# Help with an ODE

I need some help, I have this ODE but can't solve it for $$y(x)$$, I try every method I know, but with no success,please, somebody can help me?

$$(\varepsilon-x)y=y'(-x+y^2-2x^2)$$

Thanks.

• This equation only has an implict solution: $2\log y + 2\epsilon\log(x + 2 x\epsilon - y^2) - (1+2\epsilon)\log(\epsilon + 2 x\epsilon - y^2) = C$ where $C$ is an integration constant. – Winther Nov 17 '14 at 6:43

See here and here. This question is fascinating, where did you find this equation? Is it related to some geometry?

• (your two links are the same.) – Calvin Khor May 5 at 19:05
• Sorry, will edit it. – Lada Dudnikova May 5 at 19:06

Hint:

Let $$u=y^2$$ ,

Then $$\dfrac{du}{dx}=2y\dfrac{dy}{dx}$$

$$\therefore(\varepsilon-x)y=\dfrac{1}{2y}\dfrac{du}{dx}(-x+y^2-2x^2)$$

$$(y^2-2x^2-x)\dfrac{du}{dx}=2(\varepsilon-x)y^2$$

$$(u-2x^2-x)\dfrac{du}{dx}=2(\varepsilon-x)u$$

This belongs to an Abel equation of the second kind.