Where do I go wrong solving this non-homogeneous 2nd order DE? 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?
 A: Note that if you write $$y'' + by' + cy = p(x) = y'' -(u_0 + v_0)y' + u_0 v_0 y,$$ then $u_0$ and $v_0$ are the roots of $x^2 + bx + c = 0.$ To make things easier, your $u$ equals $-u_0$ and your $v$ equals $-v_0$. Then your expression in terms of $p(x)$, $u_0$ and $v_0$ reads $$y = e^{u_0 x}\int \left[ e^{(v_0-u_0)x}\int e^{-v_0 x}p(x) dx + C_0e^{(v_0 - u_0)x} \, \mathrm dx \right] \, \mathrm dx.$$ I'll leave it there so, when we get special cases like $u_0 = v_0$ we do not run into undefined terms.
Ultimate exercise: what happens when $$y'' + p(x)y' + q(x)y = r(x)?$$
A: Looks about right, except for the case where $u=v$.  You might be able to clean it up a bit.  For starters, $C_0$ has no real significance, so you can declare a new constant $C_2=\frac {C_0}{u-v}$.  That really messy part is your particular solution.
Not sure if it will work, but maybe integration by parts can clean it up a bit.
$$r=\int e^{vx}p(x)dx,dr=e^{vx}p(x)dx$$
$$ds=e^{(u-v)x}dx,s=\frac{e^{(u-v)x}}{u-v}$$
$$y=\frac{e^{-ux}e^{(u-v)x}}{u-v}\int e^{vx}p(x)dx-\frac{e^{-ux}}{u-v}\int e^{(u-v)x}e^{vx}p(x)dx+C_2e^{-vx}+C_1e^{-ux}$$
$$y=\frac{e^{-vx}}{u-v}\int e^{vx}p(x)dx-\frac{e^{-ux}}{u-v}\int e^{ux}p(x)dx+C_1e^{-ux}+C_2e^{-vx}$$
Again, if $u=v$, that changes the homogenous solution slightly as
$$\int e^{(u-v)x}dx=\int dx=x$$
I'll leave it to someone else to explain whether or not this would work when $u$ and $v$ are complex as that's not my area of expertise.
