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Find the Inverse Laplace of 1/(s^3)

  1. Nov 7, 2013 #1
    Find the Inverse Laplace of 1/(s^3)

    is there some special rule for cube?

    The answer is t^2/2

    Looking at the Laplace Table t^n looks similar but its not it exactly.


    What should I do?
     
    Last edited: Nov 7, 2013
  2. jcsd
  3. Nov 7, 2013 #2

    LCKurtz

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    You mean find the inverse transform.

    So what does the table give you for ##t^n##? Can you modify it?
     
  4. Nov 7, 2013 #3
    Yep Inverse sorry, ##t^n## , n = 1,2,3,..... is (n!)/(s^n+1)

    2/s^2 +1 ?
     
  5. Nov 7, 2013 #4

    HallsofIvy

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    Are you asking for the "Laplace transform" or the "Inverse Laplace transform"? The standard notation uses "t" for the function and "s" for its Laplace transform.

    A table of Laplace transforms, such as the one at http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf, will tell you that the Laplace transform of [itex]t^n[/itex], for n a positive integer, is [itex]n!/s^{n+1}[/itex].

    So the inverse Laplace transform of [itex]1/s^3= (1/2)(2/s^3)=(1/2)(2!/s^(2+1)[/itex] is [itex](1/2)t^2[/itex].

    I just saw your response. I think you are misreading the table. It is not "[itex]s^n+ 1[/itex]", it is [itex]s^{n+ 1}[/itex].
     
  6. Nov 7, 2013 #5

    LCKurtz

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    Halls, don't you think we should have let him figure out that step?
     
  7. Nov 7, 2013 #6
    Thanks,

    where did the 1/2 come from?
     
  8. Nov 8, 2013 #7

    HallsofIvy

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    If you cannot see that then you should not be taking this course. Look at your table of Laplace transforms again.
     
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