How to solve this ODE?

• AdrianZ
Ce^{y^2/x}=-2/xe^{y^2/x}=-2CxIn summary, the given equation is a Bernoulli differential equation which can be solved by rewriting it in terms of a new variable u and using the integrating factor μ=e^(2y^2/x). The final solution is given by e^(y^2/x)=-2Cx.

Homework Statement

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

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?

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.

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.
$$-x^{-2}+(2/y) x^{-1}=(2/y^3 )$$
$$u=x^{-1}:du/dx=-x^{-2}$$
$${du/dx} {dx/dy}+(2/y)u=(2/y^3 )$$
$$du/dy+(2/y)u=(2/y^3 )$$
$$μ=e^{∫2dy/y}=y^2$$
$$uy^2=∫y^2 (2/y^3 )dy$$
$$Ce^{y^2/x}=y^2$$