## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 17 - Review - Exercises: 1

#### Answer

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

#### Work Step by Step

The given differential equation is $4y’’-y=0$ The auxiliary equation is $4r^{2}-1=0$ This implies $4r^{2}=1$ Thus, $r=±\frac{1}{2}$ are the roots. The general solution is $y(x)=c_{1}e^{\frac{x}{2}}+c_{2}e^{\frac{-x}{2}}$

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