# Rewriting solution in terms of hyperbolic trigs

I have to find the inverse laplace transform of:

$\mathcal{L}^{-1}(\frac{s}{-8+2s+s^2})$

I found it was

$\frac{2}{3}e^{-4t}+\frac{1}{3}e^{2t}$

But the question I'm asked is, determine $A,B,C,D$ such that $e^{At}(Bcosh(Ct)+Dsinh(Ct))$ is a solution of the inverse laplace transform.

I have no idea how to proceed. I've tried multiple things to no avail.

Any help will be greatly appreciated. Thanks.

Edit: I know hyperbolic trigs can be rewritten in terms of exponentials, but I can't figure out how to use this to my advantage.

Use the fact that $$e^{At}(B\cosh Ct + D\sinh CT) = \tfrac{B+D}{2}e^{(A+C)t} + \tfrac{B-D}{2}e^{(A-C)t}$$ Compare this to your expression. The coefficients give you $$\left\{ \begin{array}{cc}\tfrac{B+D}{2}=\tfrac23 \\ \tfrac{B-D}{2}=\tfrac13 \end{array}\right.$$ The exponents give you $$\left\{ \begin{array}{cc}A+C=-4 \\ A-C=2 \end{array}\right.$$ From these systems, you can easily see that $$A=-1,B=1, C=-3, D=\tfrac13$$ so that you can write your solution as $$e^{-t}(\cosh 3t - \tfrac13\sinh 3t)$$