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: 1

Answer

y = C1$e^{3x}$ + C2$e^{-2x}$

Work Step by Step

Question: y"-y'-6y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}-r-6)e^{rx}$=0 The corresponding characteristic equation is: $r^{2}-r-6=0$ Factoring gives: $(r-3)(r+2) = 0$ r=3 and r=-2 Adding the constants gives the following general solution: y = $C_1e^{3x}$ + $C_2e^{-2x}$
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