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: 14

Answer

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

Work Step by Step

$$\eqalign{ & ty'\left( t \right) - 6y = 18 \cr & t\frac{{dy}}{{dt}} - 6y = 18 \cr & t\frac{{dy}}{{dt}} = 6y + 18 \cr & {\text{separating}} \cr & \frac{{dy}}{{6y + 18}} = \frac{1}{t}dt \cr & \frac{{dy}}{{y + 3}} = \frac{6}{t}dt \cr & \int {\frac{{dy}}{{y + 3}}} = \int {\frac{6}{t}} dt \cr & \ln \left| {y + 3} \right| = 6\ln \left| t \right| + C \cr & {\text{solve for }}y \cr & {e^{\ln \left| {y + 3} \right|}} = {e^{6\ln \left| t \right|}}{e^C} \cr & y = A{t^6} - 3 \cr & {\text{initial condition }}y\left( 1 \right) = 5 \cr & 5 = A{\left( 1 \right)^6} - 3 \cr & 8 = A \cr & y = A{t^6} - 3 \cr & {\text{replacing }}A \cr & y = 8{t^6} - 3 \cr & {\text{then}} \cr & y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem}} \cr} $$
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