Answer
$y=2e^{-x}+2x-1$
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 $y=2e^{-x}+2x-1$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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
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