How to solve $xy'+y+x^4y^4e^x=0$? I've tried a lot of tricks but I'm still not able to solve the first order differential equation 
$$
xy'+y+x^4y^4e^x=0
$$
It's not an homogeneous or isobaric equation, it's not a simple linear equation and it's also not an exact equation. 
For an exact differential equation I would need that:
$$\frac{\partial x}{\partial x}=\frac{\partial(y+x^4y^4e^x)}{\partial y},$$
which is not the case.
Now I've also tried to find some sort of integrating factor $\lambda(x,y)$ to make the above equation exact. But this yielded another difficult equation:
$$
x\frac{\partial \lambda}{\partial x}+(y+x^4y^4e^x)\frac{\partial \lambda}{\partial y}+4\lambda x^4y^3e^x=0
$$
I don't know how to proceed to solve the above differential equation ? I'm thinking of substituting a series, but I'm afraid this would also complexify my problem :(. All tips (substitutions, ...) are appreciated ! 
 A: Hint
Divide everything by $y^4$ and I suppose that the change of variable $\frac{1}{y^3}=z$ will become clear. The equation should become $$3 e^x x^4-x z'+3 z=0$$
I am sure that you can take from here.
A: You write: $xy'+y= - x^4 e^xy^4$. Consider two cases:
Case 1: $y=0$. It is easy to see that $y=0$ is a solution of the equation.
Case 2: $y\ne 0$. Dividing two both sides of the equation by $y^4$, we get
$$x.y^{-4}y+y^{-3}=-x^4.e^x.$$
Put $z=y^{-3}$. Then $z'=-3y^{-4}y'$. So, $y^{-4}.y'=-\frac{1}{3}z$. Then we get
$$-\frac{1}{3}xz'+z=-x^4e^x,$$
or
$$xz'-3z=3x^4e^x,$$
or
$$z'-\frac{3}{x}z=3x^3 e^x$$
Multiplying $\frac{1}{x^3}$ two sides of the latter equation, we get
$$\frac{1}{{{x^3}}}z' - \frac{3}{{{x^4}}}z = 3{e^x},$$
or you can write in the form
$$\frac{d}{{dx}}\left( {\frac{1}{{{x^3}}}.z} \right) = 3{e^x}.$$
This leads to
$$\frac{1}{{{x^3}}}z = \int {3{e^x}dx}  = 3{e^x} + C.$$
So, we get $z=3x^3e^x+Cx^3$. But $z=y^{-3}=\frac{1}{y^3}$, we deduce 
$$y = \frac{1}{{\sqrt[3]{z}}} = \frac{1}{{\sqrt[3]{{3{x^3}{e^x} + C{x^3}}}}}.$$
