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Express Laplace Transform of y(t) in given form.

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data

    y(t) solves the following IVP

    y''(t) + 2y'(t) + 10y(t) = r(t)
    y(0) = 2
    y'(0) = 3

    r(t) =
    0 if t < 0
    t if 0 ≤ t ≤ 1
    0 if t > 1

    Demonstrate that the laplace transform of y(t) is

    Y(s) = [itex]\frac{2s+7}{s^{2}+2s+7}[/itex] + [itex]\frac{e^{-s}}{s(s^{2}+2s+7)}[/itex] + [itex]\frac{1}{s^{2}(s^{2}+2s+7)}[/itex]+[itex]\frac{e^{-s}}{s^{2}(s^{2}+2s+7)}[/itex]

    2. Relevant equations

    H(t) is the heaviside step function which is 0 when t is less than 0 and 1 when t is greater than 1, (undefined at 0.)

    Tables of common laplace transforms are available on the internet e.g. http://www.rapidtables.com/math/calculus/laplace_transform.htm

    3. Attempt at a Solution
    Okay so I use the laplace transform on both sides of the equation with
    r(t) defined as t H(t) - t H(t-1)

    When I calculate this I finish with

    [itex]\frac{2s+7}{s^{2}+2s+7}[/itex] + [itex]\frac{1}{s^{2}(s^{2}+2s+7)}[/itex]+[itex]\frac{e^{-s}}{s^{2}(s^{2}+2s+7)}[/itex]

    This is similar to the needed answer however I seem to have dropped a factor of se-s on the right side somewhere and I cannot see any mistakes in the algebra.

    Is my definition of r(t) correct and am I correct to say that the laplace transform of t H(t) - t H(t-1) is

    [itex]\frac{1}{s^{2}}[/itex] -e[itex]^{s}[/itex]([itex]\frac{1}{s^{2}}[/itex])?
  2. jcsd
  3. Jun 12, 2013 #2


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    Science Advisor
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    Gold Member

    Yes, ##r(t) = t(H(t) - H(t-1))## but you are missing a term in the transform. You can check it yourself by just calculating the integral$$
    \int_0^1 te^{-st}\, dt$$
  4. Jun 13, 2013 #3
    OK, Thanks very much, I realised that I had applied the shifting theorem to an equation of the form:
    y(t) (H(t-a)) when y(t-a) (H(t-a)) is required

    This dropped a H(t-1) term which was the missing term.
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