Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 74

Answer

$$u\left( x \right) = {e^x} - 2{e^{ - 2x}} + \frac{1}{2}$$

Work Step by Step

$$\eqalign{ & u'\left( x \right) = \frac{{{e^{2x}} + 4{e^{ - x}}}}{{{e^x}}} \cr & u\left( x \right) = \int {u'\left( x \right)} dx \cr & then \cr & u\left( x \right) = \int {\left( {\frac{{{e^{2x}} + 4{e^{ - x}}}}{{{e^x}}}} \right)} dx \cr & u\left( x \right) = \int {\left( {\frac{{{e^{2x}}}}{{{e^x}}} + \frac{{4{e^{ - x}}}}{{{e^x}}}} \right)} dx \cr & u\left( x \right) = \int {\left( {{e^x} + 4{e^{ - 2x}}} \right)} dx \cr & find{\text{ the general solution}} \cr & u\left( x \right) = {e^x} - 2{e^{ - 2x}} + C \cr & {\text{using the initial condition }}u\left( {\ln 2} \right) = 2 \cr & 2 = {e^{\ln 2}} - 2{e^{ - 2\left( {\ln 2} \right)}} + C \cr & 2 = 2 - 2\left( {\frac{1}{4}} \right) + C \cr & C = \frac{1}{2} \cr & {\text{the solution to the initial value problem is}} \cr & u\left( x \right) = {e^x} - 2{e^{ - 2x}} + \frac{1}{2} \cr} $$
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