Problem 02 | Integrating Factors Found by Inspection

$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$

$y^4 \, dx - xy \, dx + xy^3 \, dy + x^2 \, dy = 0$

$(y^4 \, dx + xy^3 \, dy) - (xy \, dx - x^2 \, dy) = 0$

$y^3(y \, dx + x \, dy) - x(y \, dx - x \, dy) = 0$

$y^3 \, d(xy) - x(y \, dx - x \, dy) = 0$
 

Divide by y³
$d(xy) - \dfrac{x}{y} \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

$d(xy) - \dfrac{x}{y} \, d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle \int d(xy) - \int \left( \dfrac{x}{y} \right) \, d \left( \dfrac{x}{y} \right) = 0$

$xy - \dfrac{1}{2}\left( \dfrac{x}{y} \right)^2 = \dfrac{c}{2}$

$xy - \dfrac{x^2}{2y^2} = \dfrac{c}{2}$

$2xy^3 - x^2 = cy^2$           answer