Answer
$y=-16+Ce^{x}$
Work Step by Step
$y'-y=16$
Integrating factor: $e^{\int-1dx}$
$e^{\int-1dx}=e^{-x}$
$e^{-x}y'-e^{-x}y=16e^{-x}$
$ye^{-x}=\int16e^{-x}dx$
$\int16e^{-x}dx=-16e^{-x}+C$
$y=-16+Ce^{x}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 6 - Differential Equations - 6.4 Exercises - Page 428: 7
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