How Do Complex Roots Transform into Trigonometric Functions in ODEs?

[koab1mjr]

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.

 

[tiny-tim]

hi koab1mjr!

if your roots are a ± ib,

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

which is the same as e^at(Ae^ibt + Be^-ibt),

or, if you prefer, e^at(Ccosbt + D sinbt),

where A B C and D can of course be complex