Answer
$3-2 \sin 2t+5 \cos 2t$
Work Step by Step
Since, $\mathcal{L}(1)=\dfrac{1}{s}$ and
$\mathcal{L}(\cos at)=\dfrac{s}{s^2+a^2}$ and $\mathcal{L}(\sin at)=\dfrac{a}{s^2+a^2}$
Here, we have $F(S)=\dfrac{8s^2-4s+12}{s(s^2+4)}$
or, $= \dfrac{3s^2+12-4s+5s^2}{s(s^2+4)}$
or, $=\dfrac{3(s^2+4)}{s(s^2+4)}- 2\times \dfrac{2s}{s(s^2+4)}+5\times \dfrac{s^2}{s(s^2+4)}$
or, $=3\mathcal{L}(1)-2\mathcal{L}(\sin 2t)+5\mathcal{L}(\cos 2t)$
or, $=\mathcal{L}(3-2 \sin 2t+5 \cos 2t)$
Therefore, the required inverse Laplace transform is:
$\mathcal{L}^{-1}[\mathcal{L}(3-2 \sin 2t+5 \cos 2t)]=3-2 \sin 2t+5 \cos 2t$
