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Laplace Transforms

  • Thread starter Turion
  • Start date
  • #1
143
2
Please ignore this thread.
 
Last edited:

Answers and Replies

  • #2
143
2
The correct solution:

7UShynu.png
 
  • #3
143
2
Actually, I think the solutions manual is wrong. Please ignore this thread.
 
  • #4
329
34
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.
 
  • #5
143
2
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.
Sorry for wasting your time, but I already posted:

Actually, I think the solutions manual is wrong. Please ignore this thread.
 

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