Answer
$f(t)=2u_0(t)+(1-t)u_1(t)$
Work Step by Step
Since $0 \leq t \lt 1$, $f(t)=2$
then we have $f_1(t)=2u_0(t)$
Since $t \geq 1$, $f(t)=3-t$
then we have $f_2(t)=(1-t)u_1(t)$
Hence,
$f(t)=f_1(t)+f_2(t)\\
=2u_0(t)+(1-t)u_1(t)$
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 - 10.10 Chapter Review - Additional Problems - Page 719: 26
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