Solve differential equation 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.

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}
