Class 12: Differential Equations

$\displaystyle \textbf{Solve the following differential equations:}$

$\displaystyle \textbf{Question 1: }~(x-1)\frac{dy}{dx}=2xy.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (x-1)\frac{dy}{dx}=2xy$
$\displaystyle \Rightarrow (x-1)\,dy=2xy\,dx$
$\displaystyle \Rightarrow \frac{2x}{x-1}\,dx=\frac{1}{y}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{2x}{x-1}\,dx=\int \frac{1}{y}\,dy$
$\displaystyle \Rightarrow \int \left(2+\frac{2}{x-1}\right)\,dx=\int \frac{1}{y}\,dy$
$\displaystyle \Rightarrow \int 2\,dx+\int \frac{2}{x-1}\,dx=\int \frac{1}{y}\,dy$
$\displaystyle \Rightarrow 2x+2\log|x-1|=\log|y|+C$

$\displaystyle \textbf{Question 2: }~(1+x^{2})\,dy=xy\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (1+x^2)\,dy=xy\,dx$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=\frac{x}{1+x^2}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int \frac{x}{1+x^2}\,dx$
$\displaystyle \text{Substituting } 1+x^2=t, \text{ we get}$
$\displaystyle dt=2x\,dx$
$\displaystyle \Rightarrow \int \frac{1}{y}\,dy=\frac{1}{2}\int \frac{1}{t}\,dt$
$\displaystyle \Rightarrow \log|y|=\frac{1}{2}\log|t|+C$
$\displaystyle \Rightarrow \log|y|=\frac{1}{2}\log|1+x^2|+C \;(\because t=1+x^2)$
$\displaystyle \Rightarrow \log|y|=\log\left(C\sqrt{1+x^2}\right)$
$\displaystyle \Rightarrow y=C\sqrt{1+x^2}$
$\displaystyle \text{Hence, } y=C\sqrt{1+x^2} \text{ is the required solution.}$

$\displaystyle \textbf{Question 3: }~\frac{dy}{dx}=(e^{x}+1)y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=(e^x+1)y$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=(e^x+1)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int (e^x+1)\,dx$
$\displaystyle \Rightarrow \log|y|=e^x+x+C$
$\displaystyle \text{Hence, } \log|y|=e^x+x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 4: }~(x-1)\frac{dy}{dx}=2x^{3}y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (x-1)\frac{dy}{dx}=2x^3y$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=\frac{2x^3}{x-1}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int \frac{2x^3}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=2\int \frac{x^3}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=2\int \frac{(x-1)(x^2+x+1)+1}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=2\int (x^2+x+1)\,dx+2\int \frac{1}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=\frac{2}{3}x^3+x^2+2x+2\log|x-1|+C$
$\displaystyle \text{Hence, } \log|y|=\frac{2}{3}x^3+x^2+2x+2\log|x-1|+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 5: }~xy(y+1)\,dy=(x^{2}+1)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle xy(y+1)\,dy=(x^2+1)\,dx$
$\displaystyle \Rightarrow (y^2+y)\,dy=\left(x+\frac{1}{x}\right)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int (y^2+y)\,dy=\int \left(x+\frac{1}{x}\right)\,dx$
$\displaystyle \Rightarrow \int y^2\,dy+\int y\,dy=\int x\,dx+\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow \frac{y^3}{3}+\frac{y^2}{2}=\frac{x^2}{2}+\log|x|+C$
$\displaystyle \text{Hence, } \frac{y^3}{3}+\frac{y^2}{2}=\frac{x^2}{2}+\log|x|+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 6: }~5\frac{dy}{dx}=e^{x}y^{4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle 5\frac{dy}{dx}=e^x y^4$
$\displaystyle \Rightarrow \frac{5}{y^4}\,dy=e^x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{5}{y^4}\,dy=\int e^x\,dx$
$\displaystyle \Rightarrow -\frac{5}{3y^3}=e^x+C$
$\displaystyle \text{Hence, } -\frac{5}{3y^3}=e^x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 7: }~x\cos y\,dy=(xe^{x}\log x+e^{x})\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle x\cos y\,dy=(xe^x\log x+e^x)\,dx$
$\displaystyle \Rightarrow \cos y\,dy=\left(e^x\log x+\frac{1}{x}e^x\right)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \cos y\,dy=\int \left(e^x\log x+\frac{1}{x}e^x\right)\,dx$
$\displaystyle \Rightarrow \sin y=\int e^x\log x\,dx+\int \frac{1}{x}e^x\,dx$
$\displaystyle \Rightarrow \sin y=e^x\log x+C$
$\displaystyle \text{Hence, } \sin y=e^x\log x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 8: }~\frac{dy}{dx}=e^{x+y}+x^{2}e^{y}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=e^{x+y}+x^2e^y$
$\displaystyle \Rightarrow \frac{dy}{dx}=e^y(e^x+x^2)$
$\displaystyle \Rightarrow e^{-y}\,dy=(e^x+x^2)\,dx$
$\displaystyle \text{Integrating on both sides, we get}$
$\displaystyle \int e^{-y}\,dy=\int (e^x+x^2)\,dx$
$\displaystyle \Rightarrow -e^{-y}=e^x+\frac{x^3}{3}+C$
$\displaystyle \Rightarrow e^{-y}=-e^x-\frac{x^3}{3}+C$
$\displaystyle \Rightarrow e^{-y}+e^x+\frac{x^3}{3}=C$

$\displaystyle \textbf{Question 9: }~x\frac{dy}{dx}+y=y^{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle x\frac{dy}{dx}+y=y^2$
$\displaystyle \Rightarrow x\frac{dy}{dx}=y^2-y$
$\displaystyle \Rightarrow \frac{1}{y^2-y}\,dy=\frac{1}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y^2-y}\,dy=\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow \int \frac{1}{y(y-1)}\,dy=\int \frac{1}{x}\,dx$
$\displaystyle \text{Let } \frac{1}{y(y-1)}=\frac{A}{y}+\frac{B}{y-1}$
$\displaystyle \Rightarrow 1=A(y-1)+By$
$\displaystyle \text{Putting } y=0, \text{ we get } 1=-A$
$\displaystyle \Rightarrow A=-1$
$\displaystyle \text{Putting } y=1, \text{ we get } 1=B$
$\displaystyle \Rightarrow \frac{1}{y(y-1)}=-\frac{1}{y}+\frac{1}{y-1}$
$\displaystyle \Rightarrow \int \frac{1}{y(y-1)}\,dy=\int \left(-\frac{1}{y}+\frac{1}{y-1}\right)\,dy$
$\displaystyle \Rightarrow -\int \frac{1}{y}\,dy+\int \frac{1}{y-1}\,dy=\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow -\log|y|+\log|y-1|=\log|x|+C$
$\displaystyle \Rightarrow \log\left|\frac{y-1}{y}\right|-\log|x|=C$
$\displaystyle \Rightarrow \log\left|\frac{y-1}{xy}\right|=C$
$\displaystyle \Rightarrow \frac{y-1}{xy}=C$
$\displaystyle \Rightarrow y-1=Cxy$
$\displaystyle \text{Hence, } y-1=Cxy \text{ is the required solution.}$

$\displaystyle \textbf{Question 10: }~(e^{y}+1)\cos x\,dx+e^{y}\sin x\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (e^y+1)\cos x\,dx+e^y\sin x\,dy=0$
$\displaystyle \Rightarrow e^y\sin x\,dy=-(e^y+1)\cos x\,dx$
$\displaystyle \Rightarrow \frac{e^y}{e^y+1}\,dy=-\frac{\cos x}{\sin x}\,dx$
$\displaystyle \Rightarrow \frac{e^y}{e^y+1}\,dy=-\cot x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{e^y}{e^y+1}\,dy=-\int \cot x\,dx$
$\displaystyle \text{Putting } e^y+1=t, \text{ we get}$
$\displaystyle dt=e^y\,dy$
$\displaystyle \Rightarrow \int \frac{1}{t}\,dt=-\int \cot x\,dx$
$\displaystyle \Rightarrow \log|t|=-\log|\sin x|+C$
$\displaystyle \Rightarrow \log|e^y+1|+\log|\sin x|=C$
$\displaystyle \Rightarrow \log|(e^y+1)\sin x|=C$
$\displaystyle \Rightarrow (e^y+1)\sin x=C$
$\displaystyle \text{Hence, } (e^y+1)\sin x=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 11: }~x\cos^{2}y\,dx=y\cos^{2}x\,dy.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle x\cos^2 y\,dx=y\cos^2 x\,dy$
$\displaystyle \Rightarrow \frac{x}{\cos^2 x}\,dx=\frac{y}{\cos^2 y}\,dy$
$\displaystyle \Rightarrow x\sec^2 x\,dx=y\sec^2 y\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int x\sec^2 x\,dx=\int y\sec^2 y\,dy$
$\displaystyle \Rightarrow x\int \sec^2 x\,dx-\int \frac{d}{dx}(x)\int \sec^2 x\,dx\,dx$
$\displaystyle =y\int \sec^2 y\,dy-\int \frac{d}{dy}(y)\int \sec^2 y\,dy\,dy$
$\displaystyle \Rightarrow x\tan x-\int \tan x\,dx=y\tan y-\int \tan y\,dy$
$\displaystyle \Rightarrow x\tan x-\log|\sec x|=y\tan y-\log|\sec y|+C$
$\displaystyle \Rightarrow x\tan x-y\tan y=\log|\sec x|-\log|\sec y|+C$
$\displaystyle \text{Hence, } x\tan x-y\tan y=\log|\sec x|-\log|\sec y|+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 12: }~xy\,dy=(y-1)(x+1)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle xy\,dy=(y-1)(x+1)\,dx$
$\displaystyle \Rightarrow \frac{y}{y-1}\,dy=\frac{x+1}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{y-1}\,dy=\int \frac{x+1}{x}\,dx$
$\displaystyle \Rightarrow \int \left(1+\frac{1}{y-1}\right)\,dy=\int \left(1+\frac{1}{x}\right)\,dx$
$\displaystyle \Rightarrow \int dy+\int \frac{1}{y-1}\,dy=\int dx+\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow y+\log|y-1|=x+\log|x|+C$
$\displaystyle \Rightarrow y-x=\log|x|-\log|y-1|+C$
$\displaystyle \text{Hence, } y-x=\log|x|-\log|y-1|+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 13: }~x\frac{dy}{dx}+\cot y=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle x\frac{dy}{dx}+\cot y=0$
$\displaystyle \Rightarrow x\frac{dy}{dx}=-\cot y$
$\displaystyle \Rightarrow \frac{1}{x}\,dx=-\frac{1}{\cot y}\,dy$
$\displaystyle \Rightarrow \frac{1}{x}\,dx=-\tan y\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{x}\,dx=-\int \tan y\,dy$
$\displaystyle \Rightarrow \log|x|=-\log|\sec y|+C$
$\displaystyle \Rightarrow \log|x|=\log|\cos y|+C$
$\displaystyle \Rightarrow x=C\cos y$
$\displaystyle \text{Hence, } x=C\cos y \text{ is the required solution.}$

$\displaystyle \textbf{Question 14: }~\frac{dy}{dx}=\frac{xe^{x}\log x+e^{x}}{x\cos y}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=\frac{xe^x\log x+e^x}{x\cos y}$
$\displaystyle \Rightarrow x\cos y\,dy=(xe^x\log x+e^x)\,dx$
$\displaystyle \Rightarrow \cos y\,dy=\left(e^x\log x+\frac{1}{x}e^x\right)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \cos y\,dy=\int \left(e^x\log x+\frac{1}{x}e^x\right)\,dx$
$\displaystyle \Rightarrow \sin y=\int e^x\log x\,dx+\int \frac{1}{x}e^x\,dx$
$\displaystyle \Rightarrow \sin y=e^x\log x+C$
$\displaystyle \text{Hence, } \sin y=e^x\log x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 15: }~\frac{dy}{dx}=e^{x+y}+e^{y}x^{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=e^{x+y}+e^y x^3$
$\displaystyle \Rightarrow \frac{dy}{dx}=e^y(e^x+x^3)$
$\displaystyle \Rightarrow (e^x+x^3)\,dx=\frac{1}{e^y}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int (e^x+x^3)\,dx=\int e^{-y}\,dy$
$\displaystyle \Rightarrow e^x+\frac{x^4}{4}=-e^{-y}+C$
$\displaystyle \Rightarrow e^x+e^{-y}+\frac{x^4}{4}=C$
$\displaystyle \text{Hence, } e^x+e^{-y}+\frac{x^4}{4}=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 16: }~y\sqrt{1+x^{2}}+x\sqrt{1+y^{2}}\frac{dy}{dx}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle y\sqrt{1+x^2}+x\sqrt{1+y^2}\frac{dy}{dx}=0$
$\displaystyle \Rightarrow x\sqrt{1+y^2}\frac{dy}{dx}=-y\sqrt{1+x^2}$
$\displaystyle \Rightarrow x\sqrt{1+y^2}\,dy=-y\sqrt{1+x^2}\,dx$
$\displaystyle \Rightarrow \frac{\sqrt{1+y^2}}{y}\,dy=-\frac{\sqrt{1+x^2}}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{\sqrt{1+y^2}}{y}\,dy=-\int \frac{\sqrt{1+x^2}}{x}\,dx$
$\displaystyle \text{Putting } 1+y^2=t^2 \text{ and } 1+x^2=u^2, \text{ we get}$
$\displaystyle 2y\,dy=2t\,dt \text{ and } 2x\,dx=2u\,du$
$\displaystyle \Rightarrow dy=\frac{t}{y}\,dt \text{ and } dx=\frac{u}{x}\,du$
$\displaystyle \Rightarrow \int \frac{t^2}{y^2}\,dt=-\int \frac{u^2}{x^2}\,du$
$\displaystyle \Rightarrow \int \frac{t^2}{t^2-1}\,dt=-\int \frac{u^2}{u^2-1}\,du$
$\displaystyle \Rightarrow \int \frac{t^2-1+1}{t^2-1}\,dt=-\int \frac{u^2-1+1}{u^2-1}\,du$
$\displaystyle \Rightarrow \int dt+\int \frac{1}{t^2-1}\,dt=-\int du-\int \frac{1}{u^2-1}\,du$
$\displaystyle \Rightarrow t+\frac{1}{2}\log\left|\frac{t-1}{t+1}\right|=-u-\frac{1}{2}\log\left|\frac{u-1}{u+1}\right|+C$
$\displaystyle \text{Substituting } t=\sqrt{1+y^2} \text{ and } u=\sqrt{1+x^2}$
$\displaystyle \Rightarrow \sqrt{1+y^2}+\frac{1}{2}\log\left|\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right|$
$\displaystyle =-\sqrt{1+x^2}-\frac{1}{2}\log\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C$
$\displaystyle \Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}\log\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|$
$\displaystyle +\frac{1}{2}\log\left|\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right|=C$
$\displaystyle \Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}\log\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+\frac{1}{2}\log\left|\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right|=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 17: }~\sqrt{1+x^{2}}\,dy+\sqrt{1+y^{2}}\,dx=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \sqrt{1+x^2}\,dy+\sqrt{1+y^2}\,dx=0$
$\displaystyle \Rightarrow \sqrt{1+x^2}\,dy=-\sqrt{1+y^2}\,dx$
$\displaystyle \Rightarrow \frac{1}{\sqrt{1+y^2}}\,dy=-\frac{1}{\sqrt{1+x^2}}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{\sqrt{1+y^2}}\,dy=-\int \frac{1}{\sqrt{1+x^2}}\,dx$
$\displaystyle \Rightarrow \log\left|y+\sqrt{1+y^2}\right|=-\log\left|x+\sqrt{1+x^2}\right|+C$
$\displaystyle \Rightarrow \log\left|y+\sqrt{1+y^2}\right|+\log\left|x+\sqrt{1+x^2}\right|=C$
$\displaystyle \Rightarrow \log\left|(y+\sqrt{1+y^2})(x+\sqrt{1+x^2})\right|=C$
$\displaystyle \Rightarrow (y+\sqrt{1+y^2})(x+\sqrt{1+x^2})=C$
$\displaystyle \text{Hence, } (y+\sqrt{1+y^2})(x+\sqrt{1+x^2})=C \text{ is the required differential equation.}$

$\displaystyle \textbf{Question 18: }~\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\frac{dy}{dx}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \sqrt{1+x^2+y^2+x^2y^2}+xy\frac{dy}{dx}=0$
$\displaystyle \Rightarrow \sqrt{(1+x^2)(1+y^2)}+xy\frac{dy}{dx}=0$
$\displaystyle \Rightarrow xy\frac{dy}{dx}=-\sqrt{(1+x^2)(1+y^2)}$
$\displaystyle \Rightarrow xy\frac{dy}{dx}=-\sqrt{1+x^2}\sqrt{1+y^2}$
$\displaystyle \Rightarrow \frac{y}{\sqrt{1+y^2}}\,dy=-\frac{\sqrt{1+x^2}}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{\sqrt{1+y^2}}\,dy=-\int \frac{\sqrt{1+x^2}}{x}\,dx$
$\displaystyle \text{Putting } 1+y^2=t \text{ and } 1+x^2=u^2$
$\displaystyle \Rightarrow 2y\,dy=dt \text{ and } 2x\,dx=2u\,du$
$\displaystyle \Rightarrow y\,dy=\frac{dt}{2} \text{ and } x\,dx=u\,du$
$\displaystyle \Rightarrow \frac{1}{2}\int \frac{dt}{\sqrt{t}}=-\int \frac{u^2}{u^2-1}\,du$
$\displaystyle \Rightarrow \sqrt{t}=-\int \left(1+\frac{1}{u^2-1}\right)\,du$
$\displaystyle \Rightarrow \sqrt{t}=-\int du-\int \frac{1}{u^2-1}\,du$
$\displaystyle \Rightarrow \sqrt{t}=-u-\frac{1}{2}\log\left|\frac{u-1}{u+1}\right|+C$
$\displaystyle \Rightarrow \sqrt{1+y^2}=-\sqrt{1+x^2}-\frac{1}{2}\log\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C$
$\displaystyle \Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}\log\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|=C$

$\displaystyle \textbf{Question 19: }~\frac{dy}{dx}=\frac{e^{x}(\sin^{2}x+\sin 2x)}{y(2\log y+1)}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=\frac{e^x(\sin^2 x+\sin 2x)}{y(2\log y+1)}$
$\displaystyle \Rightarrow y(2\log y+1)\,dy=e^x(\sin^2 x+\sin 2x)\,dx$
$\displaystyle \Rightarrow (2y\log y+y)\,dy=e^x\sin^2 x\,dx+e^x\sin 2x\,dx$
$\displaystyle \Rightarrow 2y\log y\,dy+y\,dy=e^x\sin^2 x\,dx+e^x\sin 2x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int 2y\log y\,dy+\int y\,dy=\int e^x\sin^2 x\,dx+\int e^x\sin 2x\,dx$
$\displaystyle \Rightarrow y^2\log y-\int y\,dy+\int y\,dy=e^x\sin^2 x+C$
$\displaystyle \Rightarrow y^2\log y=e^x\sin^2 x+C$

$\displaystyle \textbf{Question 20: }~\frac{dy}{dx}=\frac{x(2\log x+1)}{\sin y+y\cos y}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=\frac{x(2\log x+1)}{\sin y+y\cos y}$
$\displaystyle \Rightarrow (\sin y+y\cos y)\,dy=x(2\log x+1)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int (\sin y+y\cos y)\,dy=\int x(2\log x+1)\,dx$
$\displaystyle \Rightarrow \int \sin y\,dy+\int y\cos y\,dy=2\int x\log x\,dx+\int x\,dx$
$\displaystyle \Rightarrow -\cos y+\left[y\sin y-\int \sin y\,dy\right]=2\left(\frac{x^2}{2}\log x-\int \frac{x^2}{2}\cdot\frac{1}{x}\,dx\right)+\frac{x^2}{2}$
$\displaystyle \Rightarrow -\cos y+y\sin y+\cos y=x^2\log x-\frac{x^2}{2}+\frac{x^2}{2}+C$
$\displaystyle \Rightarrow y\sin y=x^2\log x+C$
$\displaystyle \text{Hence, } y\sin y=x^2\log x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 21: }~(1-x^{2})\,dy+xy\,dx=xy^{2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (1-x^2)\,dy+xy\,dx=xy^2\,dx$
$\displaystyle \Rightarrow (1-x^2)\,dy=xy^2\,dx-xy\,dx$
$\displaystyle \Rightarrow (1-x^2)\,dy=xy(y-1)\,dx$
$\displaystyle \Rightarrow \frac{1}{y(y-1)}\,dy=\frac{x}{1-x^2}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y(y-1)}\,dy=\int \frac{x}{1-x^2}\,dx$
$\displaystyle \text{Let } \frac{1}{y(y-1)}=\frac{A}{y}+\frac{B}{y-1}$
$\displaystyle \Rightarrow 1=A(y-1)+By$
$\displaystyle \text{Putting } y=1, \text{ we get } B=1$
$\displaystyle \text{Putting } y=0, \text{ we get } A=-1$
$\displaystyle \Rightarrow \frac{1}{y(y-1)}=-\frac{1}{y}+\frac{1}{y-1}$
$\displaystyle \Rightarrow \int \frac{1}{y(y-1)}\,dy=-\int \frac{1}{y}\,dy+\int \frac{1}{y-1}\,dy$
$\displaystyle \Rightarrow -\log|y|+\log|y-1|=\int \frac{x}{1-x^2}\,dx$
$\displaystyle \text{Putting } 1-x^2=t, \text{ we get}$
$\displaystyle -2x\,dx=dt$
$\displaystyle \Rightarrow \int \frac{x}{1-x^2}\,dx=-\frac{1}{2}\int \frac{1}{t}\,dt$
$\displaystyle \Rightarrow \int \frac{x}{1-x^2}\,dx=-\frac{1}{2}\log|1-x^2|+C$
$\displaystyle \Rightarrow -\log|y|+\log|y-1|=-\frac{1}{2}\log|1-x^2|+C$
$\displaystyle \Rightarrow \log\left|\frac{y-1}{y}\right|=-\frac{1}{2}\log|1-x^2|+C$

$\displaystyle \textbf{Question 22: }~\tan y\,dx+\sec^{2}y\,\tan x\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \tan y\,dx+\sec^2 y\tan x\,dy=0$
$\displaystyle \Rightarrow \sec^2 y\tan x\,dy=-\tan y\,dx$
$\displaystyle \Rightarrow \frac{\sec^2 y}{\tan y}\,dy=-\frac{1}{\tan x}\,dx$
$\displaystyle \Rightarrow \frac{1}{\sin y\cos y}\,dy=-\cot x\,dx$
$\displaystyle \Rightarrow \frac{2}{\sin 2y}\,dy=-\cot x\,dx$
$\displaystyle \Rightarrow 2\,\mathrm{cosec}\,2y\,dy=-\cot x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle 2\int \mathrm{cosec}\,2y\,dy=-\int \cot x\,dx$
$\displaystyle \Rightarrow \log|\tan y|=-\log|\sin x|+C$
$\displaystyle \Rightarrow \log|\tan y|+\log|\sin x|=C$
$\displaystyle \Rightarrow \log(\tan y\sin x)=C$
$\displaystyle \Rightarrow \tan y\sin x=C$

$\displaystyle \textbf{Question 23: }~(1+x)(1+y^{2})\,dx+(1+y)(1+x^{2})\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (1+x)(1+y^2)\,dx+(1+y)(1+x^2)\,dy=0$
$\displaystyle \Rightarrow (1+x)(1+y^2)\,dx=-(1+y)(1+x^2)\,dy$
$\displaystyle \Rightarrow \frac{1+x}{1+x^2}\,dx=-\frac{1+y}{1+y^2}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1+x}{1+x^2}\,dx=-\int \frac{1+y}{1+y^2}\,dy$
$\displaystyle \Rightarrow \int \frac{1}{1+x^2}\,dx+\int \frac{x}{1+x^2}\,dx=-\int \frac{1}{1+y^2}\,dy-\int \frac{y}{1+y^2}\,dy$
$\displaystyle \text{Substituting } 1+x^2=t \text{ and } 1+y^2=u, \text{ we get}$
$\displaystyle 2x\,dx=dt \text{ and } 2y\,dy=du$
$\displaystyle \Rightarrow \int \frac{1}{1+x^2}\,dx+\frac{1}{2}\int \frac{1}{t}\,dt=-\int \frac{1}{1+y^2}\,dy-\frac{1}{2}\int \frac{1}{u}\,du$
$\displaystyle \Rightarrow \tan^{-1}x+\frac{1}{2}\log|t|=-\tan^{-1}y-\frac{1}{2}\log|u|+C$
$\displaystyle \Rightarrow \tan^{-1}x+\frac{1}{2}\log(1+x^2)=-\tan^{-1}y-\frac{1}{2}\log(1+y^2)+C$
$\displaystyle \Rightarrow \tan^{-1}x+\tan^{-1}y+\frac{1}{2}\log(1+x^2)+\frac{1}{2}\log(1+y^2)=C$
$\displaystyle \Rightarrow \tan^{-1}x+\tan^{-1}y+\frac{1}{2}\log\!\big((1+x^2)(1+y^2)\big)=C$
$\displaystyle \text{Hence, } \tan^{-1}x+\tan^{-1}y+\frac{1}{2}\log\!\big((1+x^2)(1+y^2)\big)=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 24: }~\tan y\,\frac{dy}{dx}=\sin(x+y)+\sin(x-y).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \tan y\frac{dy}{dx}=\sin(x+y)+\sin(x-y)$
$\displaystyle \Rightarrow \tan y\frac{dy}{dx}=\sin x\cos y+\cos x\sin y+\sin x\cos y-\cos x\sin y$
$\displaystyle \Rightarrow \tan y\frac{dy}{dx}=2\sin x\cos y$
$\displaystyle \Rightarrow \frac{\tan y}{\cos y}\,dy=2\sin x\,dx$
$\displaystyle \Rightarrow \tan y\sec y\,dy=2\sin x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \tan y\sec y\,dy=2\int \sin x\,dx$
$\displaystyle \Rightarrow \sec y=-2\cos x+C$
$\displaystyle \Rightarrow \sec y+2\cos x=C$
$\displaystyle \text{Hence, } \sec y+2\cos x=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 25: }~\cos x\cos y\,\frac{dy}{dx}=-\sin x\sin y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \cos x\cos y\frac{dy}{dx}=-\sin x\sin y$
$\displaystyle \Rightarrow \frac{\cos y}{\sin y}\,dy=-\frac{\sin x}{\cos x}\,dx$
$\displaystyle \Rightarrow \cot y\,dy=-\tan x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \cot y\,dy=-\int \tan x\,dx$
$\displaystyle \Rightarrow \log|\sin y|=-\log|\sec x|+C$
$\displaystyle \Rightarrow \log|\sin y|=\log|\cos x|+C$
$\displaystyle \Rightarrow \sin y=C\cos x$
$\displaystyle \text{Hence, } \sin y=C\cos x \text{ is the required solution.}$

$\displaystyle \textbf{Question 26: }~\frac{dy}{dx}+\frac{\cos x\sin y}{\cos y}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}+\frac{\cos x\sin y}{\cos y}=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{\cos x\sin y}{\cos y}$
$\displaystyle \Rightarrow \frac{\cos y}{\sin y}\,dy=-\cos x\,dx$
$\displaystyle \Rightarrow \cot y\,dy=-\cos x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \cot y\,dy=-\int \cos x\,dx$
$\displaystyle \Rightarrow \log|\sin y|=-\sin x+C$
$\displaystyle \text{Hence, } \log|\sin y|=-\sin x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 27: }~x\sqrt{1-y^{2}}\,dx+y\sqrt{1-x^{2}}\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle x\sqrt{1-y^2}\,dx+y\sqrt{1-x^2}\,dy=0$
$\displaystyle \Rightarrow y\sqrt{1-x^2}\,dy=-x\sqrt{1-y^2}\,dx$
$\displaystyle \Rightarrow \frac{y}{\sqrt{1-y^2}}\,dy=-\frac{x}{\sqrt{1-x^2}}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{\sqrt{1-y^2}}\,dy=-\int \frac{x}{\sqrt{1-x^2}}\,dx$
$\displaystyle \text{Substituting } 1-y^2=t \text{ and } 1-x^2=u, \text{ we get}$
$\displaystyle -2y\,dy=dt \text{ and } -2x\,dx=du$
$\displaystyle \Rightarrow -\frac{1}{2}\int \frac{1}{\sqrt{t}}\,dt=\frac{1}{2}\int \frac{1}{\sqrt{u}}\,du$
$\displaystyle \Rightarrow -\sqrt{t}=\sqrt{u}+C$
$\displaystyle \Rightarrow \sqrt{1-x^2}+\sqrt{1-y^2}=C$
$\displaystyle \text{Hence, } \sqrt{1-x^2}+\sqrt{1-y^2}=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 28: }~y(1+e^{x})\,dy=(y+1)e^{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle y(1+e^x)\,dy=(y+1)e^x\,dx$
$\displaystyle \Rightarrow \frac{y}{y+1}\,dy=\frac{e^x}{1+e^x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{y+1}\,dy=\int \frac{e^x}{1+e^x}\,dx$
$\displaystyle \text{Substituting } 1+e^x=t, \text{ we get}$
$\displaystyle e^x\,dx=dt$
$\displaystyle \Rightarrow \int \frac{y}{y+1}\,dy=\int \frac{1}{t}\,dt$
$\displaystyle \Rightarrow \int \frac{y+1-1}{y+1}\,dy=\int \frac{1}{t}\,dt$
$\displaystyle \Rightarrow \int dy-\int \frac{1}{y+1}\,dy=\int \frac{1}{t}\,dt$
$\displaystyle \Rightarrow y-\log|y+1|=\log|t|+C$
$\displaystyle \Rightarrow y-\log|y+1|=\log|1+e^x|+C$

$\displaystyle \textbf{Question 29: }~(y+x)\,dx+(x-xy^{2})\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (y+xy)\,dx+(x-xy^2)\,dy=0$
$\displaystyle \Rightarrow y(1+x)\,dx=x(y^2-1)\,dy$
$\displaystyle \Rightarrow \frac{1+x}{x}\,dx=\frac{y^2-1}{y}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1+x}{x}\,dx=\int \frac{y^2-1}{y}\,dy$
$\displaystyle \Rightarrow \int \frac{1}{x}\,dx+\int dx=\int y\,dy-\int \frac{1}{y}\,dy$
$\displaystyle \Rightarrow \log|x|+x=\frac{y^2}{2}-\log|y|+C$
$\displaystyle \Rightarrow \log|x|+x-\frac{y^2}{2}+\log|y|=C$
$\displaystyle \text{Hence, } \log|x|+x-\frac{y^2}{2}+\log|y|=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 30: }~\frac{dy}{dx}=1-x+y-xy.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=1-x+y-xy$
$\displaystyle \Rightarrow \frac{dy}{dx}=1+y-x(1+y)$
$\displaystyle \Rightarrow \frac{dy}{dx}=(1+y)(1-x)$
$\displaystyle \Rightarrow \frac{1}{1+y}\,dy=(1-x)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{1+y}\,dy=\int (1-x)\,dx$
$\displaystyle \Rightarrow \log|1+y|=x-\frac{x^2}{2}+C$
$\displaystyle \text{Hence, } \log|1+y|=x-\frac{x^2}{2}+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 31: }~(y^{2}+1)\,dx-(x^{2}+1)\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (y^2+1)\,dx-(x^2+1)\,dy=0$
$\displaystyle \Rightarrow (y^2+1)\,dx=(x^2+1)\,dy$
$\displaystyle \Rightarrow \frac{1}{x^2+1}\,dx=\frac{1}{y^2+1}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{x^2+1}\,dx=\int \frac{1}{y^2+1}\,dy$
$\displaystyle \Rightarrow \tan^{-1}x=\tan^{-1}y+C$
$\displaystyle \Rightarrow \tan^{-1}x-\tan^{-1}y=C$
$\displaystyle \text{Hence, } \tan^{-1}x-\tan^{-1}y=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 32: }~dy+(x+1)(y+1)\,dx=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle dy+(x+1)(y+1)\,dx=0$
$\displaystyle \Rightarrow dy=-(x+1)(y+1)\,dx$
$\displaystyle \Rightarrow \frac{1}{y+1}\,dy=-(x+1)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y+1}\,dy=-\int (x+1)\,dx$
$\displaystyle \Rightarrow \log|y+1|=-\frac{x^2}{2}-x+C$
$\displaystyle \Rightarrow \log|y+1|+\frac{x^2}{2}+x=C$
$\displaystyle \text{Hence, } \log|y+1|+\frac{x^2}{2}+x=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 33: }~\frac{dy}{dx}=(1+x^{2})(1+y^{2}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=(1+x^2)(1+y^2)$
$\displaystyle \Rightarrow \frac{1}{1+y^2}\,dy=(1+x^2)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{1+y^2}\,dy=\int (1+x^2)\,dx$
$\displaystyle \Rightarrow \tan^{-1}y=x+\frac{x^3}{3}+C$
$\displaystyle \text{Hence, } \tan^{-1}y=x+\frac{x^3}{3}+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 34: }~(x-1)\frac{dy}{dx}=2x^{3}y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (x-1)\frac{dy}{dx}=2x^3y$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=\frac{2x^3}{x-1}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int \frac{2x^3}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=2\int \frac{x^3}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=2\int \frac{(x-1)(x^2+x+1)+1}{x-1}\,dx$
$\displaystyle \Rightarrow \log|y|=2\left[\int (x^2+x+1)\,dx+\int \frac{1}{x-1}\,dx\right]$
$\displaystyle \Rightarrow \log|y|=2\left(\frac{x^3}{3}+\frac{x^2}{2}+x+\log|x-1|\right)+C$
$\displaystyle \Rightarrow \log|y|=\frac{2}{3}x^3+x^2+2x+2\log|x-1|+C$
$\displaystyle \Rightarrow y=C|x-1|^2 e^{\frac{2}{3}x^3+x^2+2x}$

$\displaystyle \textbf{Question 35: }~\frac{dy}{dx}=e^{x+y}+e^{-x+y}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=e^{x+y}+e^{-x+y}$
$\displaystyle \Rightarrow \frac{dy}{dx}=e^y(e^x+e^{-x})$
$\displaystyle \Rightarrow e^{-y}\,dy=(e^x+e^{-x})\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int e^{-y}\,dy=\int (e^x+e^{-x})\,dx$
$\displaystyle \Rightarrow -e^{-y}=e^x-e^{-x}+C$
$\displaystyle \Rightarrow e^{-x}-e^{-y}=e^x+C$
$\displaystyle \text{Hence, } e^{-x}-e^{-y}=e^x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 36: }~\frac{dy}{dx}=(\cos^{2}x-\sin^{2}x)\cos^{2}y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=(\cos^2 x-\sin^2 x)\cos^2 y$
$\displaystyle \Rightarrow \frac{dy}{dx}=\cos 2x\cos^2 y$
$\displaystyle \Rightarrow \frac{1}{\cos^2 y}\,dy=\cos 2x\,dx$
$\displaystyle \Rightarrow \sec^2 y\,dy=\cos 2x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \sec^2 y\,dy=\int \cos 2x\,dx$
$\displaystyle \Rightarrow \tan y=\frac{\sin 2x}{2}+C$
$\displaystyle \text{Hence, } \tan y=\frac{\sin 2x}{2}+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 37(i): }~(xy^{2}+2x)\,dx+(x^{2}y+2y)\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (xy^2+2x)\,dx+(x^2y+2y)\,dy=0$
$\displaystyle \Rightarrow x(y^2+2)\,dx+y(x^2+2)\,dy=0$
$\displaystyle \Rightarrow x(y^2+2)\,dx=-y(x^2+2)\,dy$
$\displaystyle \Rightarrow \frac{x}{x^2+2}\,dx=-\frac{y}{y^2+2}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{x}{x^2+2}\,dx=-\int \frac{y}{y^2+2}\,dy$
$\displaystyle \Rightarrow \frac{1}{2}\int \frac{2x}{x^2+2}\,dx=-\frac{1}{2}\int \frac{2y}{y^2+2}\,dy$
$\displaystyle \Rightarrow \frac{1}{2}\log|x^2+2|=-\frac{1}{2}\log|y^2+2|+C$
$\displaystyle \Rightarrow \frac{1}{2}\log|x^2+2|+\frac{1}{2}\log|y^2+2|=C$
$\displaystyle \Rightarrow \log\!\big((x^2+2)(y^2+2)\big)=C$
$\displaystyle \Rightarrow (x^2+2)(y^2+2)=C$

$\displaystyle \textbf{Question 37(ii): }~\mathrm{cosec}\,x\log y\,\frac{dy}{dx}+x^{2}y^{2}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \mathrm{cosec}\,x\,\log y\,\frac{dy}{dx}+x^2y^2=0$
$\displaystyle \Rightarrow \mathrm{cosec}\,x\,\log y\,\frac{dy}{dx}=-x^2y^2$
$\displaystyle \Rightarrow \frac{\log y}{y^2}\,dy=-\frac{x^2}{\mathrm{cosec}\,x}\,dx$
$\displaystyle \Rightarrow \frac{\log y}{y^2}\,dy=-x^2\sin x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{\log y}{y^2}\,dy=-\int x^2\sin x\,dx$
$\displaystyle \Rightarrow -\frac{\log y}{y}+\int \frac{1}{y}\cdot\frac{1}{y}\,dy=-\left[-x^2\cos x+\int 2x\cos x\,dx\right]+C$
$\displaystyle \Rightarrow -\frac{\log y}{y}-\frac{1}{y}=-\left[-x^2\cos x+2x\sin x-2\int \sin x\,dx\right]+C$
$\displaystyle \Rightarrow -\frac{1+\log y}{y}=-\left[-x^2\cos x+2x\sin x+2\cos x\right]+C$
$\displaystyle \Rightarrow -\frac{1+\log y}{y}-x^2\cos x+2(x\sin x+\cos x)=C$

$\displaystyle \textbf{Question 38(i): }~xy\frac{dy}{dx}=1+x+y+xy.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle xy\frac{dy}{dx}=1+x+y+xy$
$\displaystyle \Rightarrow xy\frac{dy}{dx}=(1+x)(1+y)$
$\displaystyle \Rightarrow \frac{y}{1+y}\,dy=\frac{1+x}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{1+y}\,dy=\int \frac{1+x}{x}\,dx$
$\displaystyle \Rightarrow \int \frac{1+y-1}{1+y}\,dy=\int \left(\frac{1}{x}+1\right)\,dx$
$\displaystyle \Rightarrow \int dy-\int \frac{1}{1+y}\,dy=\int \frac{1}{x}\,dx+\int dx$
$\displaystyle \Rightarrow y-\log|1+y|=\log|x|+x+C$
$\displaystyle \Rightarrow y=\log|x|+\log|1+y|+x+C$
$\displaystyle \Rightarrow y=\log|x(1+y)|+x+C$
$\displaystyle \text{Hence, } y=\log|x(1+y)|+x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 38(ii): }~y(1-x^{2})\frac{dy}{dx}=x(1+y^{2}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle y(1-x^2)\frac{dy}{dx}=x(1+y^2)$
$\displaystyle \Rightarrow \frac{y}{1+y^2}\,dy=\frac{x}{1-x^2}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{1+y^2}\,dy=\int \frac{x}{1-x^2}\,dx$
$\displaystyle \text{Substituting } 1+y^2=t \text{ and } 1-x^2=u$
$\displaystyle 2y\,dy=dt \text{ and } -2x\,dx=du$
$\displaystyle \Rightarrow \frac{1}{2}\int \frac{1}{t}\,dt=-\frac{1}{2}\int \frac{1}{u}\,du$
$\displaystyle \Rightarrow \frac{1}{2}\log|t|=-\frac{1}{2}\log|u|+C$
$\displaystyle \Rightarrow \frac{1}{2}\log|1+y^2|=-\frac{1}{2}\log|1-x^2|+C$
$\displaystyle \Rightarrow \frac{1}{2}\big(\log|1+y^2|+\log|1-x^2|\big)=C$
$\displaystyle \Rightarrow \log\!\big((1+y^2)(1-x^2)\big)=C$
$\displaystyle \Rightarrow (1+y^2)(1-x^2)=C$
$\displaystyle \text{Hence, } (1+y^2)(1-x^2)=C \text{ is the required solution.}$

$\displaystyle \text{(iii): }~y\,e^{x/y}\,dx=\left(xe^{x/y}+y^{2}\right)\,dy,\ y\neq 0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle y e^{\frac{x}{y}}\,dx=(x e^{\frac{x}{y}}+y^2)\,dy$
$\displaystyle \Rightarrow y e^{\frac{x}{y}}\,dx=x e^{\frac{x}{y}}\,dy+y^2\,dy$
$\displaystyle \Rightarrow y e^{\frac{x}{y}}\,dx-x e^{\frac{x}{y}}\,dy=y^2\,dy$
$\displaystyle \Rightarrow (y\,dx-x\,dy)e^{\frac{x}{y}}=y^2\,dy$
$\displaystyle \Rightarrow \frac{y\,dx-x\,dy}{y^2}e^{\frac{x}{y}}=dy$
$\displaystyle \Rightarrow e^{\frac{x}{y}}\,d\!\left(\frac{x}{y}\right)=dy$
$\displaystyle \Rightarrow \int e^{\frac{x}{y}}\,d\!\left(\frac{x}{y}\right)=\int dy$
$\displaystyle \Rightarrow e^{\frac{x}{y}}=y+C$

$\displaystyle \text{(iv): }~(1+y^{2})\tan^{-1}x\,dx+2y(1+x^{2})\,dy=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (1+y^2)\tan^{-1}x\,dx+2y(1+x^2)\,dy=0$
$\displaystyle \Rightarrow (1+y^2)\tan^{-1}x\,dx=-2y(1+x^2)\,dy$
$\displaystyle \Rightarrow \frac{\tan^{-1}x}{1+x^2}\,dx=-\frac{2y}{1+y^2}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{\tan^{-1}x}{1+x^2}\,dx=-\int \frac{2y}{1+y^2}\,dy$
$\displaystyle \Rightarrow \frac{(\tan^{-1}x)^2}{2}=-\log|1+y^2|+C$
$\displaystyle \Rightarrow \frac{(\tan^{-1}x)^2}{2}+\log|1+y^2|=C$

$\displaystyle \textbf{Solve the following initial value problems: (39-45)}$

$\displaystyle \textbf{Question 39: }~\frac{dy}{dx}=y\tan 2x,\ y(0)=2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=y\tan 2x,\; y(0)=2$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=\tan 2x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int \tan 2x\,dx$
$\displaystyle \Rightarrow \log|y|=\frac{1}{2}\log|\sec 2x|+C$
$\displaystyle \Rightarrow y^2=C\sec 2x \quad (1)$
$\displaystyle \text{It is given that at } x=0,\; y=2$
$\displaystyle \Rightarrow C=4$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle y^2=\frac{4}{\cos 2x}$
$\displaystyle \Rightarrow y=\frac{2}{\sqrt{\cos 2x}}$
$\displaystyle \text{Hence, } y=\frac{2}{\sqrt{\cos 2x}} \text{ is the required solution.}$

$\displaystyle \textbf{Question 40: }~2x\frac{dy}{dx}=3y,\ y(1)=2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle 2x\frac{dy}{dx}=3y,\; y(1)=2$
$\displaystyle \Rightarrow \frac{2}{y}\,dy=\frac{3}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle 2\int \frac{1}{y}\,dy=3\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow 2\log|y|=3\log|x|+\log C$
$\displaystyle \Rightarrow \log|y|^2=\log|x|^3+\log C$
$\displaystyle \Rightarrow y^2=Cx^3 \quad (1)$
$\displaystyle \text{It is given that at } x=1,\; y=2$
$\displaystyle \Rightarrow C=4$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle y^2=4x^3$
$\displaystyle \text{Hence, } y^2=4x^3 \text{ is the required solution.}$

$\displaystyle \textbf{Question 41: }~xy\frac{dy}{dx}=y+2,\ y(2)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle xy\frac{dy}{dx}=y+2,\; y(2)=0$
$\displaystyle \Rightarrow \frac{y}{y+2}\,dy=\frac{1}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{y+2}\,dy=\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow \int \frac{y+2-2}{y+2}\,dy=\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow \int dy-2\int \frac{1}{y+2}\,dy=\log|x|+C$
$\displaystyle \Rightarrow y-2\log|y+2|=\log|x|+C \quad (1)$
$\displaystyle \text{It is given that at } x=2,\; y=0$
$\displaystyle \Rightarrow 0-2\log 2=\log 2+C$
$\displaystyle \Rightarrow C=-2\log 2-\log 2$
$\displaystyle \Rightarrow C=-\log 8$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle y-2\log|y+2|=\log|x|-\log 8$
$\displaystyle \Rightarrow y-2\log|y+2|=\log\!\left|\frac{x}{8}\right|$
$\displaystyle \text{Hence, } y-2\log|y+2|=\log\!\left|\frac{x}{8}\right| \text{ is the required solution.}$

$\displaystyle \textbf{Question 42: }~\frac{dy}{dx}=2e^{x}y^{3},\ y(0)=\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=2e^x y^3,\; y(0)=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1}{y^3}\,dy=2e^x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y^3}\,dy=\int 2e^x\,dx$
$\displaystyle \Rightarrow -\frac{1}{2y^2}=2e^x+C \quad (1)$
$\displaystyle \text{Given that at } x=0,\; y=\frac{1}{2}$
$\displaystyle \Rightarrow -\frac{1}{2\left(\frac{1}{2}\right)^2}=2e^0+C$
$\displaystyle \Rightarrow -2=2+C$
$\displaystyle \Rightarrow C=-4$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle -\frac{1}{2y^2}=2e^x-4$
$\displaystyle \Rightarrow \frac{1}{y^2}=8-4e^x$
$\displaystyle \Rightarrow y^2(8-4e^x)=1$
$\displaystyle \text{Hence, } y^2(8-4e^x)=1 \text{ is the required solution.}$

$\displaystyle \textbf{Question 43: }~\frac{dr}{dt}=-rt,\ r(0)=r_{0}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dr}{dt}=-rt,\; r(0)=r_0$
$\displaystyle \Rightarrow \frac{1}{r}\,dr=-t\,dt$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{r}\,dr=-\int t\,dt$
$\displaystyle \Rightarrow \log|r|=-\frac{t^2}{2}+C \quad (1)$
$\displaystyle \text{Given that } t=0,\; r=r_0$
$\displaystyle \Rightarrow \log|r_0|=C$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \log|r|=-\frac{t^2}{2}+\log|r_0|$
$\displaystyle \Rightarrow \log\!\left|\frac{r}{r_0}\right|=-\frac{t^2}{2}$
$\displaystyle \Rightarrow r=r_0 e^{-\frac{t^2}{2}}$
$\displaystyle \text{Hence, } r=r_0 e^{-\frac{t^2}{2}} \text{ is the required solution.}$

$\displaystyle \textbf{Question 44: }~\frac{dy}{dx}=y\sin 2x,\ y(0)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=y\sin 2x,\; y(0)=1$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=\sin 2x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int \sin 2x\,dx$
$\displaystyle \Rightarrow \log|y|=-\frac{\cos 2x}{2}+C \quad (1)$
$\displaystyle \text{Given that } x=0,\; y=1$
$\displaystyle \Rightarrow \log 1=-\frac{1}{2}+C$
$\displaystyle \Rightarrow C=\frac{1}{2}$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \log|y|=-\frac{\cos 2x}{2}+\frac{1}{2}$
$\displaystyle \Rightarrow \log|y|=\frac{1-\cos 2x}{2}$
$\displaystyle \Rightarrow \log|y|=\sin^2 x$
$\displaystyle \Rightarrow y=e^{\sin^2 x}$
$\displaystyle \text{Hence, } y=e^{\sin^2 x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 45:}$
$\displaystyle \text{(i): }~\frac{dy}{dx}=y\tan x,\ y(0)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=y\tan x,\; y(0)=1$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=\tan x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int \tan x\,dx$
$\displaystyle \Rightarrow \log|y|=\log|\sec x|+C \quad (1)$
$\displaystyle \text{We know that at } x=0,\; y=1$
$\displaystyle \Rightarrow \log 1=\log 1+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \log|y|=\log|\sec x|$
$\displaystyle \Rightarrow y=\sec x$
$\displaystyle \text{Hence, } y=\sec x,\; x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), \text{ is the required solution.}$

$\displaystyle \text{(ii): }~2x\frac{dy}{dx}=5y,\ y(1)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle 2x\frac{dy}{dx}=5y,\; y(1)=1$
$\displaystyle \Rightarrow \frac{2}{y}\,dy=\frac{5}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle 2\int \frac{1}{y}\,dy=5\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow 2\log|y|=5\log|x|+C \quad (1)$
$\displaystyle \text{We know that at } x=1,\; y=1$
$\displaystyle \Rightarrow 2\log 1=5\log 1+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle 2\log|y|=5\log|x|$
$\displaystyle \Rightarrow \log|y|=\frac{5}{2}\log|x|$
$\displaystyle \Rightarrow y=|x|^{\frac{5}{2}}$
$\displaystyle \text{Hence, } y=|x|^{\frac{5}{2}} \text{ is the required solution.}$

$\displaystyle \text{(iii): }~\frac{dy}{dx}=2e^{2x}y^{2},\ y(0)=-1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=2e^{2x}y^2,\; y(0)=-1$
$\displaystyle \Rightarrow \frac{1}{y^2}\,dy=2e^{2x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y^2}\,dy=2\int e^{2x}\,dx$
$\displaystyle \Rightarrow -\frac{1}{y}=e^{2x}+C \quad (1)$
$\displaystyle \text{We know that at } x=0,\; y=-1$
$\displaystyle \Rightarrow -\frac{1}{-1}=e^0+C$
$\displaystyle \Rightarrow 1=1+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle -\frac{1}{y}=e^{2x}$
$\displaystyle \Rightarrow y=-e^{-2x}$
$\displaystyle \text{Hence, } y=-e^{-2x} \text{ is the required solution.}$

$\displaystyle \text{(iv): }~\cos y\,\frac{dy}{dx}=e^{x},\ y(0)=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \cos y\frac{dy}{dx}=e^x,\; y(0)=\frac{\pi}{2}$
$\displaystyle \Rightarrow \cos y\,dy=e^x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \cos y\,dy=\int e^x\,dx$
$\displaystyle \Rightarrow \sin y=e^x+C \quad (1)$
$\displaystyle \text{We know that at } x=0,\; y=\frac{\pi}{2}$
$\displaystyle \Rightarrow 1=1+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \sin y=e^x$
$\displaystyle \Rightarrow y=\sin^{-1}(e^x)$
$\displaystyle \text{Hence, } y=\sin^{-1}(e^x) \text{ is the required solution.}$

$\displaystyle \text{(v): }~\frac{dy}{dx}=2xy,\ y(0)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=2xy,\; y(0)=1$
$\displaystyle \Rightarrow \frac{1}{y}\,dy=2x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y}\,dy=\int 2x\,dx$
$\displaystyle \Rightarrow \log|y|=x^2+C \quad (1)$
$\displaystyle \text{We know that at } x=0,\; y=1$
$\displaystyle \Rightarrow \log 1=0+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \log|y|=x^2$
$\displaystyle \Rightarrow y=e^{x^2}$
$\displaystyle \text{Hence, } y=e^{x^2} \text{ is the required solution.}$

$\displaystyle \text{(vi): }~\frac{dy}{dx}=1+x^{2}+y^{2}+x^{2}y^{2},\ y(0)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=1+x^2+y^2+x^2y^2,\; y(0)=1$
$\displaystyle \Rightarrow \frac{dy}{dx}=(1+x^2)(1+y^2)$
$\displaystyle \Rightarrow \frac{dy}{1+y^2}=(1+x^2)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{dy}{1+y^2}=\int (1+x^2)\,dx$
$\displaystyle \Rightarrow \tan^{-1}y=x+\frac{x^3}{3}+C \quad (1)$
$\displaystyle \text{We know that at } x=0,\; y=1$
$\displaystyle \Rightarrow \tan^{-1}1=0+0+C$
$\displaystyle \Rightarrow \frac{\pi}{4}=C$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \tan^{-1}y=x+\frac{x^3}{3}+\frac{\pi}{4}$
$\displaystyle \text{Hence, } \tan^{-1}y=x+\frac{x^3}{3}+\frac{\pi}{4} \text{ is the required solution.}$

$\displaystyle \text{(vii): }~xy\frac{dy}{dx}=(x+2)(y+2),\ y(1)=-1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle xy\frac{dy}{dx}=(x+2)(y+2),\; y(1)=-1$
$\displaystyle \Rightarrow \frac{y}{y+2}\,dy=\frac{x+2}{x}\,dx$
$\displaystyle \Rightarrow \frac{y+2-2}{y+2}\,dy=\frac{x+2}{x}\,dx$
$\displaystyle \Rightarrow \left(1-\frac{2}{y+2}\right)dy=\left(1+\frac{2}{x}\right)dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int\left(1-\frac{2}{y+2}\right)dy=\int\left(1+\frac{2}{x}\right)dx$
$\displaystyle \Rightarrow y-2\log|y+2|=x+2\log|x|+C \quad (1)$
$\displaystyle \text{We know that at } x=1,\; y=-1$
$\displaystyle \Rightarrow -1-2\log 1=1+2\log 1+C$
$\displaystyle \Rightarrow -1=1+C$
$\displaystyle \Rightarrow C=-2$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle y-2\log|y+2|=x+2\log|x|-2$
$\displaystyle \text{Hence, } y-2\log|y+2|=x+2\log|x|-2 \text{ is the required solution.}$

$\displaystyle \text{(viii): }~\frac{dy}{dx}=1+x+y^{2}+xy^{2},\ y(0)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=1+x+y^2+xy^2$
$\displaystyle \Rightarrow \frac{dy}{dx}=1+x+y^2(1+x)$
$\displaystyle \Rightarrow \frac{dy}{dx}=(1+x)(1+y^2)$
$\displaystyle \Rightarrow \frac{dy}{1+y^2}=(1+x)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{dy}{1+y^2}=\int (1+x)\,dx$
$\displaystyle \Rightarrow \tan^{-1}y=x+\frac{x^2}{2}+C \quad (1)$
$\displaystyle \text{Now, } \tan^{-1}0=0+0+C \; [\text{given } y=0,\; x=0]$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \tan^{-1}y=x+\frac{x^2}{2}$
$\displaystyle \Rightarrow y=\tan\!\left(x+\frac{x^2}{2}\right)$

$\displaystyle \text{(ix): }~2(y+3)-xy\frac{dy}{dx}=0,\ y(1)=-2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle 2(y+3)-xy\frac{dy}{dx}=0$
$\displaystyle \Rightarrow 2(y+3)=xy\frac{dy}{dx}$
$\displaystyle \Rightarrow \frac{2}{x}\,dx=\frac{y}{y+3}\,dy$
$\displaystyle \Rightarrow \frac{2}{x}\,dx=\frac{y+3-3}{y+3}\,dy$
$\displaystyle \Rightarrow \frac{2}{x}\,dx=\left(1-\frac{3}{y+3}\right)dy$
$\displaystyle \Rightarrow \int \frac{2}{x}\,dx=\int \left(1-\frac{3}{y+3}\right)dy$
$\displaystyle \Rightarrow 2\log x=y-3\log|y+3|+C$
$\displaystyle \Rightarrow \log x^2+\log|y+3|^3=y+C$
$\displaystyle \Rightarrow \log\!\big[(x^2)(y+3)^3\big]=y+C \quad (1)$
$\displaystyle \text{Given } x=1,\; y=-2$
$\displaystyle \Rightarrow \log\!\big[(1)^2(-2+3)^3\big]=-2+C$
$\displaystyle \Rightarrow \log 1=-2+C$
$\displaystyle \Rightarrow C=2$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \log\!\big[(x^2)(y+3)^3\big]=y+2$
$\displaystyle \Rightarrow (x^2)(y+3)^3=e^{\,y+2}$

$\displaystyle \textbf{Solve the following:}$

$\displaystyle \textbf{Question 46: }~x\frac{dy}{dx}+\cot y=0,\ \text{given that }y=\frac{\pi}{4}\ \text{when }x=\sqrt{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle x\frac{dy}{dx}+\cot y=0$
$\displaystyle \Rightarrow x\frac{dy}{dx}=-\cot y$
$\displaystyle \Rightarrow \tan y\,dy=-\frac{1}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \tan y\,dy=-\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow \log|\sec y|=-\log|x|+\log C$
$\displaystyle \Rightarrow \log\!\big(|x|\sec y\big)=\log C$
$\displaystyle \Rightarrow x\sec y=C \quad (1)$
$\displaystyle \text{Given } x=\sqrt{2},\; y=\frac{\pi}{4}$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (1), we get}$
$\displaystyle \sqrt{2}\sec\frac{\pi}{4}=C$
$\displaystyle \Rightarrow C=2$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle x\sec y=2$
$\displaystyle \Rightarrow x=2\cos y$

$\displaystyle \textbf{Question 47: }~(1+x^{2})\frac{dy}{dx}+(1+y^{2})=0,\ \text{given that }y=1\ \text{when }x=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle (1+x^2)\frac{dy}{dx}+(1+y^2)=0,\; y=1 \text{ when } x=0$
$\displaystyle \Rightarrow (1+x^2)\frac{dy}{dx}=-(1+y^2)$
$\displaystyle \Rightarrow \frac{1}{1+y^2}\,dy=-\frac{1}{1+x^2}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{1+y^2}\,dy=-\int \frac{1}{1+x^2}\,dx$
$\displaystyle \Rightarrow \tan^{-1}y=-\tan^{-1}x+C$
$\displaystyle \Rightarrow \tan^{-1}y+\tan^{-1}x=C \quad (1)$
$\displaystyle \text{Given } x=0,\; y=1$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (1), we get}$
$\displaystyle \tan^{-1}1+\tan^{-1}0=C$
$\displaystyle \Rightarrow \frac{\pi}{4}+0=C$
$\displaystyle \Rightarrow C=\frac{\pi}{4}$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle \tan^{-1}y+\tan^{-1}x=\frac{\pi}{4}$
$\displaystyle \Rightarrow \tan^{-1}\!\left(\frac{x+y}{1-xy}\right)=\frac{\pi}{4}$
$\displaystyle \Rightarrow \frac{x+y}{1-xy}=1$
$\displaystyle \Rightarrow x+y=1-xy$

$\displaystyle \textbf{Question 48: }~\frac{dy}{dx}=\frac{2x(\log x+1)}{\sin y+y\cos y},\ \text{given that }y=0\ \text{when }x=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=\frac{2x(\log x+1)}{\sin y+y\cos y}$
$\displaystyle \Rightarrow (\sin y+y\cos y)\,dy=2x(\log x+1)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int (\sin y+y\cos y)\,dy=\int 2x(\log x+1)\,dx$
$\displaystyle \Rightarrow \int \sin y\,dy+\int y\cos y\,dy=\int 2x\log x\,dx+\int 2x\,dx$
$\displaystyle \Rightarrow -\cos y+\left[y\sin y-\int \sin y\,dy\right]=2\left[\log x\int x\,dx-\int \frac{d}{dx}(\log x)\int x\,dx\,dx\right]+x^2+C$
$\displaystyle \Rightarrow -\cos y+y\sin y+\cos y=2\left[\log x\cdot\frac{x^2}{2}-\frac{x^2}{4}\right]+x^2+C$
$\displaystyle \Rightarrow y\sin y=x^2\log x-\frac{x^2}{2}+x^2+C$
$\displaystyle \Rightarrow y\sin y=x^2\log x+\frac{x^2}{2}+C \quad (1)$
$\displaystyle \text{Given } x=1,\; y=0$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (1), we get}$
$\displaystyle 0=0+\frac{1}{2}+C$
$\displaystyle \Rightarrow C=-\frac{1}{2}$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle y\sin y=x^2\log x+\frac{x^2}{2}-\frac{1}{2}$
$\displaystyle \Rightarrow 2y\sin y=2x^2\log x+x^2-1$

$\displaystyle \textbf{Question 49: }~e^{dy/dx}=x+1,\ \text{given that }y=3\ \text{when }x=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle e^{\frac{dy}{dx}}=x+1$
$\displaystyle \Rightarrow \frac{dy}{dx}=\log(x+1)$
$\displaystyle \Rightarrow dy=\log(x+1)\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int dy=\int \log(x+1)\,dx$
$\displaystyle \Rightarrow y=\log(x+1)\int dx-\int \frac{d}{dx}\{\log(x+1)\}\int dx\,dx$
$\displaystyle \Rightarrow y=x\log(x+1)-\int \frac{x}{x+1}\,dx$
$\displaystyle \Rightarrow y=x\log(x+1)-\int \left(1-\frac{1}{x+1}\right)dx$
$\displaystyle \Rightarrow y=x\log(x+1)-\int dx+\int \frac{1}{x+1}\,dx$
$\displaystyle \Rightarrow y=x\log(x+1)-x+\log|x+1|+C$
$\displaystyle \Rightarrow y=(x+1)\log|x+1|-x+C \quad (1)$
$\displaystyle \text{It is given that } x=0 \text{ and } y=3$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (1), we get}$
$\displaystyle 3=(1)\log 1-0+C$
$\displaystyle \Rightarrow C=3$
$\displaystyle \text{Therefore, substituting the value of } C \text{ in (1), we get}$
$\displaystyle y=(x+1)\log|x+1|-x+3$

$\displaystyle \textbf{Question 50: }~\cos y\,dy+\cos x\,\sin y\,dx=0,\ \text{given that }y=\frac{\pi}{2}\ \text{when }x=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \cos y\,dy+\cos x\sin y\,dx=0$
$\displaystyle \Rightarrow \cos y\,dy=-\cos x\sin y\,dx$
$\displaystyle \Rightarrow \cot y\,dy=-\cos x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \cot y\,dy=-\int \cos x\,dx$
$\displaystyle \Rightarrow \log|\sin y|=-\sin x+C$
$\displaystyle \Rightarrow \log|\sin y|+\sin x=C \quad (1)$
$\displaystyle \text{It is given that } x=\frac{\pi}{2},\; y=\frac{\pi}{2}$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (1), we get}$
$\displaystyle \log\!\left|\sin\frac{\pi}{2}\right|+\sin\frac{\pi}{2}=C$
$\displaystyle \Rightarrow 0+1=C$
$\displaystyle \Rightarrow C=1$
$\displaystyle \text{Therefore, substituting the value of } C \text{ in (1), we get}$
$\displaystyle \log|\sin y|+\sin x=1$

$\displaystyle \textbf{Question 51: }~\frac{dy}{dx}=-4xy^{2},\ \text{given that }y=1\ \text{when }x=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle \frac{dy}{dx}=-4xy^2$
$\displaystyle \Rightarrow \frac{1}{y^2}\,dy=-4x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{y^2}\,dy=-4\int x\,dx$
$\displaystyle \Rightarrow -\frac{1}{y}=-4\cdot\frac{x^2}{2}+C$
$\displaystyle \Rightarrow -\frac{1}{y}=-2x^2+C \quad (1)$
$\displaystyle \text{It is given that } x=0,\; y=1$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (1), we get}$
$\displaystyle -1=C$
$\displaystyle \Rightarrow C=-1$
$\displaystyle \text{Therefore, substituting the value of } C \text{ in (1), we get}$
$\displaystyle -\frac{1}{y}=-2x^2-1$
$\displaystyle \Rightarrow \frac{1}{y}=2x^2+1$
$\displaystyle \Rightarrow y=\frac{1}{2x^2+1}$

$\displaystyle \textbf{Question 52: }~\text{Find the equation of a curve passing through }\\ (0,0)\text{ and whose differential equation is }\frac{dy}{dx}=e^{x}\sin x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have to find the equation of the curve that passes through the point }(0,0)\text{ and whose differential equation is}$
$\displaystyle \frac{dy}{dx}=e^x\sin x$
$\displaystyle \Rightarrow dy=e^x\sin x\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int dy=\int e^x\sin x\,dx$
$\displaystyle \Rightarrow y=\int e^x\sin x\,dx \quad (1)$
$\displaystyle \Rightarrow y=e^x\int \sin x\,dx-\int \frac{d}{dx}(e^x)\int \sin x\,dx\,dx$
$\displaystyle \Rightarrow y=-e^x\cos x+\int e^x\cos x\,dx$
$\displaystyle \Rightarrow y=-e^x\cos x+\left[e^x\int \cos x\,dx-\int \frac{d}{dx}(e^x)\int \cos x\,dx\,dx\right]$
$\displaystyle \Rightarrow y=-e^x\cos x+e^x\sin x-\int e^x\sin x\,dx$
$\displaystyle \Rightarrow y=-e^x\cos x+e^x\sin x-y+C \quad \text{[using (1)]}$
$\displaystyle \Rightarrow 2y=e^x(\sin x-\cos x)+C \quad (2)$
$\displaystyle \text{The curve passes through the point }(0,0)$
$\displaystyle \text{When } x=0,\; y=0$
$\displaystyle \text{Substituting the values of } x \text{ and } y \text{ in (2), we get}$
$\displaystyle 0=1(0-1)+C$
$\displaystyle \Rightarrow C=1$
$\displaystyle \text{Substituting the value of } C \text{ in (2), we get}$
$\displaystyle 2y=e^x(\sin x-\cos x)+1$

$\displaystyle \textbf{Question 53: }~\text{For the differential equation }xy\frac{dy}{dx}=(x+2)(y+2),\\ \text{find the solution curve passing through }(1,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }$
$\displaystyle xy\frac{dy}{dx}=(x+2)(y+2)$
$\displaystyle \Rightarrow \frac{y}{y+2}\,dy=\frac{x+2}{x}\,dx$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{y}{y+2}\,dy=\int \frac{x+2}{x}\,dx$
$\displaystyle \Rightarrow \int dy-2\int \frac{1}{y+2}\,dy=\int dx+2\int \frac{1}{x}\,dx$
$\displaystyle \Rightarrow y-2\log|y+2|=x+2\log|x|+C \quad (1)$
$\displaystyle \text{This equation represents the family of solution curves of the given differential equation.}$
$\displaystyle \text{We have to find a particular member of the family, which passes through the point }(1,-1).$
$\displaystyle \text{Substituting } x=1 \text{ and } y=-1 \text{ in (1), we get}$
$\displaystyle -1-2\log|1|=1+2\log|1|+C$
$\displaystyle \Rightarrow C=-2$
$\displaystyle \text{Putting } C=-2 \text{ in (1), we get}$
$\displaystyle y-2\log|y+2|=x+2\log|x|-2$
$\displaystyle \Rightarrow y-x+2=\log\!\big\{x^2(y+2)^2\big\}$

$\displaystyle \textbf{Question 54: }~\text{The volume of a spherical balloon being inflated changes at a constant rate.} \\ \text{Initially its radius is }3\text{ units and after }3\text{ seconds it is }6\text{ units. Find the radius after }t\text{ seconds.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } r \text{ be the radius and } V \text{ be the volume of the balloon at any time } t.$
$\displaystyle \text{Then, we have}$
$\displaystyle V=\frac{4}{3}\pi r^3$
$\displaystyle \text{Given :}$
$\displaystyle \frac{dV}{dt}=-k \quad (\text{where } k>0)$
$\displaystyle \Rightarrow \frac{d}{dt}\!\left(\frac{4}{3}\pi r^3\right)=-k$
$\displaystyle \Rightarrow 4\pi r^2\frac{dr}{dt}=-k$
$\displaystyle \Rightarrow 4\pi r^2\,dr=-k\,dt$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int 4\pi r^2\,dr=-\int k\,dt$
$\displaystyle \Rightarrow \frac{4}{3}\pi r^3=-kt+C \quad (1)$
$\displaystyle \text{It is given that } t=0,\; r=3.$
$\displaystyle \text{Substituting } t=0 \text{ and } r=3 \text{ in (1), we get}$
$\displaystyle \frac{4}{3}\pi(3)^3=C$
$\displaystyle \Rightarrow C=36\pi$
$\displaystyle \text{Putting } C=36\pi \text{ in (1), we get}$
$\displaystyle \frac{4}{3}\pi r^3=-kt+36\pi \quad (2)$
$\displaystyle \text{It is also given that } t=3,\; r=6.$
$\displaystyle \text{Putting } t=3 \text{ and } r=6 \text{ in (2), we get}$
$\displaystyle \frac{4}{3}\pi(6)^3=-3k+36\pi$
$\displaystyle \Rightarrow 288\pi=-3k+36\pi$
$\displaystyle \Rightarrow k=-84\pi$
$\displaystyle \text{Putting } k=-84\pi \text{ in (2), we get}$
$\displaystyle \frac{4}{3}\pi r^3=84\pi t+36\pi$
$\displaystyle \Rightarrow r^3=63t+27$
$\displaystyle \Rightarrow r=(63t+27)^{\frac{1}{3}}$

$\displaystyle \textbf{Question 55: }~\text{In a bank principal increases at the rate of }r\%\text{ per year. Find } \\ r\text{ if Rs }100\text{ double itself in }10\text{ years }(\log_{e}2=0.6931).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } P \text{ be the principal at any instant } t.$
$\displaystyle \text{Given:}$
$\displaystyle \frac{dP}{dt}=\frac{r}{100}P$
$\displaystyle \Rightarrow \frac{dP}{P}=\frac{r}{100}\,dt$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{dP}{P}=\int \frac{r}{100}\,dt$
$\displaystyle \Rightarrow \log P=\frac{rt}{100}+C \quad (1)$
$\displaystyle \text{Initially, i.e. at } t=0,\; \text{let } P=P_0.$
$\displaystyle \text{Putting } P=P_0 \text{ in (1), we get}$
$\displaystyle \log P_0=C$
$\displaystyle \text{Putting } C=\log P_0 \text{ in (1), we get}$
$\displaystyle \log P=\frac{rt}{100}+\log P_0$
$\displaystyle \Rightarrow \log\frac{P}{P_0}=\frac{rt}{100} \quad (2)$
$\displaystyle \text{Substituting } P_0=100,\; P=2P_0=200 \text{ and } t=10 \text{ in (2), we get}$
$\displaystyle \log 2=\frac{r}{10}$
$\displaystyle \Rightarrow r=10\log 2$
$\displaystyle \Rightarrow r=10\times 0.6931$
$\displaystyle \Rightarrow r=6.931$

$\displaystyle \textbf{Question 56: }~\text{In a bank principal increases at the rate of } \\ 5\%\text{ per year. An amount of Rs }1000\text{ is deposited with this bank; how much } \\ \text{will it be worth after }10\text{ years }(e^{0.5}=1.648).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let at any instant } t, \text{ the principal be } P.$
$\displaystyle \text{Here, it is given that the principal increases at the rate of } 5\% \text{ per year.}$
$\displaystyle \frac{dP}{dt}=\frac{5P}{100}$
$\displaystyle \Rightarrow \frac{dP}{P}=\frac{1}{20}\,dt$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{dP}{P}=\int \frac{1}{20}\,dt$
$\displaystyle \Rightarrow \log P=\frac{t}{20}+\log C \quad (1)$
$\displaystyle \text{Initially, at } t=0, \text{ it is given that } P=\text{Rs }1000.$
$\displaystyle \Rightarrow \log 1000=\log C$
$\displaystyle \text{Substituting the value of } \log C \text{ in (1), we get}$
$\displaystyle \log P=\frac{t}{20}+\log 1000$
$\displaystyle \Rightarrow \log\frac{P}{1000}=\frac{t}{20}$
$\displaystyle \text{Putting } t=10, \text{ we get}$
$\displaystyle \log\frac{P}{1000}=0.5$
$\displaystyle \Rightarrow \frac{P}{1000}=e^{0.5}$
$\displaystyle \Rightarrow P=1000\times 1.648$
$\displaystyle \Rightarrow P=1648$
$\displaystyle \text{Therefore, Rs }1000 \text{ will be worth Rs }1648 \text{ after 10 years.}$

$\displaystyle \textbf{Question 57: }~\text{In a culture the bacteria count is }100000.\ \text{The number is increased by } \\ 10\%\text{ in }2\text{ hours. In how many hours will the count reach }200000,\\ \text{if the rate of growth of bacteria is proportional to the number present?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let at any time the bacteria count be } N.$
$\displaystyle \text{Given:}$
$\displaystyle \frac{dN}{dt}\propto N$
$\displaystyle \Rightarrow \frac{dN}{dt}=\lambda N$
$\displaystyle \Rightarrow \frac{1}{N}\,dN=\lambda\,dt$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int \frac{1}{N}\,dN=\int \lambda\,dt$
$\displaystyle \Rightarrow \log N=\lambda t+\log C\quad (1)$
$\displaystyle \text{Given:}$
$\displaystyle t=0,\; N=100000$
$\displaystyle \Rightarrow \log C=\log 100000$
$\displaystyle \text{Putting this value in (1), we get}$
$\displaystyle \log N=\lambda t+\log 100000$
$\displaystyle \text{Also, at } t=2,\; N=110000$
$\displaystyle \text{Substituting in (1), we get}$
$\displaystyle \log 110000=2\lambda+\log 100000$
$\displaystyle \Rightarrow \lambda=\frac{1}{2}\log\left(\frac{11}{10}\right)$
$\displaystyle \text{Substituting the values of } \lambda \text{ and } C \text{ in (1), we get}$
$\displaystyle \log N=\frac{t}{2}\log\left(\frac{11}{10}\right)+\log 100000\quad (2)$
$\displaystyle \text{When } N=200000,\; t=T$
$\displaystyle \text{Substituting in (2), we get}$
$\displaystyle \log 200000=\frac{T}{2}\log\left(\frac{11}{10}\right)+\log 100000$
$\displaystyle \Rightarrow \log 2=\frac{T}{2}\log\left(\frac{11}{10}\right)$
$\displaystyle \Rightarrow T=\frac{2\log 2}{\log\left(\frac{11}{10}\right)}$
$\displaystyle \text{Therefore, in } \frac{2\log 2}{\log\left(\frac{11}{10}\right)} \text{ hours, the count will reach } 200000.$

$\displaystyle \textbf{Question 58: }~\text{If }y(x)\text{ is a solution of the differential equation } \\ \left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x\ \text{and }y(0)=1,\ \text{then find the value of } y\!\left(\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x$
$\displaystyle \Rightarrow \frac{1}{1+y}\,dy=-\frac{\cos x}{2+\sin x}\,dx$
$\displaystyle \Rightarrow \int \frac{1}{1+y}\,dy=-\int \frac{\cos x}{2+\sin x}\,dx$
$\displaystyle \Rightarrow \log|1+y|=-\log|2+\sin x|+\log C$
$\displaystyle \Rightarrow \log\!\big((1+y)(2+\sin x)\big)=\log C$
$\displaystyle \Rightarrow (1+y)(2+\sin x)=C\quad (1)$
$\displaystyle \text{Now, } y(0)=1$
$\displaystyle \Rightarrow (1+1)(2+0)=C$
$\displaystyle \Rightarrow C=4$
$\displaystyle \text{Substituting the value of } C \text{ in (1), we get}$
$\displaystyle (1+y)(2+\sin x)=4$
$\displaystyle \Rightarrow 1+y=\frac{4}{2+\sin x}$
$\displaystyle \Rightarrow y=\frac{4}{2+\sin x}-1$
$\displaystyle \Rightarrow y\!\left(\frac{\pi}{2}\right)=\frac{4}{2+\sin\!\left(\frac{\pi}{2}\right)}-1$
$\displaystyle =\frac{4}{3}-1$
$\displaystyle =\frac{1}{3}$

$\displaystyle \textbf{Question 59: }~\text{Find the particular solution of the differential equation } \\ (1-y^{2})(1+\log x)\,dx+2xy\,dy=0,\ \text{given that }y=0\ \text{when }x=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given:}$
$\displaystyle (1-y^{2})(1+\log x)\,dx+2xy\,dy=0$
$\displaystyle \Rightarrow (1-y^{2})(1+\log x)\,dx=-2xy\,dy$
$\displaystyle \Rightarrow \left(\frac{1+\log x}{2x}\right)\,dx=-\left(\frac{y}{1-y^{2}}\right)\,dy\quad (1)$
$\displaystyle \text{Let } 1+\log x=t$
$\displaystyle \text{and } 1-y^{2}=p$
$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt \text{ and } -2y\,dy=dp$
$\displaystyle \text{Therefore, (1) becomes}$
$\displaystyle \int \frac{t}{2}\,dt=\int \frac{1}{2p}\,dp$
$\displaystyle \Rightarrow \frac{t^{2}}{4}=\frac{\log p}{2}+C\quad (2)$
$\displaystyle \text{Substituting the values of } t \text{ and } p \text{ in (2), we get}$
$\displaystyle \frac{(1+\log x)^{2}}{4}=\frac{\log(1-y^{2})}{2}+C\quad (3)$
$\displaystyle \text{At } x=1 \text{ and } y=0,\; (3) \text{ becomes}$
$\displaystyle C=\frac{1}{4}$
$\displaystyle \text{Substituting the value of } C \text{ in (3), we get}$
$\displaystyle \frac{(1+\log x)^{2}}{4}=\frac{\log(1-y^{2})}{2}+\frac{1}{4}$
$\displaystyle \Rightarrow (1+\log x)^{2}=2\log(1-y^{2})+1$
$\displaystyle \text{Or}$
$\displaystyle (\log x)^{2}+2\log x=2\log(1-y^{2})^{2}$
$\displaystyle \text{It is the required particular solution.}$