Class 12: Differential Equations – ISC Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{The order and the degree of differential equation } \\ 1+\left(\frac{dy}{dx}\right)^2=\frac{d^2 y}{dx^2} \text{ are} \quad \text{(ISC 2024)}$
$\displaystyle \text{(a) } 2 \text{ and } \frac{3}{2} \quad \text{(b) } 2 \text{ and } 3 \quad \text{(c) } 3 \text{ and } 4 \quad \text{(d) } 2 \text{ and } 1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Given, } 1+\left(\frac{dy}{dx}\right)^{2}=\frac{d^{2}y}{dx^{2}}$
$\displaystyle \text{Here, the highest order derivative is } \frac{d^{2}y}{dx^{2}}$
$\displaystyle \therefore \text{Order}=2,\ \text{Degree}=1$
$\\$

$\displaystyle \textbf{Question 2: } \text{Solve the following differential equation } \\ 2ye^{x}dx+\left(y-2xe^{x}\right)dy=0, \text{ given } x=0 \text{ and } y=1. \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } 2ye^{x/y}dx+(y-2xe^{x/y})dy=0,\ y(0)=1$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{2xe^{x/y}-y}{2ye^{x/y}}$
$\displaystyle \text{14. Given, } 2ye^{x/y}dx+(y-2xe^{x/y})dy=0,\ y(0)=1$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{2xe^{x/y}-y}{2ye^{x/y}}$
$\displaystyle \text{On putting } x=vy \text{ and } \frac{dx}{dy}=v+y\frac{dv}{dy}, \text{ we get}$
$\displaystyle v+y\frac{dv}{dy}=\frac{2yve^{v}-y}{2ye^{v}}=\frac{2ve^{v}-1}{2e^{v}}$
$\displaystyle \Rightarrow y\frac{dv}{dy}=\frac{2ve^{v}-1}{2e^{v}}-v=-\frac{1}{2e^{v}}$
$\displaystyle \Rightarrow \frac{dv}{dy}=-\frac{1}{2ye^{v}}$
$\displaystyle \Rightarrow \frac{dy}{y}=-2e^{v}dv$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{dy}{y}=\int -2e^{v}dv$
$\displaystyle \Rightarrow \log y=-2e^{v}+C$
$\displaystyle \Rightarrow \log y+2e^{x/y}=C$
$\displaystyle \text{On applying the given condition } y(0)=1, \text{ we get}$
$\displaystyle 0+2e^{0}=C \Rightarrow C=2$
$\displaystyle \Rightarrow \log y+2e^{x/y}=2$
$\\$

$\displaystyle \textbf{Question 3: } \text{Solve the following differential equation } \\ x(x^{2}-1)\frac{dy}{dx}=1, y=0, \text{ given } x=2. \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given differential equation is } x(x^{2}-1)\frac{dy}{dx}=1 \text{ with } y(2)=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x(x^{2}-1)}$
$\displaystyle \Rightarrow dy=\frac{dx}{x(x-1)(x+1)}$
$\displaystyle \text{By partial fractions, we get}$
$\displaystyle \frac{1}{x(x-1)(x+1)}=-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}$
$\displaystyle \text{On integrating, we get}$
$\displaystyle \int dy=\int \left[-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right]dx$
$\displaystyle \Rightarrow y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+K \quad (i)$
$\displaystyle \text{On applying initial condition } y(2)=0, \text{ we get}$
$\displaystyle 0=-\log 2+\frac{1}{2}\log 1+\frac{1}{2}\log 3+K$
$\displaystyle \Rightarrow K=\log 2-\frac{1}{2}\log 3$
$\displaystyle \Rightarrow K=\log\left(\frac{2}{\sqrt{3}}\right)$
$\displaystyle \text{On substituting } K \text{ in Eq. (i), we get}$
$\displaystyle y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+\log\left(\frac{2}{\sqrt{3}}\right)$
$\displaystyle \Rightarrow y=\log\left(\frac{\sqrt{x^{2}-1}}{x}\right)+\log\left(\frac{2}{\sqrt{3}}\right)$
$\\$

$\displaystyle \textbf{Question 4: } \text{Solve the differential equation } \frac{dy}{dx}=\mathrm{cosec}\,y. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{dy}{dx}=\mathrm{cosec}\,y \Rightarrow \sin y\,dy=dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \sin y\,dy=\int dx$
$\displaystyle -\cos y=x+C$
$\displaystyle \Rightarrow x+\cos y+C=0$
$\\$

$\displaystyle \textbf{Question 5: } \text{Solve the differential equation } (1+y^{2})\,dx=(\tan^{-1}y-x)\,dy. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (1+y^{2})dx=(\tan^{-1}y-x)dy$
$\displaystyle \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^{2}}=\frac{\tan^{-1}y}{1+y^{2}}$
$\displaystyle \text{This is linear differential equation } \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where } P=\frac{1}{1+y^{2}},\ Q=\frac{\tan^{-1}y}{1+y^{2}}$
$\displaystyle \text{IF}=e^{\int Pdy}=e^{\int \frac{1}{1+y^{2}}dy}=e^{\tan^{-1}y}$
$\displaystyle \text{Solution: } x\cdot IF=\int Q\cdot IF\,dy$
$\displaystyle x e^{\tan^{-1}y}=\int \frac{\tan^{-1}y}{1+y^{2}}e^{\tan^{-1}y}dy$
$\displaystyle \text{Put } t=\tan^{-1}y \Rightarrow dt=\frac{dy}{1+y^{2}}$
$\displaystyle \Rightarrow x e^{t}=\int t e^{t}dt$
$\displaystyle \text{Integrating by parts: } \int te^{t}dt=te^{t}-e^{t}+C$
$\displaystyle \Rightarrow x e^{t}=te^{t}-e^{t}+C$
$\displaystyle \Rightarrow x=t-1+Ce^{-t}$
$\displaystyle \Rightarrow x=\tan^{-1}y-1+Ce^{-\tan^{-1}y}$
$\\$

$\displaystyle \textbf{Question 6: } \text{Solve the differential equation } (x^{2}-y^{2})\,dx+2xy\,dy=0. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (x^{2}-y^{2})dx+2xy\,dy=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}$
$\displaystyle \text{This is homogeneous; put } y=vx$
$\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\displaystyle v+x\frac{dv}{dx}=\frac{v^{2}-1}{2v}$
$\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-1}{2v}-v$
$\displaystyle =\frac{v^{2}-1-2v^{2}}{2v}=-\frac{v^{2}+1}{2v}$
$\displaystyle \Rightarrow \frac{2v}{v^{2}+1}dv=-\frac{dx}{x}$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \log(v^{2}+1)=-\log x+\log C$
$\displaystyle \Rightarrow \log[(v^{2}+1)x]=\log C$
$\displaystyle \Rightarrow (v^{2}+1)x=C$
$\displaystyle \Rightarrow \left(\frac{y^{2}}{x^{2}}+1\right)x=C$
$\displaystyle \Rightarrow \frac{y^{2}+x^{2}}{x}=C$
$\\$

$\displaystyle \textbf{Question 7: } \text{Solve the differential equation } (1+x^{2})\frac{dy}{dx}=4x^{2}-2xy. \quad \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (1+x^{2})\frac{dy}{dx}=4x^{2}-2xy$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2x}{1+x^{2}}y=\frac{4x^{2}}{1+x^{2}}$
$\displaystyle \text{This is linear differential equation } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\frac{2x}{1+x^{2}},\ Q=\frac{4x^{2}}{1+x^{2}}$
$\displaystyle \text{IF}=e^{\int \frac{2x}{1+x^{2}}dx}=e^{\log(1+x^{2})}=1+x^{2}$
$\displaystyle \text{Solution of the given differential equation is}$
$\displaystyle y(1+x^{2})=\int \frac{4x^{2}}{1+x^{2}}(1+x^{2})dx+C$
$\displaystyle =\int 4x^{2}dx+C$
$\displaystyle =\frac{4x^{3}}{3}+C$
$\displaystyle \Rightarrow y(1+x^{2})=\frac{4x^{3}}{3}+C$
$\\$

$\displaystyle \textbf{Question 8: } \text{Solve the differential equation } \frac{dy}{dx}=\frac{x+y+2}{2(x+y)-1}. \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}=\frac{x+y+2}{2(x+y)-1} \quad (i)$
$\displaystyle \text{On putting } x+y=u$
$\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{du}{dx}$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{du}{dx}-1$
$\displaystyle \text{From Eq. (i), we get}$
$\displaystyle \frac{du}{dx}-1=\frac{u+2}{2u-1}$
$\displaystyle \Rightarrow \frac{du}{dx}=\frac{u+2}{2u-1}+1=\frac{3u+1}{2u-1}$
$\displaystyle \Rightarrow \frac{2u-1}{3u+1}du=dx$
$\displaystyle =\left(\frac{2}{3}-\frac{5}{3(3u+1)}\right)du=dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \left(\frac{2}{3}-\frac{5}{3(3u+1)}\right)du=\int dx$
$\displaystyle \frac{2}{3}u-\frac{5}{9}\log|3u+1|=x+C$
$\displaystyle \Rightarrow \frac{2}{3}(x+y)-\frac{5}{9}\log|3(x+y)+1|=x+C$
$\displaystyle \Rightarrow x-\frac{2}{3}x-\frac{2}{3}y+\frac{5}{9}\log|3x+3y+1|+C=0$
$\displaystyle \Rightarrow \frac{x}{3}-\frac{2y}{3}+\frac{5}{9}\log|3x+3y+1|+C=0$
$\\$

$\displaystyle \textbf{Question 9: } \text{Solve } \sin x\frac{dy}{dx}-y= \sin x\tan\frac{x}{2}. \quad \text{ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \sin x\frac{dy}{dx}-y=\sin x\tan\frac{x}{2}$
$\displaystyle \Rightarrow \frac{dy}{dx}-y\,\mathrm{cosec}\,x=\tan\frac{x}{2}$
$\displaystyle \text{This is of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-\mathrm{cosec}\,x,\ Q=\tan\frac{x}{2}$
$\displaystyle \text{Now, IF}=e^{\int Pdx}=e^{-\int \mathrm{cosec}\,x\,dx}=e^{-\log\tan\frac{x}{2}}=\frac{1}{\tan\frac{x}{2}}$
$\displaystyle \text{Solution: } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow \frac{y}{\tan\frac{x}{2}}=\int 1\,dx+C$
$\displaystyle \Rightarrow \frac{y}{\tan\frac{x}{2}}=x+C$
$\displaystyle \Rightarrow y=x\tan\frac{x}{2}+C\tan\frac{x}{2}$
$\\$

$\displaystyle \textbf{Question 10: } \text{Solve the differential equation } x\frac{dy}{dx}+y=3x^{2}-2. \quad \text{ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, differential equation is } x\frac{dy}{dx}+y=3x^{2}-2$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=3x-\frac{2}{x}$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q,\ P=\frac{1}{x},\ Q=3x-\frac{2}{x}$
$\displaystyle \text{IF}=e^{\int \frac{1}{x}dx}=e^{\log x}=x$
$\displaystyle \text{Solution: } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow yx=\int (3x^{2}-2)dx+C$
$\displaystyle =x^{3}-2x+C$
$\displaystyle \Rightarrow y=x^{2}-2+\frac{C}{x}$
$\\$

$\displaystyle \textbf{Question 11: } \text{Solve: } (x^{2}-yx^{2})\,dy+(y^{2}+xy^{2})\,dx=0. \quad \text{ISC 2017, 01}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (x^{2}-yx^{2})dy+(y^{2}+xy^{2})dx=0$
$\displaystyle \Rightarrow x^{2}(1-y)dy+y^{2}(1+x)dx=0$
$\displaystyle \Rightarrow \frac{1-y}{y^{2}}dy+\frac{1+x}{x^{2}}dx=0$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \left(\frac{1-y}{y^{2}}\right)dy+\int \left(\frac{1+x}{x^{2}}\right)dx=0$
$\displaystyle \Rightarrow \int \left(\frac{1}{y^{2}}-\frac{1}{y}\right)dy+\int \left(\frac{1}{x^{2}}+\frac{1}{x}\right)dx=0$
$\displaystyle \Rightarrow -\frac{1}{y}-\log|y|-\frac{1}{x}+\log|x|=C_{1}$
$\displaystyle \Rightarrow -\left(\frac{1}{x}+\frac{1}{y}\right)=\log|y|-\log|x|+C_{1}$
$\displaystyle \Rightarrow \frac{x+y}{xy}=\log\left|\frac{x}{y}\right|+C$
$\\$

$\displaystyle \textbf{Question 12: } \text{Solve the differential equation } \\ x^{2}dy+(xy+y^{2})\,dx=0, \text{ when } x=1 \text{ and } y=1. \quad \text{ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, differential equation is } x^{2}dy+(xy+y^{2})dx=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{xy+y^{2}}{x^{2}} \quad (i)$
$\displaystyle \text{The given differential equation is homogeneous}$
$\displaystyle \text{Put } y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\displaystyle \text{Substituting in Eq. (i), we get}$
$\displaystyle v+x\frac{dv}{dx}=-\frac{x(vx)+v^{2}x^{2}}{x^{2}}=-(v+v^{2})$
$\displaystyle \Rightarrow x\frac{dv}{dx}=-v-v^{2}-v=-2v-v^{2}$
$\displaystyle \Rightarrow \frac{dv}{2v+v^{2}}=-\frac{dx}{x}$
$\displaystyle \Rightarrow \frac{dv}{v(v+2)}=-\frac{dx}{x}$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \left(\frac{1}{2v}-\frac{1}{2(v+2)}\right)dv=-\int \frac{dx}{x}$
$\displaystyle \Rightarrow \frac{1}{2}\log|v|-\frac{1}{2}\log|v+2|=-\log|x|+C$
$\displaystyle \Rightarrow \log\left|\frac{v}{v+2}\right|=-2\log|x|+C$
$\displaystyle \Rightarrow \frac{v}{v+2}=\frac{C}{x^{2}}$
$\displaystyle \Rightarrow \frac{y/x}{(y/x)+2}=\frac{C}{x^{2}}$
$\displaystyle \Rightarrow \frac{y}{y+2x}=\frac{C}{x^{2}}$
$\displaystyle \Rightarrow yx^{2}=C(y+2x) \quad (ii)$
$\displaystyle \text{Given that } y=1 \text{ when } x=1$
$\displaystyle \Rightarrow C=\frac{1}{3}$
$\displaystyle \text{Substituting in Eq. (ii), we get}$
$\displaystyle yx^{2}=\frac{1}{3}(y+2x)$
$\displaystyle \Rightarrow 3x^{2}y=y+2x$
$\displaystyle \Rightarrow y(3x^{2}-1)=2x$
$\\$

$\displaystyle \textbf{Question 13: } \text{Solve the differential equation } \\ e^{y}\left(1-\frac{x}{y}\right)+\left(1+e^{y}\right)\frac{dx}{dy}=0, \text{ when } x=0, y=1. \quad \text{ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, differential equation is } (1+e^{x/y})\frac{dx}{dy}+e^{x/y}\left(1-\frac{x}{y}\right)=0$
$\displaystyle \Rightarrow \frac{dx}{dy}=-\frac{e^{x/y}(1-\frac{x}{y})}{1+e^{x/y}} \quad (i)$
$\displaystyle \text{Put } \frac{x}{y}=v \Rightarrow x=vy,\ \frac{dx}{dy}=v+y\frac{dv}{dy} \quad (ii)$
$\displaystyle \text{From Eqs. (i) and (ii), we get}$
$\displaystyle v+y\frac{dv}{dy}=-\frac{e^{v}(1-v)}{1+e^{v}}$
$\displaystyle \Rightarrow y\frac{dv}{dy}=-\frac{e^{v}(1-v)}{1+e^{v}}-v$
$\displaystyle =-\frac{e^{v}-ve^{v}+v+ve^{v}}{1+e^{v}}=-\frac{v+e^{v}}{1+e^{v}}$
$\displaystyle \Rightarrow \frac{dv}{v+e^{v}}=-\frac{dy}{y(1+e^{v})}$
$\displaystyle \text{Rewriting, we get}$
$\displaystyle \frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{1+e^{v}}{v+e^{v}}dv=-\int \frac{dy}{y}$
$\displaystyle \text{Put } t=v+e^{v} \Rightarrow dt=(1+e^{v})dv$
$\displaystyle \Rightarrow \int \frac{dt}{t}=-\log|y|+C$
$\displaystyle \Rightarrow \log|v+e^{v}|=-\log|y|+\log|C|$
$\displaystyle \Rightarrow \log\left|y(v+e^{v})\right|=\log|C|$
$\displaystyle \Rightarrow y(v+e^{v})=C$
$\displaystyle \Rightarrow y\left(\frac{x}{y}+e^{x/y}\right)=C$
$\displaystyle \Rightarrow x+ye^{x/y}=C \quad (iii)$
$\displaystyle \text{Also, given } x=0,\ y=1$
$\displaystyle \Rightarrow C=1$
$\displaystyle \therefore x+ye^{x/y}=1$
$\\$

$\displaystyle \textbf{Question 14: } \text{Solve the differential equation } \sin^{-1}\left(\frac{dy}{dx}\right)=x+y. \quad \text{ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, differential equation can be rewritten as}$
$\displaystyle \frac{dy}{dx}=\sin(x+y) \quad (i)$
$\displaystyle \text{On putting } x+y=v \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$
$\displaystyle \text{Eq. (i) becomes,}$
$\displaystyle \frac{dv}{dx}-1=\sin v \Rightarrow \frac{dv}{dx}=1+\sin v$
$\displaystyle \Rightarrow \frac{dv}{1+\sin v}=dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{dv}{1+\sin v}=\int dx$
$\displaystyle \int \frac{1-\sin v}{1-\sin^{2}v}dv=\int dx$
$\displaystyle \Rightarrow \int \frac{1-\sin v}{\cos^{2}v}dv=\int dx$
$\displaystyle \Rightarrow \int (\sec^{2}v-\tan v\sec v)dv=\int dx$
$\displaystyle \tan v-\sec v=x+C$
$\displaystyle \Rightarrow \tan(x+y)-\sec(x+y)=x+C$
$\\$

$\displaystyle \textbf{Question 15: } \text{Solve the following differential equation } \\ (3xy+y^{2})\,dx+(x^{2}+xy)\,dy=0. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (3xy+y^{2})dx+(x^{2}+xy)dy=0$
$\displaystyle \Rightarrow (x^{2}+xy)dy=-(3xy+y^{2})dx$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{3xy+y^{2}}{x^{2}+xy}$
$\displaystyle \text{This is a homogeneous differential equation}$
$\displaystyle \text{Put } y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\displaystyle \Rightarrow v+x\frac{dv}{dx}=-\frac{3vx^{2}+v^{2}x^{2}}{x^{2}+vx^{2}}=-\frac{3v+v^{2}}{1+v}$
$\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{3v+v^{2}}{1+v}-v$
$\displaystyle =-\frac{3v+v^{2}+v+v^{2}}{1+v}=-\frac{4v+2v^{2}}{1+v}$
$\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{2v(v+2)}{1+v}$
$\displaystyle \Rightarrow \frac{1+v}{v(v+2)}dv=-\frac{2dx}{x} \quad (i)$
$\displaystyle \text{Using partial fractions: } \frac{1+v}{v(v+2)}=\frac{1}{2v}+\frac{1}{2(v+2)}$
$\displaystyle \text{Eq. (i) becomes}$
$\displaystyle \frac{1}{2}\left(\frac{1}{v}+\frac{1}{v+2}\right)dv=-\frac{2dx}{x}$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \frac{1}{2}\log|v|+\frac{1}{2}\log|v+2|=-2\log|x|+\log|C|$
$\displaystyle \Rightarrow \log|v(v+2)|=-4\log|x|+\log|C|$
$\displaystyle \Rightarrow \log\left|v(v+2)x^{4}\right|=\log|C|$
$\displaystyle \Rightarrow v(v+2)x^{4}=C$
$\displaystyle \Rightarrow \frac{y}{x}\left(\frac{y}{x}+2\right)x^{4}=C$
$\displaystyle \Rightarrow y(y+2x)x^{2}=C$
$\\$

$\displaystyle \textbf{Question 16: } \text{Solve the differential equation } (x+1)\,dy-2x\,dx=0. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (x+1)dy-2xy\,dx=0$
$\displaystyle \Rightarrow (x+1)dy=2xy\,dx$
$\displaystyle \Rightarrow \frac{dy}{y}=\frac{2x}{x+1}dx$
$\displaystyle =2\left(1-\frac{1}{x+1}\right)dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{dy}{y}=2\int \left(1-\frac{1}{x+1}\right)dx$
$\displaystyle \log y=2\left(x-\log|x+1|\right)+C$
$\displaystyle \Rightarrow \log y=2x-2\log|x+1|+C$
$\\$

$\displaystyle \textbf{Question 17: } \text{Solve the differential equation } \log\left(\frac{dy}{dx}\right)=2x-3y. \quad \text{ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \log\left(\frac{dy}{dx}\right)=2x-3y$
$\displaystyle \Rightarrow \frac{dy}{dx}=e^{2x-3y}$
$\displaystyle \Rightarrow e^{3y}dy=e^{2x}dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int e^{3y}dy=\int e^{2x}dx$
$\displaystyle \frac{e^{3y}}{3}=\frac{e^{2x}}{2}+C$
$\\$

$\displaystyle \textbf{Question 18: } \text{Solve the following differential equation } \\ ye^{y}\,dx=(y^{3}+2xe^{y})\,dy, \text{ given that } x=0, y=1. \quad \text{ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } ye^{y}dx=(y^{3}+2xe^{y})dy$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{y^{3}+2xe^{y}}{ye^{y}}$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{y^{2}}{e^{y}}+\frac{2x}{y}$
$\displaystyle \Rightarrow \frac{dx}{dy}-\frac{2}{y}x=y^{2}e^{-y} \quad (i)$
$\displaystyle \text{This is linear differential equation in } x$
$\displaystyle \text{Comparing with } \frac{dx}{dy}+Px=Q,\text{ we get}$
$\displaystyle P=-\frac{2}{y},\quad Q=y^{2}e^{-y}$
$\displaystyle \therefore \text{IF}=e^{\int Pdy}=e^{\int -\frac{2}{y}dy}=e^{-2\log y}=\frac{1}{y^{2}}$
$\displaystyle \text{Solution is } x\cdot IF=\int Q\cdot IF\,dy+C$
$\displaystyle \Rightarrow \frac{x}{y^{2}}=\int y^{2}e^{-y}\cdot \frac{1}{y^{2}}dy+C$
$\displaystyle \Rightarrow \frac{x}{y^{2}}=\int e^{-y}dy+C$
$\displaystyle \Rightarrow \frac{x}{y^{2}}=-e^{-y}+C \quad (ii)$
$\displaystyle \text{Also, given } x=0,\ y=1$
$\displaystyle \Rightarrow 0=-e^{-1}+C \Rightarrow C=e^{-1}$
$\displaystyle \text{On substituting in Eq. (ii), we get}$
$\displaystyle \frac{x}{y^{2}}=-e^{-y}+e^{-1}$
$\displaystyle \Rightarrow x=y^{2}(e^{-1}-e^{-y})$
$\\$

$\displaystyle \textbf{Question 19: } \text{Solve the differential equation } \\ (xy^{2}+x)\,dx+(x^{2}y+y)\,dy=0. \quad \text{ISC 2012}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (xy^{2}+x)dx+(x^{2}y+y)dy=0$
$\displaystyle \Rightarrow x(y^{2}+1)dx+y(x^{2}+1)dy=0$
$\displaystyle \Rightarrow \frac{x}{x^{2}+1}dx+\frac{y}{y^{2}+1}dy=0$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{x}{x^{2}+1}dx+\int \frac{y}{y^{2}+1}dy=0$
$\displaystyle \frac{1}{2}\log(x^{2}+1)+\frac{1}{2}\log(y^{2}+1)=C$
$\displaystyle \Rightarrow \log[(x^{2}+1)(y^{2}+1)]=C$
$\displaystyle \frac{1}{2}\log|x^{2}+1|+\frac{1}{2}\log|y^{2}+1|=C$
$\displaystyle \left[\because \int \frac{2x}{x^{2}+1}dx=\log|x^{2}+1|\right]$
$\displaystyle \Rightarrow \log\sqrt{x^{2}+1}+\log\sqrt{y^{2}+1}=C$
$\displaystyle \Rightarrow \log\left(\sqrt{x^{2}+1}\sqrt{y^{2}+1}\right)=C$
$\displaystyle \Rightarrow \sqrt{x^{2}+1}\sqrt{y^{2}+1}=e^{C}$
$\displaystyle \left[\because \log a=C \Rightarrow a=e^{C}\right]$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle (x^{2}+1)(y^{2}+1)=e^{2C}$
$\displaystyle \Rightarrow (x^{2}+1)(y^{2}+1)=C'$
$\\$

$\displaystyle \textbf{Question 20: } \text{Solve the differential equation } \frac{dy}{dx}=e^{x+y}+x^{2}e^{y}. \quad \text{ISC 2011}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{dy}{dx}=e^{x+y}+x^{2}e^{y}$
$\displaystyle \Rightarrow \frac{dy}{dx}=e^{y}(e^{x}+x^{2})$
$\displaystyle \Rightarrow e^{-y}dy=(e^{x}+x^{2})dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int e^{-y}dy=\int (e^{x}+x^{2})dx$
$\displaystyle -e^{-y}=e^{x}+\frac{x^{3}}{3}+C$
$\displaystyle \Rightarrow e^{-y}=-e^{x}-\frac{x^{3}}{3}+C$
$\\$

$\displaystyle \textbf{Question 22: } \text{Solve the differential equation } (x\cos y)\,dy=e^{x}(x\log x+1)\,dx. \quad \text{ISC 2009}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (\cos y)dy=e^{x}(x\log x+1)dx$
$\displaystyle \Rightarrow \cos y\,dy=e^{x}\left(x\log x+1\right)dx$
$\displaystyle =e^{x}\left[\log x+\frac{d}{dx}(\log x)\cdot x\right]dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \cos y\,dy=\int e^{x}\log x\,dx+\int \frac{e^{x}}{x}dx$
$\displaystyle \text{Using integration by parts on } \int e^{x}\log x\,dx$
$\displaystyle \int e^{x}\log x\,dx=e^{x}\log x-\int \frac{e^{x}}{x}dx$
$\displaystyle \therefore \sin y=e^{x}\log x+C$
$\\$

$\displaystyle \textbf{Question 23: } \text{Solve the differential equation } (x+1)\frac{dy}{dx}-y=e^{3x}(x+1)^{2}. \quad \text{ISC 2008}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (x+1)\frac{dy}{dx}-y=e^{3x}(x+1)^{2}$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x+1}=e^{3x}(x+1)$
$\displaystyle \text{This is of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-\frac{1}{x+1},\ Q=e^{3x}(x+1)$
$\displaystyle \text{On comparing the above equation with } \frac{dy}{dx}+Py=Q,\text{ we get}$
$\displaystyle P=-\frac{1}{x+1},\quad Q=e^{3x}(x+1)$
$\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{-\int \frac{1}{x+1}dx}=e^{-\log|x+1|}=\frac{1}{x+1}$
$\displaystyle \text{Solution is } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow \frac{y}{x+1}=\int e^{3x}(x+1)\cdot \frac{1}{x+1}\,dx+C$
$\displaystyle \Rightarrow \frac{y}{x+1}=\int e^{3x}\,dx+C$
$\displaystyle \Rightarrow \frac{y}{x+1}=\frac{e^{3x}}{3}+C$
$\displaystyle \Rightarrow y=\frac{e^{3x}}{3}(x+1)+C(x+1)$
$\\$

$\displaystyle \textbf{Question 24: } \text{Solve the differential equation } \tan y\,dx+\sec^{2}y\tan x\,dy=0. \quad \text{ISC 2007}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \tan y\,dx+\sec^{2}y\tan x\,dy=0$
$\displaystyle \Rightarrow \tan y\,dx=-\sec^{2}y\tan x\,dy$
$\displaystyle \Rightarrow \frac{dx}{\tan x}=-\frac{\sec^{2}y}{\tan y}dy$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle -\int \cot x\,dx=-\int \frac{\sec^{2}y}{\tan y}dy$
$\displaystyle -\log|\sin x|=-\log|\tan y|+C$
$\displaystyle \Rightarrow \log|\sin x|=\log|\tan y|+C$
$\displaystyle \Rightarrow \sin x=C\tan y$
$\\$

$\displaystyle \textbf{Question 25: } \text{Solve the differential equation } \\ x(x^{2}-x^{2}y^{2})\,dy+y(y^{2}+x^{2}y^{2})\,dx=0. \quad \text{ISC 2006}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } x(x^{2}-x^{2}y^{2})dy+y(y^{2}+x^{2}y^{2})dx=0$
$\displaystyle \Rightarrow x^{3}(1-y^{2})dy+y^{3}(1+x^{2})dx=0$
$\displaystyle \Rightarrow \frac{1-y^{2}}{y^{3}}dy+\frac{1+x^{2}}{x^{3}}dx=0$
$\displaystyle \Rightarrow \left(\frac{1}{y^{3}}-\frac{1}{y}\right)dy+\left(\frac{1}{x^{3}}+\frac{1}{x}\right)dx=0$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \left(\frac{1}{y^{3}}-\frac{1}{y}\right)dy+\int \left(\frac{1}{x^{3}}+\frac{1}{x}\right)dx=0$
$\displaystyle -\frac{1}{2y^{2}}-\log|y|-\frac{1}{2x^{2}}+\log|x|=C$
$\displaystyle \Rightarrow \log\left|\frac{x}{y}\right|=C+\frac{1}{2x^{2}}+\frac{1}{2y^{2}}$
$\\$

$\displaystyle \textbf{Question 26: } \text{Solve the differential equation } \frac{dy}{dx}-e^{y+x}=e^{x-y}. \quad \text{ISC 2005}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{dy}{dx}=e^{x+y}+e^{x-y}$
$\displaystyle \Rightarrow \frac{dy}{dx}=e^{x}(e^{y}+e^{-y})$
$\displaystyle \Rightarrow \frac{dy}{e^{y}+e^{-y}}=e^{x}dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{dy}{e^{y}+e^{-y}}=\int e^{x}dx$
$\displaystyle \text{Let } e^{y}=t \Rightarrow dy=\frac{dt}{t}$
$\displaystyle \therefore \int \frac{dy}{e^{y}+e^{-y}}=\int \frac{dt}{t^{2}+1}$
$\displaystyle =\tan^{-1}t+C=\tan^{-1}(e^{y})+C$
$\displaystyle \Rightarrow \tan^{-1}(e^{y})=e^{x}+C$
$\displaystyle \Rightarrow \frac{e^{y}\,dy}{1+e^{2y}}=e^{x}dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \frac{e^{y}}{1+e^{2y}}\,dy=\int e^{x}dx$
$\displaystyle \text{Putting } e^{y}=t \Rightarrow e^{y}dy=dt$
$\displaystyle \Rightarrow \int \frac{dt}{1+t^{2}}=e^{x}+C$
$\displaystyle \Rightarrow \tan^{-1}t=e^{x}+C$
$\displaystyle \Rightarrow \tan^{-1}(e^{y})=e^{x}+C$
$\\$

$\displaystyle \textbf{Question 27: } \text{Solve the differential equation } \\ (1-x^{2})\frac{dy}{dx}-xy=x, \text{ gives } y=2, \text{ when } x=0. \quad \text{ISC 2004}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (1-x^{2})\frac{dy}{dx}-xy=x$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{x}{1-x^{2}}y=\frac{x}{1-x^{2}}$
$\displaystyle \text{On comparing with } \frac{dy}{dx}+Py=Q,\text{ we get}$
$\displaystyle P=-\frac{x}{1-x^{2}},\quad Q=\frac{x}{1-x^{2}}$
$\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int \frac{-x}{1-x^{2}}dx}=e^{\frac{1}{2}\log|1-x^{2}|}=\sqrt{1-x^{2}}$
$\displaystyle \text{Solution is } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow y\sqrt{1-x^{2}}=\int \frac{x}{1-x^{2}}\sqrt{1-x^{2}}\,dx+C$
$\displaystyle \Rightarrow y\sqrt{1-x^{2}}=\int \frac{x}{\sqrt{1-x^{2}}}\,dx+C$
$\displaystyle \text{On putting } t=\sqrt{1-x^{2}}\Rightarrow t^{2}=1-x^{2}$
$\displaystyle \Rightarrow 2t\,dt=-2x\,dx \Rightarrow t\,dt=-x\,dx$
$\displaystyle \therefore y\sqrt{1-x^{2}}=-\int dt+C$
$\displaystyle \Rightarrow y\sqrt{1-x^{2}}=-t+C$
$\displaystyle \Rightarrow y\sqrt{1-x^{2}}=-\sqrt{1-x^{2}}+C \quad (i)$
$\displaystyle \text{Given, } y=2 \text{ when } x=0$
$\displaystyle \Rightarrow 2\cdot 1=-1+C$
$\displaystyle \Rightarrow C=3$
$\displaystyle \text{On putting } C=3 \text{ in Eq. (i), we get}$
$\displaystyle y\sqrt{1-x^{2}}=-\sqrt{1-x^{2}}+3$
$\displaystyle \Rightarrow y\sqrt{1-x^{2}}+\sqrt{1-x^{2}}=3$
$\\$

$\displaystyle \textbf{Question 28: } \text{Solve the differential equation } \frac{dy}{dx}=\frac{(1+\cos^{2}x)\sin^{2}y}{(1+\sin^{2}y)\cos^{2}x}. \quad \text{ISC 2003}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{dy}{dx}=\frac{(1+\cos^{2}x)\sin^{2}y}{(1+\sin^{2}y)\cos^{2}x}$
$\displaystyle \Rightarrow \frac{(1+\sin^{2}y)}{\sin^{2}y}\,dy=\frac{(1+\cos^{2}x)}{\cos^{2}x}\,dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int (\mathrm{cosec}^{2}y+1)\,dy=\int (\sec^{2}x+1)\,dx$
$\displaystyle \Rightarrow -\cot y+y=\tan x+x+C$
$\displaystyle \Rightarrow y-x=\tan x+\cot y+C$
$\\$

$\displaystyle \textbf{Question 29: } \text{Solve } (1+y+x^{2}y)\,dx+(x+x^{3})\,dy=0. \quad \text{ISC 2003}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (1+y+x^{2}y)dx+(x+x^{3})dy=0$
$\displaystyle \Rightarrow 1+y(1+x^{2})+x(1+x^{2})\frac{dy}{dx}=0$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{x}=-\frac{1}{x(1+x^{2})}$
$\displaystyle \text{On comparing with } \frac{dy}{dx}+Py=Q,\text{ we get}$
$\displaystyle P=\frac{1}{x},\quad Q=-\frac{1}{x(1+x^{2})}$
$\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int \frac{1}{x}dx}=e^{\log|x|}=x$
$\displaystyle \text{Solution is } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow yx=\int \left[-\frac{1}{x(1+x^{2})}\right]x\,dx+C$
$\displaystyle \Rightarrow yx=-\int \frac{dx}{1+x^{2}}+C$
$\displaystyle \Rightarrow yx=-\tan^{-1}x+C$
$\\$

$\displaystyle \textbf{Question 30: } \text{Solve } e^{y}(1+x^{2})\,dy-\frac{x}{y}\,dx=0. \quad \text{ISC 2002}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } e^{y}(1+x^{2})\,dy-\frac{x}{y}\,dx=0$
$\displaystyle \Rightarrow y e^{y}dy=\frac{x}{1+x^{2}}dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int y e^{y}dy=\int \frac{x}{1+x^{2}}dx$
$\displaystyle \Rightarrow y\int e^{y}dy-\int \left[\frac{d}{dy}y\right]\left[\int e^{y}dy\right]dy=\frac{1}{2}\log(1+x^{2})+C$
$\displaystyle \Rightarrow ye^{y}-\int e^{y}dy=\log\sqrt{1+x^{2}}+C$
$\displaystyle \Rightarrow ye^{y}-e^{y}=\log\sqrt{1+x^{2}}+C$
$\displaystyle \Rightarrow e^{y}(y-1)-\log\sqrt{1+x^{2}}=C$
$\\$

$\displaystyle \textbf{Question 31: } \text{Solve } \cos^{2}x\frac{dy}{dx}+y=\tan x. \quad \text{ISC 2002}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \cos^{2}x\frac{dy}{dx}+y=\tan x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{\cos^{2}x}=\frac{\tan x}{\cos^{2}x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\sec^{2}x\,y=\tan x\sec^{2}x$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q,\text{ we get}$
$\displaystyle P=\sec^{2}x,\quad Q=\tan x\sec^{2}x$
$\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int \sec^{2}x\,dx}=e^{\tan x}$
$\displaystyle \text{Solution is } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow y e^{\tan x}=\int \tan x\sec^{2}x\,e^{\tan x}dx+C$
$\displaystyle \text{Put } t=\tan x \Rightarrow dt=\sec^{2}x\,dx$
$\displaystyle \Rightarrow y e^{t}=\int t e^{t}dt+C$
$\displaystyle \text{Integrating by parts: } \int t e^{t}dt=te^{t}-e^{t}+C$
$\displaystyle \Rightarrow y e^{t}=te^{t}-e^{t}+C$
$\displaystyle \Rightarrow y=t-1+Ce^{-t}$
$\displaystyle \Rightarrow y=\tan x-1+Ce^{-\tan x}$
$\\$

$\displaystyle \textbf{Question 32: } \text{Solve } \frac{dy}{dx}+\frac{2x}{1+x^{2}}y= \frac{1}{(1+x^{2})^{2}}. \quad \text{ISC 2001}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{dy}{dx}+\frac{2x}{1+x^{2}}y=\frac{1}{(1+x^{2})^{2}}$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q,\text{ we get}$
$\displaystyle P=\frac{2x}{1+x^{2}},\quad Q=\frac{1}{(1+x^{2})^{2}}$
$\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int \frac{2x}{1+x^{2}}dx}=e^{\log(1+x^{2})}=1+x^{2}$
$\displaystyle \text{Solution is } y\cdot IF=\int Q\cdot IF\,dx+C$
$\displaystyle \Rightarrow y(1+x^{2})=\int \frac{1}{(1+x^{2})^{2}}(1+x^{2})dx+C$
$\displaystyle \Rightarrow y(1+x^{2})=\int \frac{dx}{1+x^{2}}+C$
$\displaystyle \Rightarrow y(1+x^{2})=\tan^{-1}x+C$
$\\$

$\displaystyle \textbf{Question 33: } \text{Solve } (x-y-2)\,dx-(2x-2y-3)\,dy=0. \quad \text{ISC 2001}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (x-y-2)dx-(2x-2y-3)dy=0$
$\displaystyle \Rightarrow (x-y-2)-[2(x-y-2)+1]\frac{dy}{dx}=0$
$\displaystyle \text{On putting } x-y-2=t$
$\displaystyle \Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}$
$\displaystyle \Rightarrow \frac{dy}{dx}=1-\frac{dt}{dx}$
$\displaystyle \text{Therefore, given differential equation becomes}$
$\displaystyle t-(2t+1)\left(1-\frac{dt}{dx}\right)=0$
$\displaystyle \Rightarrow t-2t-1+(2t+1)\frac{dt}{dx}=0$
$\displaystyle \Rightarrow -t-1+(2t+1)\frac{dt}{dx}=0$
$\displaystyle \Rightarrow (2t+1)\frac{dt}{dx}=t+1$
$\displaystyle \Rightarrow dx=\frac{2t+1}{t+1}dt$
$\displaystyle \Rightarrow dx=\left(2-\frac{1}{t+1}\right)dt$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int dx=\int \left(2-\frac{1}{t+1}\right)dt$
$\displaystyle \Rightarrow x=2t-\log|t+1|+C$
$\displaystyle \Rightarrow x=2(x-y-2)-\log|x-y-1|+C$
$\displaystyle \Rightarrow 2y-x+4+\log|x-y-1|=C$
$\\$

$\displaystyle \textbf{Question 34: } \text{Solve } \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}. \quad \text{ISC 2001}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$
$\displaystyle \Rightarrow \frac{dy}{dx}=(e^{x}+x^{2})e^{-y}$
$\displaystyle \Rightarrow e^{y}dy=(e^{x}+x^{2})dx$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int e^{y}dy=\int (e^{x}+x^{2})dx$
$\displaystyle \Rightarrow e^{y}=e^{x}+\frac{x^{3}}{3}+C$
$\\$

$\displaystyle \textbf{Question 35: } \text{Solve } x(x-y)\,dy+y^{2}\,dx=0. \quad \text{ISC 2000}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } x(x-y)\frac{dy}{dx}+y^{2}=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{y^{2}}{x(x-y)}$
$\displaystyle =-\frac{y^{2}}{x^{2}-xy}$
$\displaystyle \text{Since degrees are same, it is homogeneous; put } y=vx$
$\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\displaystyle v+x\frac{dv}{dx}=-\frac{v^{2}x^{2}}{x^{2}(1-v)}=-\frac{v^{2}}{1-v}$
$\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{2}}{1-v}-v$
$\displaystyle =-\frac{v^{2}+v(1-v)}{1-v}=-\frac{v}{1-v}$
$\displaystyle \Rightarrow \frac{1-v}{v}dv=-\frac{dx}{x}$
$\displaystyle \text{On integrating both sides, we get}$
$\displaystyle \int \left(\frac{1}{v}-1\right)dv=-\int \frac{dx}{x}$
$\displaystyle \Rightarrow \log|v|-v=-\log|x|+C$
$\displaystyle \Rightarrow \log|v|+\log|x|-v=C$
$\displaystyle \Rightarrow \log|y|- \frac{y}{x}=C$
$\\$