Answer
See below
Work Step by Step
Given:
$$y'+x^2y=x^3y^3\\\frac{dy}{dx}+x^2y=x^3y^3$$
This is a Bernoulli equation where
$p(x)=x^2\\q(x)=x^3\\n=3$
Divide both sides by $y^3$ we have:
$y^{-3}\frac{dy}{dx}+x^2y^{-2}=x^3$
Substitute $u=y^{-2} \rightarrow \frac{du}{dx}=-2y^{-3}\frac{dy}{dx}\\
\rightarrow -\frac{1}{2}\frac{du}{dx}+x^2u=x^3\\
\rightarrow \frac{du}{dx}-2x^2u=-2x^3$
Integrating factor for this will be: $I(x)=e^{-2x^2dx}=e^{-\frac{2}{3}x^3}$
The equation becomes: $\frac{d}{dx}(ue^{-\frac{2}{3}x^3})=-2x^3e^{-\frac{2}{3}x^3}$
Integrate both sides: $ue^{-\frac{2}{3}x^3}=\int-2x^3e^{-\frac{2}{3}x^3}dx$
Substitute back $u=y^{-2}$ we have: $y^{-2}=e^{\frac{2}{3}x^3}\int -2x^3e^{-\frac{2}{3}x^3}dx$
The solution to this is, $y(x)=\pm \frac{1}{\sqrt e^{\frac{2}{3}x^3}\int -2x^3e^{-\frac{2}{3}x^3}dx}$
