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Laplace D.E Problem

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data

    2r7c9wy.png

    2. Relevant equations

    Laplace's transforms.

    3. The attempt at a solution

    Okay, so applying the Laplace on both sides yields:

    [itex]s^2Y(s) - sy(0) - y'(0) + 2sY(s) - 2y(0) + 10Y(s) = ? + 13 / s - 1[/itex]

    Is [itex] e^{-2s} / s[/itex] the Laplace [itex]\delta(t - 2)[/itex]? Our instructor gave us the result of the Lapace but he did not prove it. The only thing he gave us was that the Laplace of [itex]f(t - a)\delta(t - a) = e^{-as}F(s)[/itex]. Can someone point me on how to prove this? It seems our instructor told us that we need to know the proof without him giving it to us. I know that this unit step function is defined to be 0 when 0 <= t < 2 and 1 when t >= 2.
     
  2. jcsd
  3. May 26, 2012 #2

    vela

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    Are you sure ##\delta(t)## is the unit step function? It typically denotes the Dirac delta function.
     
  4. May 27, 2012 #3
    Yes yes. I mixed up the function. Sorry about that.

    So is the Laplace of [itex]\delta(t - 2)[/itex] typically [itex]e^{-2s} / s[/itex]?
     
  5. May 27, 2012 #4

    vela

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    Which function are you talking about? If it's the delta function, then no, that's not correct. If it's the unit step, then that's right.
     
  6. May 27, 2012 #5
    The delta function. Then is it simply [itex]e^{-2s}[/itex]?

    If so, how to prove it starting with [itex]\int^\infty_0 e^{-st}\delta(t - a)dt = e^{as}[/itex]? The proof is apparently required for our exam yet our instructor failed to provide the solution.
     
  7. May 27, 2012 #6

    vela

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    What's the defining property of the Dirac delta function?
     
  8. May 27, 2012 #7
    ..that the [itex]\int^{a+e}_{a-e} f(t)\delta(t - a)dt = f(a)[/itex] for [itex]e > 0[/itex]?

    I appreciate your patience but I'm relatively 'new' to this concept.
     
  9. May 27, 2012 #8

    vela

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    Yup. Just apply that to the Laplace transform integral you have.
     
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