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Math Help - Laplace Transform

  1. #1
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    Laplace Transform

    Question: Consider the differential equation 5y''' + 4y'' + 1y' - 3y = 0 with  y(0) = -3, y'(0) = 4, y''(0) = -3 . Taking the Laplace transform and solving for  L(y) yields  L(y) = F(s) where  F(s) = .

     5\mathfrak{L}\{y'''\} +4\mathfrak{L}\{y''\} +\mathfrak{L}\{y'\} -3\mathfrak{L}\{y\} = \mathfrak{L}\{0\}

     5s^{3}Y(s) -5s^{2}y(0) -5sy'(0) -5y''(0)) +4s^{2}Y(s) - 4sy(0) -4y'(0) + sY(s) -y(0) -3Y(s) = 0

     5s^{3}Y(s) + 15s^{2} -20s +15 +4s^{2}Y(s) +12s -16 + sY(s) + 3 - 3Y(s) = 0

     (5s^{3} + 4s^{2} + s - 3)Y(s)  + 15s^{2} -8s + 2 = 0

    Y(s) = \frac{\-15s^{2}}{5s^{3} + 4s^{2} + s - 3} + \frac{\ 8s}{5s^{3} + 4s^{2} + s - 3} - \frac{\ 2}{5s^{3} + 4s^{2} + s - 3}

    It seems that after this step, the denominator doesn't factor with integers, so I am stuck at this point. I know to use partial fraction decomposition to find the  F(s) . Any help is appreciated!

    Thank you!
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  2. #2
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    The first term on the RHS should be negative. Here is WolframAlpha's solution to the roots. They're not very pretty. You can get an analytical solution, but it's very messy, because you're solving a cubic and the roots are, as you say, not anything obvious. I'd recommend going with the approximation there.
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  3. #3
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    Thanks for the reply, so are you sure that the only way is to use approximation?
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  4. #4
    A Plied Mathematician
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    No, you can get an exact answer. However, it's very messy, and probably useless for all practical purposes. I think you (or your boss if you were working in industry) would prefer having an answer for which you have a good intuitive feel. The decimal approximation gives you that, whereas the analytical solution is so involved that I doubt anyone would have a good feel for how the solution behaved.
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