|Aug7-12, 01:55 PM||#1|
Complexs Roots of an ODE
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.
|Aug7-12, 03:07 PM||#2|
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
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