Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 17 - Second-Order Differential Equations - 17.1 Second-Order Linear Equations - 17.1 Exercises - Page 1200: 23

Answer

$y=\frac{1}{7}e^{4x-4}-\frac{1}{7}e^{3-3x}$

Work Step by Step

$y''-y'-12y=0$ Use auxiliary equation $r^{2}-r-12=0$ $r=\frac{1±\sqrt (1^{2}-4(1)(-1))}{2(1)}$ $r=\frac{1±\sqrt 49}{2}$ $r_{1}=4$ $r_{2}=-3$ $y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$ $y=c_{1}e^{4x}+c_{2}e^{-3x}$ $y(1)=0$ $y(1)=0=c_{1}e^{4x}+c_{2}e^{-3x}$ $y'(1)=1$ $y'(1)=1=4c_{1}e^{4x}-3c_{2}e^{-3x}$ $c_{1}e^{4x}+c_{2}e^{-3x}=0$ $4c_{1}e^{4x}-3c_{2}e^{-3x}=1$ Solve the system of equations: $c_{2}=-c_{1}e^{7}$ $4c_{1}e^{4}+3c_{1}e^{4}=1$ $c_{1}=\frac{1}{7}e^{-4}$ $c_{2}=-\frac{1}{7}e^{3}$ Hence, $y=\frac{1}{7}e^{4x-4}-\frac{1}{7}e^{3-3x}$
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