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Integration via complex exponential

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant 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.


    3. 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?
     
  2. jcsd
  3. Sep 8, 2013 #2

    vanhees71

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    2016 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!
     
  4. Sep 8, 2013 #3
    Yeah looking back i see the trick is just to square the ((1/-3i_2))
     
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