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

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    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?
    Last edited by Lancet; July 16th 2011 at 05:22 PM.
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