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Integration via complex exponential

  • #1

Homework Statement



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

[itex]\frac{d^2f}{dt^2}[/itex] = 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 [ [itex]\frac{-3-2i}{13}[/itex] . e-3(cos2t+isin2t)
= Re [ [itex]\frac{e^-3t}{13}[/itex] .( -3 cost t + sin 2t - i(3sin 2t + 2 cos 2t) +C

= [itex]\frac{e^-3t}{13}[/itex](-3 cos 2t + sin 2t) +C

What is next to do or is there a trick in an earlier step?
 

Answers and Replies

  • #2
vanhees71
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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!
 
  • #3
Yeah looking back i see the trick is just to square the ((1/-3i_2))
 

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