Answer to Question #228823 in Differential Equations for frost

Question

Solve the following homogeneous differential equation 2x³ y^' = y(2x² − y²) 

Answer 1

$2x^3y'=2x^2y-y^3\\ \frac{dy}{dx}=\frac{2x^2y-y^3}{2x^3}\\ \text{put\,} y=vx\\ v+x\frac{dv}{dx}=\frac{2v-v^3}{2}\\ v+x\frac{dv}{dx}=v-\frac{v^3}{2}\\ x\frac{dv}{dx}=-\frac{v^3}{2}\\ 2\frac{dv}{v^3}=-\frac{dx}{x}\\ \frac{-1}{v^2}=-lnx+c\\ v^2=\frac{-1}{c-lnx}\\ \frac{y^2}{x^2}=\frac{1}{lnx+C}\space(put\space C=-c)\\ \text{This is the required solution.} \\$