Question

Find particular solution:

$(x-y)\frac{dy}{dx}=(x+2y)$, given $y=0$ when $x=1$

Answer 1

Given,

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

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

substitute $y=vx\to \frac{dy}{dx}=v+x\frac{dv}{dx}$

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

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

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

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

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

integrating on both sides, we get,

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

$\int \frac{2-2v}{v^2+v+1}dv=2\log x+c$

$\int \frac{2-2v}{v^2+v+1}dv=\log x^2+c$

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

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

substitute $y=0,x=1$

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

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

$c=2\sqrt{3}\cdot \frac{\pi }{6}=\frac{\pi }{\sqrt{3}}$

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

$\therefore 6{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\sqrt{3}\log \left|x^2\frac{y^2+yx+x^2}{x^2}\right|+\pi$