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: 9

Answer

$y = \sqrt x {\text{ is not a solution}}$

Work Step by Step

$$\eqalign{ & y = \sqrt x ,{\text{ }}xy' - y = 0 \cr & {\text{Differentiate the given function}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt x } \right] \cr & \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} \cr & or \cr & y' = \frac{1}{{2\sqrt x }} \cr & \cr & {\text{Substitute }}y{\text{ and }}y'{\text{ into the given differential equation}} \cr & \underbrace {{\text{ }}xy' - y = 0}_ \downarrow \cr & {\text{ }}x\left( {\frac{1}{{2\sqrt x }}} \right) - \sqrt x = 0 \cr & {\text{ }}\frac{x}{{2\sqrt x }} - \sqrt x = 0 \cr & {\text{ }}2\sqrt x - \sqrt x = 0 \cr & {\text{ }}\sqrt x \ne 0 \cr & {\text{The statement is false}}{\text{, then }}y = \sqrt x {\text{ is not a solution}} \cr & {\text{of the given differential equation}}{\text{.}} \cr} $$
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