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, $y=e^x(-10e^{-x}+C)=-10+Ce^x$
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 - Review Exercises - Page 432: 49
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