Question

Solve the following differentiate:
$x(x-y)dy+y^{2}dx=0$.

Answer 1

Given,

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

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

$\frac{1}{x^2}\frac{dx}{dy}-\frac{1}{xy}=-\frac{1}{y^2}$

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

$\frac{dv}{dy}+\frac{v}{y}=-\frac{1}{y^2}$

$I.F=e^{\int \frac{1}{y}}dy=y$

$vy=\int -\frac{1}{y^2}ydy+c$

$vy=-\log y+\log c$

$vy=\log \frac{c}{y}$

$e^{vy}=\frac{c}{y}$

$e^{-\frac{y}{x}}=\frac{c}{y}$