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