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Laplacetransform an Integral, problem with the inverstransform!

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Hey! I have tried to solve this problem but I get stuck when it comes to the inverstransforming. Anyway here is the problem and my attempt to a solution:

    Solve [tex]f(t)=2\int_{0}^{t}sin(9u)f'(t-u)du+sin9t,t\geq 0[/tex] for f(t)

    3. The attempt at a solution

    Laplacetransforming:

    [tex]F(s)=\frac{18}{s^{2}+9^{2}}sF(s) + \frac{9}{s^{2}+9^{2}}[/tex]

    Solving F(s)
    [tex]F(s)=\frac{9}{s^{2}+9^{2}-18s}[/tex]

    I dont know what to do now, with the inverstransform? And the F(s) feels wrong..Is my Laplacetransforming correct, I have tried to use the convolution theorem..but I dont know if I used it correct..

    Thanks! :smile:
     
  2. jcsd
  3. Oct 2, 2011 #2

    HallsofIvy

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    Did you notice that [itex]s^2- 18s+ 9^2= (s- 9)^2[/itex]?
     
  4. Oct 2, 2011 #3
    Yes, But its something wrong bcs in the end f(t) should be sin9t.

    And this answer will give me 9te^(9t).

    Do you know what am I doing wrong?
     
  5. Oct 2, 2011 #4

    vela

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    Your answer is correct; 9te9t satisfies the original equation. Did you copy the problem correctly?
     
  6. Oct 2, 2011 #5
    Really?? :D hmm, yes but I tried the problem in wolframalpha.com and the answer was sin9t. I maybe did write the problem wrong in wolfram..

    Thank you for the answer! :D
     
  7. Oct 2, 2011 #6
    Sorry but.. How does 9te^(9t) satisfies the answer.
    Im trying to derivate it f'(t-u) and put that in to the original equation. But I get a strange answer.
     
  8. Oct 2, 2011 #7

    vela

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    Show us your work. When I do what you describe, it works out.
     
  9. Oct 2, 2011 #8
    Okey :)

    [tex]f'(t-u)= 9e^{t-u}(9(t-u)+1)[/tex]

    [tex]f(t)=2\int_{0}^{t}(sin9u)(9e^{t-u}(9(t-u)+1)du + sin9t[/tex]

    [tex]f(t)=\frac{2}{9}(cos9t-1)+ 2(e^{9t}(t-\frac{1}{9})+\frac{1}{9})+sin9t[/tex]
     
  10. Oct 2, 2011 #9

    vela

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    You dropped the factor of 9 in the exponent.
     
  11. Oct 2, 2011 #10
    How will sin and cos go away?
     
  12. Oct 2, 2011 #11

    vela

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    They end up canceling out if you integrate everything correctly. If you're just doing this to check the answer, why don't you just evaluate the integrals using Wolfram Alpha?
     
  13. Oct 3, 2011 #12
    No, I wanted to try it by my self. And now I did :)
    Thank you so much for the help, I really appreciate it!
     
  14. Oct 3, 2011 #13

    vela

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    Good work! I totally understand wanting to crank it out by hand. There's nothing better than practice to learn how to calculate quickly and how to avoid making mistakes.
     
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