## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 17 - Section 17.1 - Second-Order Linear Equations - 17.1 Exercise - Page 1160: 4

#### Answer

y = $C_1e^{3x}$ + $C_2e^{-4x}$

#### Work Step by Step

Question: y"+y'-12y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}+r-12)e^{rx}$=0 The corresponding characteristic equation is: $r^{2}+r-12=0$ Factoring gives: $(r-3)(r+4) = 0$ r=3 and r=-4 Adding the constants gives the following general solution: y = $C_1e^{3x}$ + $C_2e^{-4x}$

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