Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 17 - Section 17.1 - Second-Order Linear Equations - 17.1 Exercise: 8

Answer

y = $C_{1}e^{x}$ + $C_{2}e^{-x}$

Work Step by Step

Question: y=y" We write: y"-y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}-1)e^{rx}=0$ The corresponding characteristic equation is: $r^{2}-1=0$ Solving the equation gives: $r^{2}=1$ $r=\frac{+}{-}1$ Adding the constants gives the following general solution: y = $C_{1}e^{x}$ + $C_{2}e^{-x}$
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