# How to solve this ODE?

1. Nov 4, 2011

1. The problem statement, all variables and given/known data
$$2xy'(x-y^2)+y^3=0$$
2. Relevant equations

3. 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?

2. Nov 4, 2011

### 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.

3. Nov 4, 2011

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