Solve the differential equation $\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$ Solve the differential equation $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$\frac{dy}{dx}=\frac{1}{y^3}\frac{\left(1+2y^4\right)}{1+\frac{4x}{y^4}}$$
Multiplying both sides with $\frac{1}{y^5}$ we get:
$$\frac{1}{y^5}\frac{dy}{dx}=\frac{1}{y^8}\frac{y^4(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}=\frac{1}{y^4}\frac{(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}$$
Now letting $$\frac{1}{y^4}=t$$ we get
$$\frac{-1}{4}\frac{dt}{dx}=\frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
 A: 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 $W(z)\exp(W(z)) = W(z\exp(z)) = z$,
$$y = \left( \frac{4x}{W(kx\exp(-8x))} \right)^\frac{1}{4}$$
A: $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)\frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is : 
$$x(y)=\frac{y^4}{2y^4+1}\ln(c\:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=\left(\frac{4x}{\text{W}(C\:xe^{-8x})}\right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
