How to solve this ODE?

[AdrianZ]

Homework Statement

$$2xy'(x-y^2)+y^3=0$$

Homework Equations

The Attempt at a Solution

What kind of an equation is that? I first thought that might be a Bernoulli differential equation with respect to x but I failed to convert it that form. I also checked if the equation could have single variable integrating factors but the answer is negative. How can I solve it?

 

[jackmell]

I would approach it parametrically:

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

so:

$$\frac{dy}{dt}=-y^3$$

$$\frac{dx}{dt}=2x(x-y^2)$$

Note the first is de-coupled so solve that one and substitute it into the second one. Maybe though there is something I'm missing and would allow an easier approach.

 

[AdrianZ]

Thanks for your help jackmell but I already solved it as a Bernoulli equation with respect to x. I'll write down my answer so that it might be useful for someone else.
$$2xy^' (x-y^2 )+y^3=0$$
$$2x(x-y^2 )+y^3 x^'=0$$
$$x^'+(2x^2-2xy^2)/y^3 =0$$
$$x^'+(-2/y)x=(-2/y^3 ) x^2$$

so it's a Bernoulli equation and can be solved easily.
$$-x^{-2}+(2/y) x^{-1}=(2/y^3 )$$
$$u=x^{-1}:du/dx=-x^{-2}$$
$${du/dx} {dx/dy}+(2/y)u=(2/y^3 )$$
$$du/dy+(2/y)u=(2/y^3 )$$
$$μ=e^{∫2dy/y}=y^2$$
$$uy^2=∫y^2 (2/y^3 )dy$$
$$Ce^{y^2/x}=y^2$$