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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.

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  • $\begingroup$ 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. $\endgroup$ – Winther Nov 17 '14 at 6:43
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See here and here. This question is fascinating, where did you find this equation? Is it related to some geometry?

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

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