Inverse Laplace Transforms Problem 2

Spoolx
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Homework Statement


f(s) = 6/s^2-9


Homework Equations


I think
f(t) = (1/b-a)(e^-at-e^-bt)


The Attempt at a Solution


Replace 6/s^2-9 with 6/(s-3)(s+3)
a=-3
b=3

Plug in
(1(6)/3-(-3))(e^-(-3)t-e^-3t)

Final Result
e^3t-e^-3t
 
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Spoolx said:

Homework Statement


f(s) = 6/s^2-9


Homework Equations


I think
f(t) = (1/b-a)(e^-at-e^-bt)


The Attempt at a Solution


Replace 6/s^2-9 with 6/(s-3)(s+3)
a=-3
b=3

Plug in
(1(6)/3-(-3))(e^-(-3)t-e^-3t)

Final Result
e^3t-e^-3t

It's easy enough for you to check it yourself. Take the transform of your answer and see it it gives what you started with.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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