Integration via complex exponential

[tylersmith7690]

Homework Statement

Using the complex exponential, nd the most general function f such that

$\frac{d^2f}{dt^2}$ = e^-3t cos 2t , t all real numbers.

Homework Equations

I'm having a lot of trouble with this question, my thinking is to integrate once and then one more time to undo the second derivative. However I am wondering if there is a trick to do this as i know there is a trick to doing it when say finding the 45th derivative of say e^t cos 5t.

The Attempt at a Solution

d/dt = ∫ e^-3t cos (2t) dt

= e^-3t . Re(e^2ti
= Re ∫ e<sup>(-3+2i)[t/SUP]
= Re[ 1/(-3+2i) . e(-3+2i)t ] + C
= Re [ 1/(-3+2i) . (-3-2i/-3-2i) . e(-3+2i)t ] + C
= Re [ $\frac{-3-2i}{13}$ . e-3(cos2t+isin2t)
= Re [ $\frac{e^-3t}{13}$ .( -3 cost t + sin 2t - i(3sin 2t + 2 cos 2t) +C

= $\frac{e^-3t}{13}$(-3 cos 2t + sin 2t) +C

What is next to do or is there a trick in an earlier step?</sup>

 

[vanhees71]

You just have to integrate twice. The friendly advice is to express the cosine with complex exponentials. Then the task is pretty simple, as you'll see immediately!

 

[tylersmith7690]

Yeah looking back i see the trick is just to square the ((1/-3i_2))