Answer
$2 e^{2t} \cos (3t)+\dfrac{4}{3} e^{2t}\sin (3t) $
Work Step by Step
Since, $F(s)=\dfrac{2s}{s^2-4s+13}=\dfrac{2(s-2)}{(s-2)^2+3^2}+\dfrac{4}{(s-2)^2+3^2}$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [\dfrac{2(s-2)}{(s-2)^2+3^2}+\dfrac{4}{(s-2)^2+3^2} ]$
Now, apply the first shifting Theorem.
$f(t)= 2 e^{2t} \cos (3t)+\dfrac{4}{3} e^{2t}\sin (3t) $
