Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.4 Exercises: 7

Answer

$y=-16+Ce^{x}$

Work Step by Step

$y'-y=16$ Integrating factor: $e^{\int-1dx}$ $e^{\int-1dx}=e^{-x}$ $e^{-x}y'-e^{-x}y=16e^{-x}$ $ye^{-x}=\int16e^{-x}dx$ $\int16e^{-x}dx=-16e^{-x}+C$ $y=-16+Ce^{x}$
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