# complex exponentials.

• Sep 7th 2013, 04:29 PM
tylersmith7690
complex exponentials.
Using the complex exponential, find the most general function f such that

f''(x) = e-3tcos (2t) , for t element of reals.

Relevant equations:

f''(x) is second derivative.

Do i tackle this problem by taking the anti derivative of both sides, then doing it one more time to get
f(x) = ............
• Sep 7th 2013, 05:25 PM
HallsofIvy
Re: complex exponentials.
Yes, and using the fact that $e^{-3t}cos(2t)$ is the real part of $e^{(-3+ 2i)t}$ (that is the "complex exponential") makes it very easy.
• Sep 8th 2013, 06:43 PM
tylersmith7690
Re: complex exponentials.
I still dont really get it is there any other way to help explain it? i did the first derivative and got down to
1/13*e^(-3t) *( -3 cos2t + sin2t) + c

What would be the next step or have i gone about it the hard way? thanks for your help.
• Sep 8th 2013, 07:45 PM
tylersmith7690
Re: complex exponentials.
yeah it is easy all i had to do was square the (1/(-3+2i)) which is like integrating twice. I hope that's what you mean to do.