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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.

2 Answers 2

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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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    Oh sorry. I though of $z=x-2y$ as the OP noted. Excuses and +1.2012-10-31