$y' = x - 2y$
$\dfrac{dy}{dx} + 2y = x$ → linear in y
Hence,
$P = 2$
$Q = x$
Integrating factor:
$\displaystyle e^{\int P \, dx} = e^{\int 2 \, dx}$
$\displaystyle e^{\int P \, dx} = e^{2x}$
Thus,
$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx} \, dx + C$
$\displaystyle ye^{2x} = \int xe^{2x} \, dx + c$
Using integration by parts
$u = x$, $du = dx$
$dv = e^{2x} \, dx$, $v = \frac{1}{2}e^{2x}$
$\displaystyle ye^{2x} = \frac{1}{2}xe^{2x} - \frac{1}{2}\int e^{2x} \, dx + c$
$ye^{2x} = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + \frac{1}{4}c$
Multiply by 4e-2x
$4y = 2x - 1 + ce^{-2x}$ answer
