## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 17 - Section 17.1 - Second-Order Linear Equations - 17.1 Exercise - Page 1160: 22

#### Answer

$y=2e^{2.5x}-8xe^{2.5x}$

#### Work Step by Step

$4y''-20y'+25y=0$ Use auxiliary equation $4r^{2}-20r+25=0$ $(2r-5)(2r-5)=0$ $r=\frac{5}{2}$ $y=c_{1}e^{(2.5)x}+c_{2}xe^{(2.5)x}$ $y(0)=2$ $y(0)=2=c_{1}e^{(2.5)x}+c_{2}xe^{(2.5)x}$ $2=c_{1}e^{0}+0c_{2}e^{0}$ $2=c_{1}$ $y'(0)=-3$ $y'(0)=-3=\frac{5}{2}c_{1}e^{0}+\frac{5}{2}c_{2}(0)e^{0}+e^{0}c_{2}$ $-3=\frac{5}{2}c_{1}+c_{2}$ $-3=\frac{5}{2}(2)+c_{2}$ $c_{2}=-8$ $y=2e^{2.5x}-8xe^{2.5x}$

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