Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises - Page 605: 10

Answer

$ z = -ln(e^t + C)$

Work Step by Step

$ dz/dt + e^{t+z} = 0$ This can be rewritten as: $ dz/dt = -e^{t+z}$ $ dz/dt = -e^te^z $ Now let's separate z and t: $ e^{-z} dz = -e^t dt $ Now let's integrate both sides: $\int e^{-z} dz = \int -e^t dt $ $-e^{-z} = -e^t + C $ To isolate z, we can add ln() on both sides: $ ln(-e^{-z}) = ln(-e^t + C)$ $-z = ln(-e^t + C)$ $ z = -ln(-e^t + C)$
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