Can an Inverse Laplace transform have 2 different answers?

Question

I have the following Inverse Laplace transform to complete:

$$\mathcal{L}^{-1} \left\{\dfrac{6}{s^2-2s-8}\right\}$$

I used complete the square to solve, giving the following result:

$$\dfrac{6}{(s-1)^2-9} = \dfrac{2 \times 3}{(s-1)^2-9}$$

This fits with the table [$\sinh bt$] and using first shift theorem [$e^{at}$]

Resulting in: [$2e^t \sinh 3t$] or [$e^t 2\sinh 3t$]

(can't remember if the constant goes in front of the first or second term, if someone can correct me on this please do!)

However....

The example has the solution solved with partial fractions as attached giving the following answer:

[$e^{4t} - e^{-2t}$]

Have I gone wrong or is the question a bad one??

Many thanks for help

Inverse Laplace as done in textbook is shown in figure below:

Answer 1

Note that $$\sinh 3t = \frac{e^{3t}-e^{-3t}}{2}$$ Hence, $$2e^t \sinh 3t = e^{4t} - e^{-2t}$$

Answer 2

No, both the Laplace and the inverse transform is unique.

You didn't follow through in using that you matched the transform of $\sinh bt$ you have that by definition $e^t 2\sinh 3t = e^t 2(e^{3t}-e^{-3t})/2 = e^{4t}-e^{-2t}$.