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Another laplace differential equation.

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    question is attached

    2. Relevant equations
    L(1) = 1 /s
    L(t^n) = n!/s^(n+1)
    L{f '(t)} = s F(s) - f(0)
    L{f ''(t)} = s^2 F(s) - s f(0) - f '(0)

    3. The attempt at a solution

    okay so i laplace transformed both sides and i got:

    (s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s
    Y(s) [s^2 - 5s - 6] + 5s - 30 = -84exp(-5s) / s
    Y(s) = ((-exp(-5s)(5s(s-6)*exp(5s) +84)) / (s^3 - 5s^2 - 6s)

    but thats so long and complicated that im wondering if i've stuffed up because I can't even seem to find the inverse laplace of the function.
     

    Attached Files:

    Last edited: Mar 24, 2012
  2. jcsd
  3. Mar 24, 2012 #2
    In your first line, I think I found a problem:

    (s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s

    The laplace transform of the forcing function -84H(t-5), assuming H is the unit step function, that laplace transform should be different.

    The table I have claims that:
    [tex]L{f(t)U(t-a)) = e^{-as}L(F(t+a))[/tex]

    Since yout f(t) is basically 1, if you just pull out the constant -84. the laplace of which is 1/s, so the laplace of f(t+a) should be 1/(s+a) I believe.
     
  4. Mar 24, 2012 #3
    laplace of f(t+a) isnt 1/(s+a) i think.
     
  5. Mar 24, 2012 #4
    I'm looking at the table right now.

    The laplace transform of f(t)U(t-1) is e^(-as) times the laplace transform of f(t+a).

    Oh wait, I see what you mean.
     
  6. Mar 24, 2012 #5
    okay so that would mean laplace of -84 H(t-5) = -84exp(-5s) / (s+5) yea?
     
  7. Mar 24, 2012 #6
    That's how I interpret it.
     
  8. Mar 24, 2012 #7
    so my working out now is: (s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s+5

    therefore Y(s) = -exp(-5s)(5(s-6)(s+5)*exp(5s) +84) / (s-6)(s+1)(s+5)
    making y(t) = -1/11exp(-5(t+6))*[3(4exp(11t) - 11exp(4t+35) + 7exp(55) H(t-5)+55 exp(4t+30)] which is way to long to be right.
     
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