6.4 First Shifting Theorem: the \(s\)-Shift
If \(f(t)\) has the transform \(F(s)\), where \(s>\alpha \), then \(e^{at}f(t)\) has the transform \(F(s -a)\), \(\hspace {0.2cm} s-a>\alpha \): thus, if \(\mathcal {L}\{f(t)\}= F(s)\), then \[\mathcal {L}\{e^{at}f(t)\} = F(s-a)\]
From this, we get \(\displaystyle {\mathcal {L}^{-1}\{F(s-a)\} = e^{at} f(t)}\)
By means of FST, we have \(\displaystyle {\mathcal {L}\{e^{at}t^n\} = \frac {n!}{(s-a)^{n+1}}}\) since \(\displaystyle {\mathcal {L}\{t^n\} = \frac {n!}{s^{n+1}}}\)
\(\implies \hspace {0.5cm}\displaystyle { \mathcal {L}\{e^{at} \cos \omega t\}=\frac {s - a}{(s-a)^2 + \omega ^2}}\) since \(\displaystyle {\mathcal {L}\{\cos \omega t\}=\frac {s}{s^2 + \omega ^2}}\)
Solve the IVP \(\hspace {0.5cm} y'' - 2y' + y = e^t + t,\hspace {0.5cm} y(0) = 1,\hspace {0.5cm} y'(0) = 0\)
Solution.
\(\displaystyle {\mathcal {L}\{y'' - 2y' + y\}= \mathcal {L}\{e^t + t\}}\). Let \(Y(s) = \mathcal {L}\{y\}\)
\[s^2Y -sY(0) - Y'(0) - 2(sY- Y(0)) + Y = \frac {1}{s -1} + \frac {1}{s^2} \]
\[\implies \hspace {0.5cm} s^2Y - s -2s Y + Y + 2 = \frac {1}{s - 1} + \frac {1}{s^2}\]
\[(s^2 -2s + 1)Y = \frac {1}{s - 1} + \frac {1}{s^2} + s-2\]
\begin {align*} Y(s) & = \frac {1}{(s -1)^3} + \frac {1}{s^2 (s-1)^2} + \frac {s-2}{(s-1)^2}\\\\ Y(s) & = \frac {1}{(s -1)^2} + \frac {s - 1}{(s -1)^2} - \frac {1}{(s - 1)^2} + \frac {2s + 1}{s^2} - \frac {1}{s-1} + \frac {1}{(s - 1)^2}\\\\ \implies \hspace {0.5cm} Y(s) & = \frac {1}{(s - 1)^3} + \frac {2}{s} + \frac {1}{s^2} + \frac {1}{s-1} \end {align*}
\begin {align*} \therefore \hspace {0.5cm} y(t) & = \mathcal {L}^{-1}\Bigg [\frac {1}{(s - 1)^3} + \frac {2}{s} + \frac {1}{s^2} + \frac {1}{s - 1}\Bigg ]\\\\ & = \frac {1}{2}e^tt^2 + 2 + t - e^t\\\ \end {align*}
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.