Question

Solve y' + 2xy = 2x^3, y(0) = 1

Answer 1

Multiply both sides by $e^{x^2}$, then you get

$e^{x^2}+2xe^{x^2}y=2x^3e^{x^2}$
$\left(e^{x^2}y\right)'=2x^3e^{x^2}$
$e^{x^2}y=\displaystyle2\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}y=e^{x^2}(x^2-1)+C$
$y=x^2-1+Ce^{-x^2}$

Given that $y(0)=1$, you have

$1=-1+Ce^0\implies C=2$

so that the particular solution is

$y=x^2-1+2e^{-x^2}$