Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.9 Introduction to Differential Equations - 7.9 Exercises: 37

Answer

$$y = \sqrt {{e^t} - 1} $$

Work Step by Step

$$\eqalign{ & y'\left( t \right) = \frac{{{e^t}}}{{2y}} \cr & \frac{{dy}}{{dt}} = \frac{{{e^t}}}{{2y}} \cr & {\text{separate}} \cr & 2ydy = {e^t}dt \cr & {\text{integrate both sides}} \cr & \int {2ydy} = \int {{e^t}dt} \cr & {y^2} = {e^t} + C \cr & {\text{the initial condition }}y\left( {\ln 2} \right) = 1{\text{ implies that}} \cr & {\left( 1 \right)^2} = {e^{\ln 2}} + C \cr & 1 = 2 + C \cr & C = - 1 \cr & {y^2} = {e^t} - 1 \cr & y = \sqrt {{e^t} - 1} \cr} $$
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