I'll just show you my work and you can tell me where I career off the road and burst into flames.
$y''+by'+cy=p(x)$
set $b=u+v$ and $c=uv$, and rewrite
$y''+(u+v)y'+(uv)y=p(x)$
$y''+ uy' + vy' + (uv)y=p(x)$
$(y'' + uy') + v(y' + uy) = p(x)$
$\frac{d}{dx}(y'+uy) + v(y' + uy) = p(x)$
set $z=y'+uy$, and rewrite
$\frac{dz}{dx} + vz = p(x)$
Solve this 1st-order DE for $z$
($e^{vx}z)' = e^{vx}p(x)$
$e^{vx}z = \int e^{vx}p(x) dx + C_0$
$z = e^{-vx}\int e^{vx}p(x) dx + C_0e^{-vx}$
Sub. this back into the 1st-order DE
$y'+uy = e^{-vx}\int e^{vx}p(x) dx + C_0e^{-vx}$
Solve this 1st-order DE for $y$:
$(e^{ux}y)' = e^{ux}e^{-vx}\int e^{vx}p(x) dx + C_0e^{-vx}e^{ux}$
$(e^{ux}y)' = e^{(u-v)x}\int e^{vx}p(x) dx + C_0e^{(u-v)x}$
$e^{ux}y = \int \left[ e^{(u-v)x}\int e^{vx}p(x) dx + C_0e^{(u-v)x} \right] dx$
$e^{ux}y = \int \left[ e^{(u-v)x}\int e^{vx}p(x) dx \right] dx + C_0\int e^{(u-v)x} dx$
$e^{ux}y = \int \left[ e^{(u-v)x}\int e^{vx}p(x) dx \right] dx + \frac{C_0}{u-v}e^{(u-v)x} + C_1$
$y = e^{-ux}\int \left[ e^{(u-v)x}\int e^{vx}p(x) dx \right] dx + e^{-ux}\frac{C_0}{u-v}e^{(u-v)x} + C_1e^{-ux}$
$y = e^{-ux}\int \left[ e^{(u-v)x}\int e^{vx}p(x) dx \right] dx + e^{-vx}\frac{C_0}{u-v} + C_1e^{-ux}$
Okay, two questions: is this correct and can I simplify any further?