Solve $y' = x^4y+x^4y^4$ Solve the differential equation $$y' = x^4y+x^4y^4.$$
I'm not sure how to deal with the $x^4y^4$ term. So far I have only encountered differential equations where the exponent of $y$ was at most one.
Could someone please share a general strategy/hint?
(This is test preparation, not homework.)
 A: You may want to take all the $y$ terms on one side as follows:
\begin{align*}
\frac{1}{y(1+y^3)}\frac{dy}{dx} & = x^4\\
\int \frac{1}{y(1+y^3)} \, dy & = \int x^4 \, dx\\
\end{align*}
The integral on the left can be dealt with partial fractions. Hopefully you can take it from here.
A: $$\frac{y'}{y+y^4}=x^4\tag{1}$$
and since
$$\int\frac{dz}{z+z^4}=\frac{1}{3}\log\frac{z^3}{1+z^3}\tag{2}$$
by partial fraction decomposition, by integrating both terms in $(1)$ we get:
$$\log\left(1+\frac{1}{y^3}\right)=-\frac{3}{5}x^5+C,\tag{3}$$
hence:
$$ 1+\frac{1}{y^3}= K e^{-\frac{3}{5}x^5},$$

$$ y= \frac{1}{\sqrt[3]{K e^{-\frac{3}{5}x^5}-1}}.\tag{4}$$

A: Solve the separable equation:
$$y'=x^4y^4+x^4y$$
Siimplify:
$$y'=x^4(y^4+y)$$
divide by $y+y^4$
$$\frac{\frac{dy}{dx}}{y+y^4}dx=x^4dx$$
Integrate both sides wrt x:
$$\int\frac{\frac{dy}{dx}}{y+y^4}dx=\int x^4dx$$
Evaluate the integral:
$$-\frac13\ln(y^3+1)+\ln y=\frac{x^5}5+\text{constant}$$
Solve for y:[Spoiler:Answer]

 $$y=-\frac{\exp(x^5/5)+c}{(\exp(3/5(x^5+c))-1)^{1/3}}$$

