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Homework Help: Laplace Transform problem

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Why I am getting wrong answer related to this Laplace Transforms problem?

    According to the book "Higher Engineering Mathematics 6th edition by John O Bird" page no. 583 one should get

    (e[itex]^{-st}[/itex]/(s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))(a sin at - s cos at)

    2. Relevant equations

    ∫e[itex]^{-st}[/itex]cos at dt

    3. The attempt at a solution

    u = e[itex]^{-st}[/itex]

    du = -se[itex]^{-st}[/itex] dt

    Let dv = cos at dt

    v = (sin at)/ a

    Integrating by parts

    ∫e[itex]^{-st}[/itex]cos at dt =

    (e[itex]^{-st}[/itex] sin at / a) + (s/a)∫e[itex]^{-st}[/itex]sin at dt

    = (e[itex]^{-st}[/itex] sin at / a) + (s/a)[(-e[itex]^{-st}[/itex] cos at / a) - (s/a)∫e[itex]^{-st}[/itex]cos at dt]

    = (e[itex]^{-st}[/itex] sin at / a) - (s/a[itex]^{2}[/itex] )(e[itex]^{-st}[/itex] cos at) - s[itex]^{2}[/itex]/a[itex]^{2}[/itex])∫e[itex]^{-st}[/itex]cos at dt]


    (1 + (s[itex]^{2}[/itex]/a[itex]^{2}[/itex]))∫e[itex]^{-st}[/itex]cos at dt =

    (e[itex]^{-st}[/itex] sin at / a) - (s/a[itex]^{2}[/itex])(e[itex]^{-st}[/itex] cos at)

    = (e[itex]^{-st}[/itex]/a[itex]^{2}[/itex])(a sin at - s cos at)

    ∫e[itex]^{-st}[/itex]cos at dt =

    (a[itex]^{2}[/itex]/(a[itex]^{2}[/itex] + s[itex]^{2}[/itex]))((e[itex]^{-st}[/itex]/a[itex]^{2}[/itex])(a sin at - s cos at))

    = ((e[itex]^{-st}[/itex]/ (s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))(a sin at - s cos at)
    Last edited: Dec 27, 2013
  2. jcsd
  3. Dec 27, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    There is no way you could get the answer that you claim the book obtains: the Laplace transform of a function f(t) will not have a "t" in it, since t has be "integrated out". I hope you realize that you need to take a limit!
  4. Dec 27, 2013 #3
    The integration limits on the LT integration are from 0 to ∞. Try these limits and see what you get. It's a definite integral, not an indefinite integral.
  5. Dec 27, 2013 #4
    Applying limits on the last step was not a problem. I was getting wrong answer that is there was a mistake in sign at one place where Integration by parts is done second time. After fixing it I got the right answer which I have modified in post 1. After applying the limits I got the right answer.

    e[itex]^{-s\infty}[/itex] becomes 0


    e[itex]^{-s * 0}[/itex] becomes 1

    sin 0 = 0

    cos 0 = 1

    First term of the equation after applying infinity becomes 0. Remaining is applying 0 as limit but a - sign appears before the equation.

    - [ e[itex]^{-s * 0}[/itex] / (s[itex]^{2}[/itex] + a[itex]^{2}[/itex])(a sin a (0) - s cos a (0))

    = - (1 / (s[itex]^{2}[/itex] + a[itex]^{2}[/itex])) ( - s)

    = (s / (s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))

    Am I right?
    Last edited: Dec 27, 2013
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