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I would like to solve this DE, but I don't know how.

$y' = (x-2y)^4 + \dfrac{1}{2} $

Thank you for your help in advance.

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up vote 3 down vote accepted

Let $v = x-2y$. We then have that $$\dfrac{dv}{dx} = 1 - 2 \dfrac{dy}{dx}$$ Hence, the differential equation becomes $$\dfrac{1-v'}2 = v^4 + \dfrac12$$ $$v' = -2v^4$$ Hence, $$\dfrac{dv}{v^4} = -2dx \implies \dfrac{v^{-3}}{-3} = -2x + c$$ Hence, $$v^3 = \dfrac1{6x+k}$$ i.e. $$v = \dfrac1{(6x+k)^{1/3}}$$ Hence, \begin{align} x - 2y & = \dfrac1{(6x+k)^{1/3}}\\ y & = \dfrac{x}2-\dfrac1{2(6x+k)^{1/3}} \end{align}

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Let $z=2y-x$. Then $z~'=2z^4$, which is easier to solve.

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@BabakSorouh, I think there is no negative sign. :) – 23rd Oct 31 '12 at 19:06
Oh sorry. I though of $z=x-2y$ as the OP noted. Excuses and +1. – Babak S. Oct 31 '12 at 19:23

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