Question

Find the general solution of $y^{2}dx+(x^2-xy+y^2)dy=0$.

Answer 1

$y^{2}dx+(x^2-xy+y^2)dy=0$

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

$\frac{dy}{dx}=\frac{-(y/x{)}^2}{1-\left(y/x\right)+(y/x{)}^2}$

Momogeneous equation,

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

$\frac{dy}{dx}=\frac{d}{dx}\left(vx\right)$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

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

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

$\frac{-v^2-v(1-v+v^2)}{1-v+v^2}$

$=\frac{v^2-v+v^2-v^3}{1-v+v^2}=\frac{v(1+v^2)}{1-v+v^2}$

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

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

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

integrating

$-lnx=-{\tan }^{-1}v+lnv+c$

$\therefore lnv+lnx-{\tan }^{-1}v=c$

$lnvx-{\tan }^{-1}v=c$

$lny-{\tan }^{-1}\frac{y}{x}=c$ [by placing $v=\frac{y}{x}$]