# Thread:

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?
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