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Laplace Transforms

  1. Oct 3, 2013 #1
    Please ignore this thread.
     
    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2
    The correct solution:

    7UShynu.png
     
  4. Oct 3, 2013 #3
    Actually, I think the solutions manual is wrong. Please ignore this thread.
     
  5. Oct 3, 2013 #4
    It would have helped if you stated Theorem 4.3.3. That said, we could try it without any special theorem. We have

    ##Lf = \int_0^{\infty} e^{-st}f(t)dt ##

    which you can split up into the sum of integrals from (0,a) (a,2a) ... ([n-1]a,na) (na,t), where na < t < (n+1)a.

    Each of these is easy enough to evaluate, but be careful to separate out the case where n is even and n is odd (why?). I'll bet you can factor some stuff out of the resulting sum.

    Why don't you try this out? And why don't you quote theorem 4.3.3 -- maybe it's an easier way to do things.
     
  6. Oct 3, 2013 #5
    Sorry for wasting your time, but I already posted:

     
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