substitution

Problem 01
$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
 

Solution 01
$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

Substitute,
$3(3x^2 + v^2x^2) \, dx - 2vx^2 (v \, dx + x \, dv) = 0$

$3(3 + v^2)x^2 \, dx - 2vx^2 (v \, dx + x \, dv) = 0$
 

Divide by x²,
$3(3 + v^2) \, dx - 2v (v \, dx + x \, dv) = 0$

$9 \, dx + 3v^2 \, dx - 2v^2 \, dx - 2vx \, dv = 0$

$9 \, dx + v^2 \, dx - 2vx \, dv = 0$

$(9 + v^2) \, dx - 2vx \, dv = 0$