Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 426: 50

Answer

$${\text{The function satisfies the given differential equation}}$$

Work Step by Step

$$\eqalign{ & {\text{Let the differential equation be}}\frac{{dy}}{{dx}} = {e^{ - y}} \cr & {\text{Separate the variables}} \cr & {e^{ - y}}dy = dx \cr & {\text{Integrate both sides with respect to }}x \cr & - {e^{ - y}} = x + C \cr & {\text{With }}y = - 2{\text{ and }}x = 0 \cr & - {e^{ - \left( { - 2} \right)}} = 0 + C \cr & C = - {e^2} \cr & {\text{then}}{\text{, }} \cr & - {e^{ - y}} = x - {e^2} \cr & {e^{ - y}} = {e^2} - x \cr & {\text{Solving for }}y \cr & \ln {e^{ - y}} = \ln \left( {{e^2} - x} \right) \cr & y = - \ln \left( {{e^2} - x} \right) \cr & {\text{The function satisfies the given differential equation}} \cr} $$
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