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 - Page 589: 13

Answer

$$y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem }}$$

Work Step by Step

$$\eqalign{ & y'\left( t \right) - 2y = 20 \cr & \frac{{dy}}{{dt}} - 2y = 20 \cr & \frac{{dy}}{{dt}} = 2y + 20 \cr & {\text{separating}} \cr & \frac{{dy}}{{2y + 20}} = dt \cr & \frac{{dy}}{{y + 10}} = 2dt \cr & \int {\frac{{dy}}{{y + 10}}} = \int 2 dt \cr & \ln \left| {y + 10} \right| = 2t + C \cr & {\text{solve for }}y \cr & y + 10 = {e^{2t + C}} \cr & y = A{e^{2t}} - 10 \cr & {\text{initial condition }}y\left( 0 \right) = 6 \cr & 6 = A{e^{2\left( 0 \right)}} - 10 \cr & A = 16 \cr & y = A{e^{2t}} - 10 \cr & {\text{replacing }}A \cr & y = 16{e^{2t}} - 10 \cr & {\text{then}} \cr & y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem}} \cr} $$
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