Answer
$\dfrac{3e^{-s}}{(s+2)^2+9}$
Work Step by Step
We are given that $f(t)=e^{-2(t-1)}\sin 3 (t -1)u_{1} (t)$
Now,$F(s)= L[e^{-2(t-1)}\sin 3 (t -1)u_{1} (t)]\\
=e^{-s}L[e^{-2t} \sin 3t]\\
= \dfrac{3e^{-s}}{(s+2)^2+9}$
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 704: 10
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