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

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

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

which you can get the characteristic equation like so:

$\displaystyle Ar^2 + Br + C = 0$

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

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

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

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

However, if you have an Euler equation like this:

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

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

Can you just do a substitution like

$\displaystyle t = x + D$

...and revert it back at the end?