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Extra negative in in Laplace transform solution

  1. Nov 4, 2009 #1
    I have the correct solution of this problem ([tex]cos(t)+u_{3\pi}[1-cos(t-3\pi)][/tex]). As you can tell from my solution below I am off by a extra negative sign on the cos(t) coefficient. I have had this same issue on several other problems, so I believe I must be missing something on the procedure. Any guidance would be much appreciated!

    Solve IVP [tex]y'' + y = u_{3\pi}(t) \left\{y(0)=1, y'(0)=0[/tex]

    My solution:
    [tex]s^{2}Y(s) + sY(0) + Y'(0) + Y(s) = \frac{e^{-3\pi s}}{s}[/tex]
    [tex]Y(s)(s^{2} +1) + s = \frac{e^{-3\pi s}}{s}[/tex]
    [tex]Y(s) = \frac{e^{-3\pi s}}{s(s^{2}+1)} - \frac{s}{(s^{2}+1)}[/tex]

    [tex]H(t) = \frac{1}{s(s^{2}+1)}[/tex]
    [tex]h(t) = Y(H(t))[/tex]

    [tex]y = u_{3\pi}h(t-3\pi}) - cos(t)[/tex]

    [tex]h(t) = \frac{1}{s} - \frac{s}{s^{2}+1}[/tex]
    [tex]h(t) = 1 - cos(t)[/tex]

    so ...

    [tex]-cos(t) + u_{3\pi}[1-cos(t-3\pi)][/tex]


    I have seen in other examples (http://tutorial.math.lamar.edu/Classes/DE/IVPWithStepFunction.aspx , example 2) which have similar setups, and I noticed that when the little h is introduce (the inverse transformation to the [tex]u_{3\pi}[/tex] term) the sign of the [tex]cos(t)[/tex] term changes but I do not understand why.

    Thank you so much for help and guidance you can provide!

    - Andrew Balmos
     
  2. jcsd
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