Answer
y=C_{1}sinx-C_{2}cosx
y'=C_{1}cosx+C_{2}sinx
y''=-C_{1}sinx+C_{2}cosx
Work Step by Step
y''+y=(-C_{1}sinx+C_{2}cosx)+(C_{1}sinx-C_{2}cosx)=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 6 - Differential Equations - 6.1 Exercises - Page 403: 5
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