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: 25

Answer

$$y = 7{e^{3t}} + 2$$

Work Step by Step

$$\eqalign{ & y'\left( t \right) = 3y - 6 \cr & \frac{{dy}}{{dt}} = 3y - 6 \cr & {\text{separate the variables}} \cr & \frac{{dy}}{{3y - 6}} = dt \cr & {\text{integrate both sides}} \cr & \int {\frac{{3dy}}{{3y - 6}}} = 3\int {dt} {\text{ }} \cr & {\text{ln}}\left| {3y - 6} \right| = 3t + c \cr & {\text{solve for }}y \cr & {e^{{\text{ln}}\left| {3y - 6} \right|}} = {e^{3t + c}} \cr & 3y - 6 = {e^c}{e^{3t}} \cr & 3y = {e^c}{e^{3t}} + 6 \cr & y = \frac{{{e^c}{e^{3t}}}}{3} + \frac{6}{3} \cr & {\text{set }}\frac{{{e^c}}}{3} = C \cr & y = C{e^{3t}} + 2 \cr & {\text{the initial condition }}y\left( 0 \right) = 9{\text{ implies that}} \cr & 9 = C{e^0} + 2 \cr & 9 = C + 2 \cr & C = 7 \cr & {\text{so the solution of the initial value problem is}} \cr & y = 7{e^{3t}} + 2 \cr} $$
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