Answer
See below
Work Step by Step
We are given that $F(s)=\frac{e^{-5s}}{s^2+16}$
Now,$f(t)= L^{-1}[\frac{e^{-5s}}{s^2+16}]\\
=\frac{1}{4}L^{-1}[e^{-5s}(L(\sin 4t))]\\
=\frac{1}{4}\sin 4(t-5)u_5(t)$
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 704: 21
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