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