So I've got to find the general solution to the following differential equation:

$\displaystyle (3y^2)y' + y^3 = e^-^x$

I've been hinted that it is a problem that can be put into a form of a Bernoulli equation, but I can't get it right.

Bernoulli Equations look like this:

$\displaystyle (dy/dx) + P(x)y = Q(x)y^n$

This is what I've tried:

$\displaystyle (3y^2)(dy/dx) + y^3 = e^-^x$

$\displaystyle (dy/dx) + y/3 = (1/(3y^2))(e^-^x)$

In it's current state, I can't get it right.