Solve $\frac{{dy}}{{dx}} + 2xy = y$

Given that, $\frac{{dy}}{{dx}} + 2xy = y$

$\Rightarrow$ $\frac{{dy}}{{dx}} + 2xy - y = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} + (2x - 1)y = 0$

which is a linear differential equation.
On comparing it with $\frac{{dy}}{{dx}} + Py = Q,$

we get

$P = (2x - 1),$ $Q = 0$
$IF = {e^{\int P dx}} = {e^{\int {(2x - 1)} dx}}$

$= {e^{\left( {\frac{{2{x^2}}}{2} - x} \right)}} = {e^{{x^2} - x}}$

The complete solution is

$y \cdot {e^{{x^2} - x}} = \int Q \cdot {e^{{x^2} - x}}dx + C$

$\Rightarrow$ $y \cdot {e^{{x^2} - x}} = 0 + C$

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