Math Help - How do you get the characteristic equation of trickier Euler equations?

1. How do you get the characteristic equation of trickier Euler equations?

When dealing with 2nd order ODE's with constant coefficients, you have this:

$Ay'' + By' + Cy = 0$

which you can get the characteristic equation like so:

$Ar^2 + Br + C = 0$

Now, when dealing with Euler equations, you have this:

$Ax^2y'' + Bxy' + Cy = 0$

Here's where things get dicey.... When you have it in this nice form, you can get the characteristic equation like so:

$Ar(r - 1) + Br + C = 0$

However, if you have an Euler equation like this:

$A(x + D)^2y'' + B(x + D)y' + Cy = 0$

...now it's a lot nastier. How do you deal with Euler equations like this?

Can you just do a substitution like

$t = x + D$

...and revert it back at the end?