Apply First Shifting Theorem to find inverse Laplace transform
Solution :
First Shifting Theorem: First Shifting Theorem States that If \[L\left[f\left(t\right)\right]=\ F\left(s\right)\] Then \[ L\left[e^{\alpha t}f\left(t\right)\right]=\ F\left(s\ -\ \alpha\right),\ s\ >\ a\ +\ \alpha \]
(i)
Let \[f\left(t\right)=1,\ \ \alpha=3\] Then by First Shifting Theorem \[L\left[e^{3t}1\right]=\frac{1}{\left(s\ -3\right)}\] \[L^{-1}\left[\frac{1}{s-3}\right]=e^{3t}1=e^{3t}\] Therefore \[L^{-1}\left[\frac{4}{s-3}\right]=e^{3t}\]
(ii)
We shall find inverse Laplace transform of 5/(s+1)^4 Let \[f\left(t\right)=t^3,\ \ \alpha=-1\] Then by First Shifting Theorem \[L\left[e^{-t}t^3\right]=\frac{24}{\left(s+1\right)^4}\] \[L^{-1}\left[\frac{5}{\left(s+1\right)^4}\right]=\frac{5}{24}e^{-t}t^3\]
(iii)
\[f\left(t\right)=\sin{5t},\ \alpha=2\] \[L\left[\sin{5t}\right]=\frac{5}{s^2+25}\] \[L^{-1}\left[\frac{5}{s^2+25}\right]=\sin{5t} \] And by First Shifting Theorem \[L\left[e^{2t}\sin{5t}\right]=\frac{5}{\left(s-2\right)^2+25}\] Therefore \[L^{-1}\left[\frac{1}{{(s-2)}^2+25}\right]=\frac{e^{3t}t^3}{5}\]
(iv)
Since \[f\left(t\right)=\cos{t},\ \alpha=-4 \] \[L\left[\cos{t}\right]=\frac{s}{s^2+1^2}\] \[L^{-1}\left[\frac{s}{s^2+1}\right]=\cos{t}\] And by by First Shifting Theorem \[L\left[e^{-4t}\cos{t}\right]=\frac{s+4}{\left(s+4\right)^2+1^2}\] Therefore \[L^{-1}\left[\frac{s+4}{{(s}^2+8s+37)}\right]=e^{-4t}\cos{t}\]