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Differential Equations with Discontinuous Forcing Functions

  1. Nov 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the given initial value problem:
    [itex] y'' + y = u(t-\pi) - u(t-2 \pi) [/itex]
    [itex] y(0) = 0 [/itex]
    [itex] y'(0) = 1 [/itex]

    2. Relevant equations


    3. The attempt at a solution
    First I took the Laplace transform of both sides:
    [itex] \mathcal{L}[y'' + y ] = \mathcal{L}(u(t-\pi)) - \mathcal{L}(u(t-2 \pi)) [/itex]
    [itex] (s^{2}Y(s) - sy(0) - y'(0)) + Y(s) = \frac{e^{-\pi s}}{s} - \frac{e^{-2 \pi s}}{s} [/itex]
    [itex] s^{2}Y(s) - 1 + Y(s) = \frac{e^{-\pi s}-e^{-2 \pi s}}{s} + 1 [/itex]
    [itex] Y(s)(s^{2}+1) = \frac{e^{-\pi s}-e^{-2 \pi s}+s}{s} [/itex]
    [itex] Y(s) = \frac{e^{-\pi s}-e^{-2 \pi s} + s}{s(s^{2}+1)} [/itex]
    At this point I get stuck, I tried doing a partial fraction decomp, but that didn't seem to get me anywhere closer to the solution..
     
  2. jcsd
  3. Nov 24, 2014 #2

    LCKurtz

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    I haven't checked your algebra, but assuming it is correct you need to use the formula$$
    \mathcal Lf(t-a)u(t-a) = e^{-as}\mathcal L(f(t)) = e^{-as}F(s)$$So if you have a transform ##F(s)## that is multiplied by ##e^{-as}## all you have to do is invert the ##F(s)## to ##f(t)## then truncate and translate it go get ##f(t-a)u(t-a)##. So just break up your problem into three fractions and work them separately.
     
  4. Nov 24, 2014 #3
    ah, ok.. So I'll have something like:
    [itex] Y(s) = \frac{e^{-\pi s}}{s(s^{2}+1)} - \frac{e^{-2\pi s}}{s(s^{2}+1)} + \frac{1}{s^{2}+1} [/itex]
    Then I would do a decomposition on this last fraction and then apply the inverse Laplace?
     
  5. Nov 24, 2014 #4

    LCKurtz

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    Yes. Use the formula on each of the first two and do the third one by itself. Then add the three answers.
     
  6. Nov 24, 2014 #5
    ok...I'm getting a lot of terms so I wanted to make sure I'm doing this correctly. For my decomposition I found:
    [itex] \frac{1}{s(s^{2}+1)} = \frac{1}{s} - \frac{s}{s^{2}+1} [/itex]
    Therefore, I found the following:
    [itex] Y(s) = \frac{e^{-\pi s}}{s} - \frac{se^{-\pi s}}{s^{2}+1} - \frac{e^{-2\pi s}}{s} + \frac{se^{-2\pi s}}{s^{2}+1} - \frac{1}{s^{2}+1} [/itex]
    So after applying the inverse Laplace I get:
    [itex] y(t) = u(t - \pi)-\cos (t-\pi)u(t-\pi) - u(t -2\pi)+\cos (t-2\pi)u(t-2\pi)+\sin t [/itex]
    The above answer can probably be simplified, but I'll worry about that later :)
     
    Last edited: Nov 24, 2014
  7. Nov 24, 2014 #6

    LCKurtz

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    That's the idea. Hopefully it's starting to look more like your given solution.
     
  8. Nov 24, 2014 #7
    It definitely looks close, the given solution is:
    [itex] \sin t + u(t-\pi)(1-\cos (t-\pi)) - u(t-2\pi)(1-\cos (t -2\pi)) [/itex]
    I need to check and make sure my signs are okay, but I think I've got the hang of it, thanks for the help!
     
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