Solve the differential equation $\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$

This is an "almost" exact equation. Write it in $A(x,y)\text dx + B(x,y)\text dy=0$ form

$$(-y-2y^5)\text dx + (4x+y^4)\text dy = 0$$

Then, look for a function $\mu(y)$ such that

$$ (-y-2y^5)\mu(y)\text dx + (4x+y^4)\mu(y)\text dy = 0 $$

Let $\mu(y) = y^{-5}$. Then, the equation becomes

$$ (-y^{-4}-2)\text dx + (4xy^{-5}+y^{-1})\text dy = 0 \\ \text d\left[ -xy^{-4} - 2x + \log y \right] = 0$$

Integrating, we get

$$ -xy^{-4} - 2x + \log y = C $$

Solving for $x$ gives us

$$ x = \frac{y^4}{2y^4+1}\log(cy) $$

Solving for $y$ in terms of the function $W(z)$ such that $W(z)\exp(W(z)) = W(z\exp(z)) = z$,

$$y = \left( \frac{4x}{W(kx\exp(-8x))} \right)^\frac{1}{4}$$