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: 33

Answer

$$y = - 2\ln \left( {\frac{1}{2}\cos t + C} \right)$$

Work Step by Step

$$\eqalign{ & y'\left( t \right) = {e^{y/2}}\sin t \cr & \frac{{dy}}{{dt}} = {e^{y/2}}\sin t \cr & {\text{separate}} \cr & \frac{{dy}}{{{e^{y/2}}}} = \sin tdt \cr & {\text{integrate both sides}} \cr & \int {\frac{{dy}}{{{e^{y/2}}}}} = \int {\sin t} dt \cr & \int {{e^{ - y/2}}dy} = \int {\sin t} dt \cr & - 2{e^{ - y/2}} = - \cos t + C \cr & {\text{solve for }}y \cr & 2{e^{ - y/2}} = \cos t + C \cr & {e^{ - y/2}} = \frac{1}{2}\cos t + C \cr & \ln {e^{ - y/2}} = \ln \left( {\frac{1}{2}\cos t + C} \right) \cr & - \frac{y}{2} = \ln \left( {\frac{1}{2}\cos t + C} \right) \cr & y = - 2\ln \left( {\frac{1}{2}\cos t + C} \right) \cr} $$
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