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?

$$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)\frac{dx}{dy}-4x=y^4$$
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.
