Question

Solve the differential equation: $x(x-y)dy+y^{2}dx=0$

Answer 1

Given differential equation is $x(x-y)dy+y2dx=0$

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

Putting $y=vx$

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

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

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

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

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

On integrating both sides

$\log x=v-\log v+c$

Putting $v=\frac{y}{x}$

$\log y=\frac{y}{x}+c$

On $e$ both sides

$e^{\log y}=e^{y/x+c}$

$y=c{e}^{y/x}$