x = vy

Problem 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Solution 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Let
$x = vy$

$dx = v \, dy + y \, dv$
 

$vy^2(v \, dy + y \, dv) - (v^2y^2 + 3y^2) \, dy = 0$

$vy^2(v \, dy + y \, dv) - y^2(v^2 + 3) \, dy = 0$

$v(v \, dy + y \, dv) - (v^2 + 3) \, dy = 0$

$v^2 \, dy + vy \, dv - v^2 \, dy - 3 \, dy = 0$

$vy \, dv - 3 \, dy = 0$

$v \, dv - \dfrac{3 \, dy}{y} = 0$

$\displaystyle \int v \, dv - 3 \int \dfrac{dy}{y} = 0$