Question

Find $\frac{dy}{dx}$
$(x^3-2y^3)dx+3x{y}^{2}dy=0$

Answer 1

Consider the given equation.

$(x^3-2y^3)dx+3x{y}^{2}dy=0$

$(x^3-2y^3)dx=-3x{y}^{2}dy$

$\frac{dy}{dx}=-\frac{x^3-2y^3}{3x{y}^2}$

$\frac{dy}{dx}=\frac{2y^3-x^3}{3x{y}^2}$

Hence, this is the answer.