University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 402: 49



Work Step by Step

Here, $y=\int (-2e^{-x}+c) dx$ Now, we have $y=2e^{-x}+cx+c'$ Apply the initial conditions $y(0)=1$ We get $c'=-1$ and $y'(0)=0 \implies c=2$ Thus, we have $y=2e^{-x}+2x-1$
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