6.4 First Shifting Theorem: the \(s\)-Shift
Theorem 6.4.1 (First shifting theorem). If \(f(t)\) has the transform \(F(s)\) for \(s>\alpha \), then \(e^{at}f(t)\) has the transform \(F(s -a)\) for \(s-a>\alpha \): \[\mathcal {L}\{e^{at}f(t)\} = F(s-a),\qquad \text {equivalently}\qquad \mathcal {L}^{-1}\{F(s-a)\} = e^{at} f(t) .\]
Proof. \(F(s - a) = \int _0^\infty e^{-(s-a)t}f(t)\,dt = \int _0^\infty e^{-st}\big (e^{at}f(t)\big )\,dt = \mathcal {L}\{e^{at}f(t)\}\). □
From the table, for example, \[\mathcal {L}\{e^{at}t^n\} = \frac {n!}{(s-a)^{n+1}} ,\qquad \mathcal {L}\{e^{at} \cos \omega t\}=\frac {s - a}{(s-a)^2 + \omega ^2} ,\qquad \mathcal {L}\{e^{at}\sin \omega t\}=\frac {\omega }{(s-a)^2 + \omega ^2} .\]
Example 6.4.2. Solve the initial value problem \(y'' - 2y' + y = e^t + t\), \(y(0) = 1\), \(y'(0) = 0\).
Solution. Transforming, with \(Y = \mathcal {L}\{y\}\), \[\big (s^2Y - s\cdot 1 - 0\big ) - 2\big (sY - 1\big ) + Y = \frac {1}{s -1} + \frac {1}{s^2},\] so \[(s - 1)^2Y = \frac {1}{s - 1} + \frac {1}{s^2} + s - 2 ,\qquad Y(s) = \frac {1}{(s -1)^3} + \frac {1}{s^2 (s-1)^2} + \frac {s-2}{(s-1)^2} .\] By partial fractions, \[\frac {1}{s^2(s-1)^2} = \frac {2}{s} + \frac {1}{s^2} - \frac {2}{s - 1} + \frac {1}{(s-1)^2},\qquad \frac {s - 2}{(s-1)^2} = \frac {1}{s - 1} - \frac {1}{(s-1)^2},\] so \[Y(s) = \frac {1}{(s - 1)^3} + \frac {2}{s} + \frac {1}{s^2} - \frac {1}{s-1} .\] By Theorem 6.4.1, \(\mathcal {L}^{-1}\left \{\frac {1}{(s-1)^3}\right \} = e^t\frac {t^2}{2}\), and therefore \[y(t) = \frac {1}{2}t^2e^t + 2 + t - e^t .\] As a check, \(y(0) = 2 - 1 = 1\).
Solution. Completing the square, \(s^2 - 4s + 13 = (s - 2)^2 + 9\), and \(3s - 1 = 3(s - 2) + 5\), so \[\frac {3s - 1}{(s-2)^2 + 9} = 3\cdot \frac {s - 2}{(s-2)^2 + 9} + \frac 53\cdot \frac {3}{(s-2)^2 + 9},\] whose inverse transform is \(e^{2t}\left (3\cos 3t + \frac 53\sin 3t\right )\).
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