\[ -e^{-x^2} x+y'(x)+2 x y(x)=0 \] ✓ Mathematica : cpu = 0.0389584 (sec), leaf count = 30
\[\left \{\left \{y(x)\to \frac {1}{2} e^{-x^2} x^2+c_1 e^{-x^2}\right \}\right \}\] ✓ Maple : cpu = 0.007 (sec), leaf count = 18
\[y \relax (x ) = \left (\frac {x^{2}}{2}+c_{1}\right ) {\mathrm e}^{-x^{2}}\]
Hand solution
\begin {equation} \frac {dy}{dx}+2xy\relax (x) =e^{-x^{2}}x \tag {1} \end {equation}
Integrating factor \(\mu =e^{\int 2xdx}=e^{x^{2}}\). Hence (1) becomes
\begin {align*} \frac {d}{dx}\left (e^{x^{2}}y\relax (x) \right ) & =e^{x^{2}}e^{-x^{2}}x\\ \frac {d}{dx}\left (e^{x^{2}}y\relax (x) \right ) & =x \end {align*}
Integrating both sides
\begin {align*} e^{x^{2}}y\relax (x) & =\frac {x^{2}}{2}+C\\ y\relax (x) & =e^{-x^{2}}\left (\frac {x^{2}}{2}+C\right ) \end {align*}