I am asked to find the inverse Laplace transform of:
$$F(s)=\frac{2s^2+s+13}{(s-1)((s+1)^2+4)}$$
I did the partial fraction decomposition by hand and got:
$$F(s)=\frac{2}{s-1}-\frac{3}{(s+1)^2+4}$$
Mathematica's Apart command returned the same answer.
Apart[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4))]
(* 2/(-1 + s) - 3/(5 + 2 s + s^2) *)
The answer can be written in the form:
$$F(s)=2\cdot\frac{1}{s-1}-\frac32\cdot\frac{2}{(s+1)^2+4}$$
I can then produce the inverse Laplace transform by hand:
$$f(t)=2e^t-\frac32e^{-t}\sin 2t$$
However, when I try to check my answer in Mathematica, I get the following response:
InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s,
t]
(* 2 E^t + 3/4 I E^((-1 - 2 I) t) (-1 + E^(4 I t)) *)
They are the same answer:
2 E^t + 3/4 I E^((-1 - 2 I) t) (-1 + E^(4 I t)) ==
2 Exp[t] - 3/2*Exp[-t] Sin[2 t] // Simplify
(* True *)
But I am wondering how my hand calculated answer can be returned without the complex form.
