Answer
$\frac{1}{2}e^{3(t-2)}(t-2)^2u_2(t)$
Work Step by Step
We are given that $F(s)=\frac{e^{-2s}}{(s-3)^3}$
Now,$f(t)= L^{-1}[\frac{e^{-2s}}{(s-3)^3}]\\
=\frac{1}{2}L^{-1}[e^{-2s}(L(e^{3t}t^2))]\\
=\frac{1}{2}e^{3(t-2)}(t-2)^2u_2(t)$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.7 The Second Shifting Theorem - Problems - Page 705: 22
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