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Complexs Roots of an ODE

  1. Aug 7, 2012 #1
    Hi All

    I am rusty with my my math and got stumped with a straight forward question regarding vibrations and complex roots.

    I have a 2nd order ODE

    x'' +4 x' + 16 x = some forcing funciton

    This turns out complex roots. I go through the run around of solving this and I get a complete solution complementary plust specific. My question is in going from the general form of an 2nd order ODE C1exp At+ C2exp-Bt = x wheere A and B are imaginary to the trig representation. I still have an i. On some websites you will see the trig representation without an i. Is that folded into C2?

    I want to keep the i but not sure why. I vaguely remember it having to do with how you wish to express the motion or something like that if it was a vibration problem. I also remembered if you looked at it as vectors on the complex plane the presense of the i was the same as rotating 90 degrees CCW.

    Any math gurus out there can get the dust betweeen my ears out it would be much apprecaited.
     
  2. jcsd
  3. Aug 7, 2012 #2

    tiny-tim

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    hi koab1mjr! :smile:

    if your roots are a ± ib,

    then your solutions are Ae(a+ib)t + Be(a-ib)t,

    which is the same as eat(Aeibt + Be-ibt),

    or, if you prefer, eat(Ccosbt + D sinbt),

    where A B C and D can of course be complex :wink:
     
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