac{x y+2 x}{3}\right)^{2}+2$Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x}=(x+2)(y+2) ; p t(1,-1)$Solution: $\begin{aligned} &x y \frac{d y}{d x}=(x+2)(y+2) \\\\ &\Rightarrow \frac{y}{y+2} d y=\frac{x+2}{x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{y}{y+2} d y=\int \frac{x+2}{x} d x \\\\ &\int \frac{y+2-2}{y+2} d y=\int\left(1+\frac{2}{x}\right) d x \\\\ &\Rightarrow \int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x \end{aligned}$$\begin{aligned} &\Rightarrow y-2 \log |y+2|=x+2 \log |x|+c \\\\ &\Rightarrow y=x+2 \log |x|+2 \log |y+2|+c \\\\ &\Rightarrow y=x+2 \log |x(y+2)|+c \\\\ &\Rightarrow y=x+\log (x y+2 x)^{2}+c \end{aligned}$ ..............(1)
It passes through pt(-1,1)
$\begin{aligned} &1=-1+\log |(-1)(1)+2(-1)|^{2}+c \\\\ &\Rightarrow 2=\log (-1-2)^{2}+c \\\\ &\Rightarrow 2-\log 9=c \end{aligned}$Put in (1) we get
$\begin{aligned} &\Rightarrow y=x+\log (x y+2 x)^{2}+2-\log 9 \\\\ &\Rightarrow y=x+\log \left(\frac{x y+2 x}{3}\right)^{2}+2 \\\\ &{\left[\therefore \log m-\log n=\log \frac{m}{n}\right]} \end{aligned}$Differential Equations exercise 21.7 question 54
Answer: $(63 t+27)^{\frac{1}{3}}$Hint: Separate the terms of x and y and then integrate them.
Given: The volume of a spherical balloon being inflated changes at a constant rate. If initially it radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds
Solution: Let V be the volume of spherical balloon
Now it is being inflated changes at a constant rate
$\because \frac{d v}{d t}=k$where k is any constant ……………….(*)
New volume of spherical balloon
$=\frac{4}{3} \pi r^{3}$, r is radius
$\begin{aligned} &\frac{d v}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}\\\\ &\Rightarrow k=4 \pi r^{2} \frac{d r}{d t}\\\\ &\Rightarrow d t=\frac{4 \pi}{k} r^{2} d r \end{aligned}$Integrating both sides
$\begin{aligned} &\int d t=\frac{4 \pi}{k} \int r^{2} d r\\\\ &\Rightarrow t=\frac{4 \pi}{k}\left[\frac{r^{3}}{3}\right]+c\\\\ &\Rightarrow \frac{r^{8}}{3}=\frac{k}{4 \pi} t+c \end{aligned}$ ...........(1)
Now given conditions are:
When
$t = 0; r = 3$ and when
$t = 3, r = 6$We have to find r at
$t = t$Put in (1) we get
$\begin{aligned} &\text { At } t=0, r=3 \Rightarrow \frac{3^{8}}{3}=\frac{k}{4 \pi} 0+c \Rightarrow 9=c \\ \end{aligned}$ ?By(1)
$\frac{r^{8}}{3}=\frac{k}{4 \pi} t+9$ ..................(2)
Now at
$\mathrm{t}=3, \mathrm{r}=6 \Rightarrow \frac{6^{3}}{3}=\frac{k}{4 \pi} 3+9$$\Rightarrow \frac{6 \times 6 \times 6}{3}-9=\frac{3 k}{4 \pi}$$\begin{aligned} &\Rightarrow 72-9=\frac{3 k}{4 \pi} \Rightarrow \frac{63 \times 4 \pi}{3}=k \\ &\Rightarrow k=84 \pi \end{aligned}$Put in (2)
$\begin{aligned} &\frac{r^{3}}{3}=\frac{84 \pi}{4 \pi} t+9 \Rightarrow \frac{r^{3}}{3}=21 t+9 \\\\ &\Rightarrow r^{3}=63 t+27 \\\\ &\Rightarrow r=(63 t+27)^{\frac{1}{3}} \end{aligned}$which is our required radius.
Differential Equations exercise 21.7 question 55
Answer: $6.931$Hint: Separate the terms of x and y and then integrate them.
Given: In a bank principal increases at the rate of r% per year. We have to find the value of r if Rs.100 double itself in 10 years (log 2=0.6931)
Solution: Let Principal = P and rate = r
As principal increases at the rate of r% w.r.t time i.e. per year.
$\begin{aligned} &\therefore \frac{d P}{d t}=\frac{r}{100} \times P \\\\ &\Rightarrow \frac{d P}{P}=\frac{r}{100} d t \end{aligned}$Integrating both sides
$\Rightarrow \log P=\frac{r}{100} t+c$ ..................(1)
Suppose initially
$t = 0,P = P_{0}$$\begin{aligned} &\Rightarrow \log P_{0}=\frac{r}{100} 0+c \\\\ &\Rightarrow c=\log P_{0} \end{aligned}$Put in (1)
$\begin{aligned} &\Rightarrow \log P=\frac{r}{100} t+\log P_{0} \\\\ &\log P-\log P_{0}=\frac{r}{100} t \end{aligned}$ ..............(2)
According to given
When
$t = 10, P = 2 * 100, P_{0}=100$By (2)
$\begin{aligned} &\log 200-\log 100=\frac{r}{100} 10 \\\\ &\Rightarrow \log \left(\frac{200}{100}\right)=\frac{r}{10} \\\\ &\Rightarrow \log 2=\frac{r}{10} \Rightarrow 10 \log _{e} 2=r \\ &\Rightarrow r=10 \times 0.6931=6.931 \\ &\Rightarrow r=6.931 \end{aligned}$Differential Equations exercise 21.7 question 56 maths
Answer: $1648$Hint: Separate the terms of x and y and then integrate them.
Given: In a bank principal increases increases at the rate of 5% per year.
An amount of Rs.1000 is deposited with this bank, how much will it worth after 10 years.(e0.5 =1.648)
Solution: Let P be the Principal
As Principal increases at the rate of 5% w.r.t t
$\begin{aligned} &\therefore \frac{d P}{d t}=\frac{5}{100} \times P \\\\ &\Rightarrow \frac{d P}{P}=\frac{1}{20} d t \end{aligned}$Integrating both sides
$\int \frac{d P}{P}=\frac{1}{20} \int 1 d t$ ................(1)
Now initially P
0=1000; after 10 years P= P
10 also initially t =0 and after 10 years t = 10
By (1)
$\begin{aligned} &\int_{P_{0}}^{P_{0}} \frac{d P}{P}=\frac{1}{20} \int_{0}^{+} 1 d t \\\\ &\Rightarrow[\log |P|]_{1000}^{P_{0}}=\frac{1}{20}[t]_{0}^{10} \\\\ &\Rightarrow \log P_{10}-\log 1000=\frac{1}{20}(10-0) \end{aligned}$$\begin{aligned} &\Rightarrow\left[\log \left|\frac{P_{10}}{1000}\right|\right]=\frac{1}{2} \\\\ &{\left[\log m-\log n=\log \frac{m}{n}\right]} \\\\ &\Rightarrow\left|\frac{P_{10}}{1000}\right|=e^{\frac{1}{2}} \end{aligned}$$\begin{aligned} &{\left[\begin{array}{l} \log _{a} e=x \\ a=e^{x} \end{array}\right]} \\\\ &\Rightarrow P_{10}=1000(1.648) \\\\ &{\left[e^{\frac{1}{2}}=1.648\right]} \\\\ &\Rightarrow P_{10}=1648 \end{aligned}$Principal after 10 years = 1648
Differential Equations exercise 21.7 question 57
Answer: $\frac{2 \log 2}{\log \frac{11}{10}}$Hint: Separate the terms of x and y and then integrate them.
Given: In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. We have to find, in how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
Solution: Let the count of bacteria be N
As the rate of growth of bacteria is proportional to the no. present
$\begin{aligned} &\therefore \frac{d N}{d t} \alpha N \\\\ &\Rightarrow \frac{d N}{d t}=K N \\\\ &\Rightarrow \frac{d N}{N}=K d t \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{d N}{N}=K \int d t\\\\ &\Rightarrow \log |N|=K t+c \end{aligned}$ .................(1)
Initially
$N=100000;\; t=0$By (1)
$\begin{aligned} &\log 100000=K(0)+c \\\\ &\Rightarrow c=\log 100000 \end{aligned}$Put in (1) we get
$\log N=K t+\log 100000$ ...................(2)
According to given
When
$\begin{aligned} \mathrm{t}=2, \mathrm{~N} &=10 \% \text { of } 100000+100000 \\ &=10000+100000=110000 \end{aligned}$$\begin{aligned} &\operatorname{By}(2) \log 110000=\mathrm{K}(2)+\log 100000 \\\\\ &\log 110000=K(2)+\log 100000 \\\\ &\Rightarrow \log \frac{110000}{100000}=K(2) \\\\ &{\left[\log m-\log n=\log \frac{m}{n}\right]} \end{aligned}$$\begin{aligned} &{\left[\log m-\log n=\log \frac{m}{n}\right]} \\\\ &\Rightarrow K=\frac{1}{2} \log \frac{11}{10} \\\\ &\therefore \log N=\frac{1}{2} \log \frac{11}{10} t+\log 100000 \end{aligned}$ ....................(3)
Now we have to find in how many hours i.e t
1 ; N=200000
By (3)
$\begin{aligned} &\log 200000=\frac{1}{2}\left(\log \frac{11}{10}\right) t^{1}+\log 100000 \\\\ &\Rightarrow \log \left|\frac{200000}{100000}\right|=\frac{1}{2} \log \left(\frac{11}{10}\right) t^{1} \\\\ &\Rightarrow \log 2=\frac{1}{2} \log \left(\frac{11}{10}\right) t^{1} \end{aligned}$$\begin{aligned} &\Rightarrow 2 \log 2=\log \left(\frac{11}{10}\right) t^{1} \\\\ &\Rightarrow \frac{2 \log 2}{\log \left(\frac{11}{10}\right)}=t^{1} \\\\ &\Rightarrow t^{1}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \end{aligned}$Differential Equations exercise 21.7 question 58
Answer: $\frac{1}{3}$Hint: Separate the terms of x and y and then integrate them.
Given: IF
$y(x)$ is a solution of the differential equation
$\left(\frac{2+\sin x}{1+y}\right) \frac{d y}{d x}=-\cos x \text { and } y(0)=1$ then find
$y\left ( \frac{\pi }{2} \right )$Solution: $\begin{aligned} &\left(\frac{2+\sin x}{1+y}\right) \frac{d y}{d x}=-\cos x \\\\ &\Rightarrow \frac{d y}{1+y}=\frac{-\cos x}{2+\sin x} d x \end{aligned}$Integrating both sides
$\begin{aligned} &\Rightarrow \int \frac{d y}{1+y}=-\int \frac{\cos x}{2+\sin x} d x \\\\ &\Rightarrow \log |1+y|=-\log |2+\sin x|+\log c \\\\ &{\left[\begin{array}{l} \text { Put } \\ \sin x=t \\ \cos x d x=d t \\ \int \frac{1}{t} d t=\log |t|+c \end{array}\right]} \end{aligned}$$\begin{aligned} &\Rightarrow \log |y+1|+\log |2+\sin x|=\log c \\\\ &\Rightarrow \log |(y+1)(2+\sin x)|=\log c \\\\ &\Rightarrow(y+1)(2+\sin x)=c \end{aligned}$ ..........(1)
According to given:
$y(0) = 1\; at\; x=0, y = 1$$\text { (2) }(2+\sin 0)=c \Rightarrow c=4$Put in (1)
$\Rightarrow(y+1)(2+\sin x)=4$ .............(2)
Now we have to find
$y\left ( \frac{\pi }{2} \right )$i.e. value of y at
$x= \frac{\pi }{2}$Put in (2)
$\begin{aligned} &(y+1)\left(2+\sin \frac{\pi}{2}\right)=4 \\\\ &\Rightarrow(y+1)(2+1)=4 \\\\ &{\left[\therefore \sin \frac{\pi}{2}=1\right]} \end{aligned}$$\begin{aligned} &\Rightarrow 3(y+1)=4 \\\\ &\Rightarrow 3 y+3=4 \Rightarrow 3 y=1 \Rightarrow y=\frac{1}{3} \\\\ &y\left(\frac{\pi}{2}\right)=\frac{1}{3} \end{aligned}$Differential Equations exercise 21.7 question 59
Answer: $(1+\log x)^{2}=\log \left(1-y^{2}\right)^{2}+1$Hint: Separate the terms of x and y and then integrate them.
Given: $\left(1-y^{2}\right)(1+\log x) d x+2 x y d y=0 \text { given that } y=0 \text { when } x=1$Solution: $\begin{aligned} &\left(1-y^{2}\right)(1+\log x) d x=-2 x y d y \\\\ &\Rightarrow \frac{1+\log x}{x} d x=\frac{-2 y}{1-y^{2}} d y \end{aligned}$Integrating both sides
$\begin{aligned} &\int \frac{1+\log x}{x} d x=\int \frac{-2 y}{1-y^{2}} d y \ldots \ldots \ldots \ldots . .(*) \\ &\Rightarrow I_{1}=I_{2} \text { where } I_{1}=\int \frac{1}{x}(1+\log x) d x \text { and } \\\\ &I_{2}=\int \frac{-2 y}{1-y^{2}} d y \\\\ &\therefore I_{1}=\int \frac{1}{x}(1+\log x) d x \end{aligned}$Put
$\begin{aligned} &1+\log x=t \\\\ &\frac{1}{x} d x=d t \\\\ &=\int t d t \\\\ &=\frac{t^{2}}{2}+c=\frac{(1+\log x)^{2}}{2}+c \end{aligned}$Now,
$I_{2}=\int \frac{-2 y}{1-y^{2}} d y$Put
$\begin{aligned} &1-y^{2}=t \\\\ &-2 y d y=d t \\\\ &=\int \frac{d t}{t} \\\\ &=\log |t|+c \\\\ &=\log \left|1-y^{2}\right|+c \end{aligned}$Put the values in (*) we get
$\frac{(1+\log x)^{2}}{2}=\log \left|1-y^{2}\right|+c$ ................(1)
Now according to given
$y = 0$ when
$x = 1$$\begin{aligned} &\therefore \frac{(1+\log 1)^{2}}{2}=\log |1-0|+c \\\\ &\Rightarrow \frac{(1+0)^{2}}{2}=\log (1)+c \\\\ &{[\log 1=0]} \end{aligned}$$\begin{aligned} &\Rightarrow 1=2[(0)+c] \\\\ &\Rightarrow 1=2 c \Rightarrow c=\frac{1}{2} \end{aligned}$Put in (1) we get
$\begin{aligned} &\Rightarrow \frac{(1+\log x)^{2}}{2}=\log \left|1-y^{2}\right|+\frac{1}{2} \\\\ &\Rightarrow(1+\log x)^{2}=2 \log \left(1-y^{2}\right)+1 \\\\ &\Rightarrow(1+\log x)^{2}=\log \left(1-y^{2}\right)^{2}+1 \end{aligned}$
The Differential Equation chapter 21 in mathematics of class 12 has about eleven exercises. The seventh exercise, ex 21.7, has 72 questions, with some of them having subparts in the textbook. RD Sharma solutions The concept of these questions revolves around solving the differential equations, initial value problems, equations in variable separable form and word problems using differentiation. This exercise has questions only in the Level 1 category. Yet, the importance of RD Sharma Class 12 Chapter 21 Exercise 21.7 solution book has not been reduced.
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