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