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