Answer
See below
Work Step by Step
Since, $y(s)=\dfrac{1}{s-2} +\dfrac{6}{(s-2)^3}-\dfrac{2}{(s-2)^2}$
The inverse Laplace transform of function can be expressed as:
$y(t)=L^{-1} [\dfrac{1}{s-2} +\dfrac{6}{(s-2)^3}-\dfrac{2}{(s-2)^2}$
Now, apply the first shifting Theorem.
$f(t)= e^{2t} -2te^{2t} +3t^2e^{2t} $
