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: 24

Answer

$$y = C{e^{2t}} - 3$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dt}} = 2y + 6 \cr & {\text{separate the variables}} \cr & \frac{{dy}}{{2y + 6}} = dt \cr & {\text{integrate both sides}} \cr & \int {\frac{{2dy}}{{2y + 6}}} = 2\int {dt} {\text{ }} \cr & {\text{ln}}\left| {2y + 6} \right| = 2t + c \cr & {\text{solve for }}y \cr & {e^{{\text{ln}}\left| {2y + 6} \right|}} = {e^{2t + c}} \cr & 2y + 6 = {e^c}{e^{2t}} \cr & 2y = {e^c}{e^{2t}} - 6 \cr & y = \frac{{{e^c}{e^{2t}}}}{2} - \frac{6}{2} \cr & {\text{set }}\frac{{{e^c}}}{2} = C \cr & y = C{e^{2t}} - 3 \cr} $$
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