Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 9 - Section 9.1 - Modeling with Differential Equations - 9.1 Exercises - Page 610: 7

Answer

$y = \frac{2}{3}{e^x} + {e^{ - 2x}}{\text{ is a solution}}$

Work Step by Step

$$\eqalign{ & y = \frac{2}{3}{e^x} + {e^{ - 2x}},{\text{ }}y' + 2y = 2{e^x} \cr & {\text{Differentiate the given function}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{2}{3}{e^x}} \right] + \frac{d}{{dx}}\left[ {{e^{ - 2x}}} \right] \cr & \frac{{dy}}{{dx}} = \frac{2}{3}\left( {{e^x}} \right) + \left( { - 2{e^{ - 2x}}} \right) \cr & \frac{{dy}}{{dx}} = \frac{2}{3}{e^x} - 2{e^{ - 2x}} \cr & or \cr & y' = \frac{2}{3}{e^x} - 2{e^{ - 2x}} \cr & \cr & {\text{Substitute }}y{\text{ and }}y'{\text{ into the given differential equation}} \cr & \underbrace {y' + 2y = 2{e^x}}_ \downarrow \cr & \left( {\frac{2}{3}{e^x} - 2{e^{ - 2x}}} \right) + 2\left( {\frac{2}{3}{e^x} + {e^{ - 2x}}} \right) = 2{e^x} \cr & \frac{2}{3}{e^x} - 2{e^{ - 2x}} + \frac{4}{3}{e^x} + 2{e^{ - 2x}} = 2{e^x} \cr & \frac{6}{3}{e^x} = 2{e^x} \cr & 2{e^x} = 2{e^x} \cr & {\text{The statement is true}}{\text{, then }}y = \frac{2}{3}{e^x} + {e^{ - 2x}}x{\text{ is a solution}} \cr & {\text{of the given differential equation}}{\text{.}} \cr} $$
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