Find the general solution to the differential equation. Primes denote derivatives with respect to x.
(3y^2)y' + y^3 = e^-x
I beleive this is a Bernoulli Equation but I am still having trouble.
Note that the derivative of $\displaystyle y^3 $ is
$\displaystyle \frac{d}{dx}y^3=3y^2y'$
try the substitution $\displaystyle u=y^3 \implies \frac{du}{dx}=3y^2\frac{dy}{dx}$
So we get the equation $\displaystyle \frac{du}{dx}+u=e^{-x}$
This can be solved with an integrating factor.
Happy Hunting
TES
$\displaystyle \frac{du}{dx}+1\cdot u =e^{-x}$
$\displaystyle p(x)=1 \implies e^{\int 1 dx}=e^x$
$\displaystyle
e^x\frac{du}{dx}+e^xu=1 \iff \frac{d}{dx}\left[ e^x \cdot u\right]=1
$
Now integrating both sides you get
$\displaystyle e^x\cdot u=x+c \implies u=xe^{-x}+ce^{-x} \implies y^3=xe^{-x}+ce^{-x}$