Answer
$\frac{1}{s^3-5s^2+8s+4}$
Work Step by Step
Given: $f(t)=e^t\\
g(t)=te^{2t}$
Obtain: $L[f * g]=L[f]L[g]\\
=L[e^t]L[e^{2t}]\\
=\frac{1}{s-1}.\frac{s}{(s-2)^2}\\
=\frac{1}{(s-1)(s-2)^2}\\
=\frac{1}{s^3-5s^2+8s+4}$
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: 15
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
