Question

Find the particular solution of the differential equation $(x-y)\frac{dy}{dx}=x+2y$, given that when $x=1,y=0$.

Answer 1

Given $(x-y)\frac{dy}{dx}=x+2y$

Let $y=vx\Rightarrow \frac{dy}{dx}=x\frac{dy}{dx}+v$

$\frac{dy}{dx}=\frac{x+2y}{x+y}=\frac{x+2vx}{x-yx}=\frac{1+2v}{1-v}$

$\Rightarrow x\frac{dv}{dx}+v=\frac{1+2v}{1-v}$

$x\frac{dv}{dx}=\frac{1+2v}{1-v}-v=\frac{1+2v-v+v^2}{1-v}$

$x\frac{dv}{dx}=\frac{v^2+v+1}{1-v}$

$dv\left(\frac{v-1}{v^2+v+1}\right)=\frac{-dx}{x}$

$dv\left(\frac{(v+\frac{1}{2})-\frac{3}{2}}{{(v+\frac{1}{2})}^2+\frac{3}{4}}\right)=\frac{-dx}{x}$

A integrating on both sides

$\int \frac{v+\frac{1}{2}}{{(v+\frac{1}{2})}^2+\frac{3}{4}}-\frac{3}{2}\int \frac{1}{{(v+\frac{1}{2})}^2+\frac{3}{4}}=-\int \frac{dx}{x}$

$\Rightarrow \frac{1}{2}\log |v^2+v+1|-\frac{3}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{v+\frac{1}{2}}{\sqrt{3}/2}\right)=-\log |x|+c$

$\Rightarrow$ Put $v=y/x$

$\frac{1}{2}\log |\frac{y^2}{x^2}+\frac{y}{x}+1|-\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=-\log 1+1+c$

$y=0,x=1$

$\frac{1}{2}\log |0+0+1|-\sqrt{3}{\tan }^{-1}\left(\frac{2\left(0\right)+1}{\sqrt{3}}\right)=-\log \left(1\right)+c$

$-\sqrt{3}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=c$

$\Rightarrow c=-\sqrt{3}\times \frac{\pi }{6}=\frac{-\pi }{2\sqrt{3}}$