I am trying to find the inverse laplace transform of $(s^2+4) \over (s-2)(s+2)$.
The solution is $ {2\over(s-2)} - {2\over(s+2)} + 1 $.
But I can't figure out how to break it up so I can find the solution algebraically.
i.e $ (s^2+4)/((s-2)(s+2)) = As+B/(s-2)+C/(s+2) $ etc. What terms should I use?
On
$$\begin{align}\frac{s^2+4}{s^2-4}&=\frac{s^2-4+8}{s^2-4}\\&= 1+\frac{8}{s^2-4}\\&=1+2\frac{4}{(s+2)(s-2)}\end{align}$$
Obviously $4=(s+2)-(s-2)$
substituting, we have $$1+2\frac{(s+2)-(s-2)}{(s+2)(s-2)}$$
$$1+2\frac{(s+2)}{(s+2)(s-2)}-2\frac{(s-2)}{(s+2)(s-2)}$$
Cancelling,
$$\require{cancel}{1+2\frac{\cancel{(s+2)}}{\cancel{(s+2)}(s-2)}-2\frac{\cancel{(s-2)}}{(s+2)\cancel{(s-2)}}}$$
And thus, we have
$$1+\frac{2}{s-2}-\frac{2}{s+2}$$
Write ${s^2+4 \over (s-2)(s+2)} = A + {B\over(s-2)} + {C\over(s+2)}$
Multiply through by $(s-2)(s+2)$ and equate coefficients.
Since the Laplace transform is linear (I believe), you can use the known ones for constants and $ 1/s $.
Alternatively, just write $$ {s^2+4 \over (s-2)(s+2)} = {(s^2-4)+8 \over s^2-4} = 1 + {8 \over s^2-4}$$
and we know that (correct me if I'm wrong) the inverse Laplace transform of $ 1/(s^2 - b^2) $ is ${1 \over b}\sinh(bt)$.
From this, we get the final answer along with the inverse Laplace transform of 1.
Hope this helps! :)