# 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 im 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(e2ti
= Re ∫ e(-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?

Gold Member
2021 Award
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))