Answer
$y(x)$ = $\frac{1}{2}+Ce^{-x^{2}}$
Work Step by Step
let
$α(x)$ = $e^{x^{2}}$
$[α(x)y]'$ = $[e^{x^{2}}y]'$ = $2xe^{2}y+e^{x^{2}}y'$ = $e^{x^{2}}
(2xy+y')$ = $α(x)(2xy+y')$
multiply both sides of the differential equation $y'+2xy = x$ by $α(x)$, we obtain
$α(x)(y'+2xy )$ = $xα(x)$ = $xe^{x^{2}}$
$α(x)(y'+2xy )$ = $[α(x)y]'$
$[α(x)y]$ = $\int{xe^{x^{2}}dx}$ = $\frac{1}{2}e^{x^{2}}+C$
$y(x)$ = $\frac{1}{2}+Ce^{-x^{2}}$
