The general Solution of differential equation $\frac{{dy}}{{dx}} = {e^{\frac{{{x^2}}}{2}}} + xy$ is

Given that, $\frac{{dy}}{{dx}} = {e^{{x^2}/2}} + xy$

$\Rightarrow$ $\frac{{dy}}{{dx}} - xy = {e^{{x^2}/2}}$

Here, $P = - x,Q = {e^{{x^2}/2}}$

$\therefore$ $IF = {e^{\int - xdx}} = {e^{ - {x^2}/2}}$

The general Solution is
$y \cdot {e^{ - {x^2}/2}} = \int {{e^{ - {x^2}/2}}} - {e^{{x^2}/2}}dx + C$

$\Rightarrow$ $y{e^{ - {x^2}/2}} = \int 1 dx + C$

$\Rightarrow$ $y \cdot {e^{ - {x^2}/2}} = x + C$

$\Rightarrow$ $y = x{e^{{x^2}/2}} + C{e^{ + {x^2}/2}}$

$\Rightarrow$ $y = (x + C){e^{{x^2}/2}}$