Inverse Laplace Transform

  • Thread starter Jrlinton
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  • #1
Jrlinton
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Homework Statement


Y=(8s-4)/(s²-4)

Homework Equations




The Attempt at a Solution


I rearranged the right side as:
8*(s/(s²-2²))-2*(2/(s²-2²))

Using the Laplace transform chart given in the class I was able to identify these as the transforms of hyperbolic sine and hyperbolic cosine making the inverse Laplace:
y=8cosh(2t)-2sinh(2t)
I checked this answer by working backwards and it all checks out but the worked out problem given by the instructor states the final answer as:
y=3e^(2t)-5e(-2t)=3[e][/2t]-5[e[/-2t]

I am wondering if these two solutions are equivalent?
 

Answers and Replies

  • #2
Jrlinton
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Correction: The given solution was y=3e^(2t)+5e(-2t).
I understand how he got this by using partial fractions, I would just like to know if the two functions are equivalent or if there was a mistake.
 
  • #3
Ray Vickson
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Correction: The given solution was y=3e^(2t)+5e(-2t).
I understand how he got this by using partial fractions, I would just like to know if the two functions are equivalent or if there was a mistake.

You do not need to ask us; you can work it out for yourself---and you should: that is how you will learn. Just expand sinh and cosh in terms of exp and see what you get.
 

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