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IVP Laplace Transform Problem - Tricky Inverse Laplace Transform

  1. Jan 29, 2012 #1
    IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

    1. The problem statement, all variables and given/known data

    Solve x"+x'+x=1, given x(0)=x'(0)=0


    2. Relevant equations



    3. The attempt at a solution

    Plugged in transforms: s2*Y(s)-s*y(0)-y'(0)+s*Y(s)-y(0)+Y(s)=1/s
    Plugged in initial value points, simplified to Y(s)=1/((s2+s+1)*s)
    Partial Fractions led me to Y(s)=1/s+ (-s-1)/(s^2+s+1)

    I get stuck at finding an inverse transform for the second term. You can't complete the square for the denominator, right?
     
  2. jcsd
  3. Jan 29, 2012 #2
    Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

    You can still do a partial fraction expansion even if the roots of the denominator aren't real.
     
    Last edited: Jan 29, 2012
  4. Jan 29, 2012 #3

    vela

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    Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

    Why not?
     
  5. Jan 30, 2012 #4
    Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

    If I complete the square, I get the denominator is (s+1/2)2+3/4, and the numerator is -(s+1), and I don't know what inverse laplace transform this would be.
     
  6. Jan 30, 2012 #5

    vela

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    Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

    Hint: s+1 = (s+1/2) + 1/2
     
  7. Jan 30, 2012 #6
    Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

    oh right, I didn't think to break it up like that.

    so you end up with 1/s-[(s+1/2)/((s+1/2)2+3/4)+1/2 / ((s+1/2)2+3/4), which becomes 1-e-t/2cos3t/4-√3/3 * e-t/2sin√3/2

    thanks!
     
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