Answer
$2e^{-t} \cos 3t-\dfrac{5}{3}e^{-t} \sin 3t$
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 $F(S)=\dfrac{2s-3}{s^2+2s+10}$
or, $= \dfrac{2s+2-5}{s^2+2s+10}$
or, $=2 [\dfrac{s+1}{(s+1)^2+(3)^2}]-\dfrac{5}{3} [\dfrac{3}{(s+1)^2+(3)^2}] $
or, $=\mathcal{L}(2e^{-t} \cos 3t-\dfrac{5}{3}e^{-t} \sin 3t)$
Therefore, the required inverse Laplace transform is:
$\mathcal{L}^{-1}[\mathcal{L}(2e^{-t} \cos 3t-\dfrac{5}{3}e^{-t} \sin 3t)]=2e^{-t} \cos 3t-\dfrac{5}{3}e^{-t} \sin 3t$
