If \(Y\left ( s\right ) \) has form \(\frac{s}{s^{2}+as+b}\) where roots of quadratic are complex, then complete the square. Write \(s^{2}+as+b=\left ( s+A\right ) ^{2}+B\) and find \(A,B\). Then\begin{align*} \frac{s}{s^{2}+as+b} & =\frac{s}{\left ( s+A\right ) ^{2}+B}\\ & =\frac{s+A-A}{\left ( s+A\right ) ^{2}+B}\\ & =\frac{s+A}{\left ( s+A\right ) ^{2}+B}-A\frac{1}{\left ( s+A\right ) ^{2}+B}\\ & =\frac{\tilde{s}}{\tilde{s}^{2}+B}-A\frac{1}{\tilde{s}^{2}+B}\\ & =\frac{\tilde{s}}{\tilde{s}^{2}+B}-\frac{A}{\sqrt{B}}\frac{\sqrt{B}}{\tilde{s}^{2}+B} \end{align*}
And now use tables. Due to shifting, multiply result by \(e^{-Bt}\). So inverse Laplace of the above is \(e^{-Bt}\left ( \cos \sqrt{B}t-\frac{A}{\sqrt{B}}\sin \sqrt{B}t\right ) \)
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