Why Does the Solution of Underdamped Systems Include a Sine Term?

[tomizzo]

I'm trying to do some refreshing of differential equations featuring damped systems. Specifically, I have a question regarding the differential equation solution to an under damped system involving complex roots.

Referring to the attached pdf, an under damped system will yield a complex conjugate pair of roots. I am curious as to why the basic real solution features a sine term (refer to second attachment). If I remember Euler's formula correctly, the sine term is always imaginary and is not featured in the real solution. However, this document states otherwise... I believe I have a fundamental misconception regarding this topic.

Any idea why the negative imaginary conjugate yields a sine term in the time domain?

 

[HallsofIvy]

If you have complex numbers are solutions to your characteristic equations, say, a+ bi and a- bi, then the solution to the corresponding differential equation is of the form $$Ae^{(a+ bi)t}+ Be^{(a- bi)t}= Ae^{at}e^{ibt}+ Be^{at}e^{-ibt}= e^{at}(Ae^{ibt}+ Be^{-ibt})$$$$= e^{at}(Acos(bt)+ Ai sin(bt)+ Bcos(bt)- Bsin(bt))= e^{at}((A+ B)cos(bt)+ i(A- B)sin(bt))$$

There is, in fact, an "i" multiplying A- B. HOWEVER, "A" and "B" themselves are complex numbers, not real numbers. As long as we have only real number values for the initial or boundary values, because this has to solve a problem involving only real numbers, we must have A+ B real and A- B imaginary (so that i(A- B) is real). That is the same as saying that A and B must be conjugate complex numbers.

(If you had a differential equation in which the initial values or boundary values involved complex numbers themselves, then you could not assume the coefficients are real- but in that case, you would be better off leaving the solutions as $e^{(a+ bi)t}$ and $e^{(a- bi)t}$.)