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: 30

Answer

$y=4e^{\frac{1}{2}x}-2xe^{\frac{1}{2}x}$

Work Step by Step

$4y''-4y'+y=0$ Use auxiliary equation $4r^{2}-4r+1=0$ $r^{2}-r+\frac{1}{4}=0$ $(r-\frac{1}{2})^{2}=0$ $r=\frac{1}{2}$ $y=c_{1}e^{r_{1}x}+c_{2}xe^{r_{2}x}$ $y=c_{1}e^{(\frac{1}{2})x}+c_{2}xe^{(\frac{1}{2})x}$ $y(0)=4$ $4=c_{1}e^{(\frac{1}{2})(0)}+c_{2}(0)e^{(\frac{1}{2})(0)}$ $4=c_{1}(1)+0$ $c_{1}=4$ $y(2)=0$ $0=c_{1}e^{(\frac{1}{2})(2)}+c_{2}(2)e^{(\frac{1}{2})(2)}$ $0=c_{1}e+2c_{2}e$ $c_{1}e=-2c_{2}e$ $c_{1}=-2c_{2}$ $4=-2c_{2}$ $c_{2}=-2$ $y=4e^{\frac{1}{2}x}-2xe^{\frac{1}{2}x}$
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