Answer
$2 \cosh (2t)-\dfrac{3}{2} \sinh(2t)$
Work Step by Step
Since, $\mathcal{L}(\sinh at)=\dfrac{a}{s^2-a^2}$ and $\mathcal{L}(\cosh at)=\dfrac{s}{s^2-a^2}$
Here, we have $F(S)=\dfrac{2s-3}{s^2-4}$
or, $=2 \times \dfrac{s}{s^2-4}-\dfrac{3}{2} \times \dfrac{2}{s^2-4}$
or, $=\mathcal{L}[2 \cosh (2t)-\dfrac{3}{2} \sinh(2t)] $
Therefore, the required inverse Laplace transform is:
$\mathcal{L}^{-1}[\mathcal{L}[2 \cosh (2t)-\dfrac{3}{2} \sinh(2t)]]=2 \cosh (2t)-\dfrac{3}{2} \sinh(2t)$
