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(Inverse) Laplace Transforms

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]y'' + 2y' + 2y = t^2 + 4t [/tex]
    [tex]y(0) = 0[/tex]
    [tex]y'(0) = -1[/tex]

    2. Relevant equations
    [tex]L(y) = Y [/tex]
    [tex]L(y') = sY - y(0) [/tex]
    [tex]L(y'') = s^2 Y - s y(0) - y'(0) [/tex]
    [tex]L(t^n) = \frac {n!}{s^{n+1}}[/tex]
    [tex]L(a*b) = AB[/tex]
    [tex]a*b = L^{-1}(AB)[/tex]

    3. The attempt at a solution

    Take the Laplace transform of both sides.

    [tex]L(y'' + 2y' + 2y) = L(t^2 + 4t) [/tex]

    By linearity:

    [tex]L(y'') + 2L(y') + 2L(y) = L(t^2) + 4L(t) [/tex]

    Substitute:

    [tex](s^2 Y - s y(0) - y'(0)) + 2(sY - y(0)) + 2Y = \frac{2}{s^3} + \frac{4}{s^2} [/tex]

    Plug in initial conditions:

    [tex](s^2 Y + 1) + 2(sY) + 2Y = \frac{2}{s^3} + \frac{4}{s^2} [/tex]

    Solve for Y:

    [tex] Y (s^2 + 2s + 2) -1 = \frac{2}{s^3} + \frac{4}{s^2} [/tex]

    [tex] Y = \frac {\frac{2}{s^3} + \frac{4}{s^2} + 1}{s^2 + 2s + 2} [/tex]

    Now I must take the inverse Laplace transform. This is where I get confused.

    Rewrite:


    [tex] Y = \frac{1}{s^2 + 2s + 2} (\frac{2}{s^3} + \frac{4}{s^2} + 1)[/tex]

    Distribute:

    [tex] Y = \frac{1}{s^2 + 2s + 2} \frac{2}{s^3} + \frac{1}{s^2 + 2s + 2} \frac{4}{s^2} + \frac{1}{s^2 + 2s + 2} [/tex]

    Completing the square:

    [tex] Y = \frac{1}{(s+1)^2 + 1} \frac{2}{s^3} + \frac{1}{(s+1)^2 + 1} \frac{4}{s^2} + \frac{1}{(s+1)^2 + 1} [/tex]

    Now to take the inverse Laplace transform of both sides:

    [tex] y = L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3}) + L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2}) + L^{-1}(\frac{1}{(s+1)^2 + 1}) [/tex]

    [tex] y = L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3}) + L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2}) + e^{-t}sin(t) [/tex]

    So now I need to do the separate inverse transforms (via convolutions):

    [tex]L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3})[/tex]

    [tex]= e^{-t}sin(t) * t^2[/tex]

    [tex]= \int_0^t (e^{-(t-v)}sin(t-v))(v^2)) dv [/tex]

    [tex]L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2})[/tex

    [tex]= e^{-t}sin(t) * 4t [/tex]

    [tex]= \int_0^t (e^{-(t-v)}sin(t-v))(4v) dv [/tex]

    I don't know how to figure those convolution integrals. Any ideas?
    Alternatively, is there a better way to approach the problem?
     
  2. jcsd
  3. Apr 19, 2009 #2
    Perhaps you might have better luck using partial fractions, instead of completing the square... I don't know for sure since I didn't do it, but it's a suggestion you might want to look into.
     
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