Answer
See below
Work Step by Step
Given: $f(t)=e^{2t}\\
g(t)=1$
Obtain: $L[f * g]=L[f]L[g]\\
=L[e^{2t}]L[1]\\
=\frac{1}{s-2}.\frac{1}{s}\\
=\frac{1}{s^2-2s}$
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 - 10.9 The Convolution Integral - Problems - Page 716: 8
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