$(2\mathrm{.}1$#$33)$ Solve the initial value problem.

$y^{\text{'}}+2xy=x,y(0)=3$

$p(x)=2x=>x=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}2xdx}=e^{x^2}$

$e^{x^2}y={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{x^2}xdx$