Answer to Question #229242 in Differential Equations for samsam

Question

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

Answer 1

$\displaystyle 2x^3y' = y(2x^2 - y^2) \\ \textsf{Substitute}\,\,\, y = vx\\ 2x^3(xv' + v) = 2x^2\cdot vx - v^3 x^3\\ 2(xv' + v) = 2v - v^3\\ 2xv' = -v^3\\ 2x \frac{\mathrm{d}v}{\mathrm{d}x} = -v^3\\ \frac{\mathrm{d}v}{v^3} = -\frac{\mathrm{d}x}{2x}\\ -\int \frac{2\mathrm{d}v}{v^3} = \int\frac{\mathrm{d}x}{x}\\ \frac{1}{v^2} = \ln{x} + C\\ v^2 = \frac{1}{\ln{x} + C}\\ \frac{y^2}{x^2} = \frac{1}{\ln{x} + C}\\ \implies y^2 = \frac{x^2}{\ln{x} + C}$