Answer
0
Work Step by Step
$y=e^{3x}+e^{-3x}$
$y’=3e^{3x}-3e^{-3x}$
$y’’=9e^{3x}+9e^{-3x}$
$y’’-9y=9e^{3x}+9e^{-3x}-9(e^{3x}+9^{-3x})$
$=9e^{3x}+9e^{-3x}-9e^{3x}-9e^{-3x}$
$=0$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 70
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