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Laplace transform IVP 2nd order

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    y''+6'+10y=0
    y(0)=2
    y'(0)=1

    2. Relevant equations



    3. The attempt at a solution

    Laplace everything and I get
    s^2*Y(s)-2s-1+6s*Y(s)-12+10Y(s)=0

    isolate Y(s)
    Y(s)=(2s+13)/(s^2+6s+10)

    split into 2 terms, bottom can be rearranged by completing the square

    2s/[(s+3)^2+1^1] + 13/[(s+3)^2+1^1]

    inverse laplace

    y(t)=2e^-3t*cost + 13e^-3t*sint

    both wolfram and the answer key say the last term should be 7 instead of 13, but i don't see where I made a mistake
     
  2. jcsd
  3. Feb 16, 2012 #2
    [tex]L^{-1}\frac s{(s+b)^2+a^2}=e^{-bx}(\cos ax-\frac ba\sin ax)[/tex]
     
  4. Feb 16, 2012 #3

    vela

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    The Laplace transform of ##\cos t## is ##\frac{s}{s^2+1}##, so when you replace s by s+3, you get ##\frac{s+3}{(s+3)^2+1}##. Compare that to what you said was the Laplace transform of ##e^{-3t}\cos t##.
     
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