Calculus 10th Edition

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

Chapter 6 - Differential Equations - Review Exercises: 49

Answer

$y=Ce^x-10$

Work Step by Step

$y'-y=10$ Here, $P(x)=-1$ and $Q(x)=10$ Therefore, $u(x)=e^{\int-1dx}=e^{-x}$ So, $y=\frac{1}{e^{-x}}\int 10 e^{-x}dx$ Or, $y=e^x(-10e^{-x}+C)=-10+Ce^x$
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