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 - Page 1160: 29

Answer

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

Work Step by Step

$y''=y'$ Use auxiliary equation $r^{2}=r$ $r(r-1)=0$ $r_{1}=0$ $r_{2}=1$ $y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$ $y=c_{1}+c_{2}e^{x}$ $y(0)=1$ $1=c_{1}+c_{2}e^{0}$ $c_{1}=1-c_{2}$ $y(1)=2$ $2=c_{1}+c_{2}e^{1}$ $2=c_{1}+c_{2}e$ $2=(1-c_{2})+c_{2}e$ $2=1-c_{2}+c_{2}e$ $1=c_{2}(e-1)$ $c_{2}=\frac{1}{e-1}$ $y=\frac{e-2}{e-1}+\frac{e^{x}}{e-1}$
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