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: 27

Answer

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

Work Step by Step

$$\eqalign{ & y'\left( t \right) = - 2y - 4 \cr & \frac{{dy}}{{dt}} = - 2y - 4 \cr & {\text{separate the variables}} \cr & \frac{{dy}}{{ - 2y - 4}} = dt \cr & {\text{integrate both sides}} \cr & \int {\frac{{ - 2dy}}{{ - 2y - 4}}} = - 2\int {dt} {\text{ }} \cr & {\text{ln}}\left| { - 2y - 4} \right| = - 2t + c \cr & {\text{solve for }}y \cr & {e^{{\text{ln}}\left| { - 2y - 4} \right|}} = {e^{ - 2t + c}} \cr & - 2y - 4 = {e^c}{e^{ - 2t}} \cr & - 2y = {e^c}{e^{ - 2t}} + 4 \cr & y = \frac{{{e^c}{e^{ - 2t}}}}{{ - 2}} + \frac{4}{{ - 2}} \cr & {\text{set }}\frac{{{e^c}}}{{ - 2}} = C \cr & y = C{e^{ - 2t}} - 2 \cr & {\text{the initial condition }}y\left( 0 \right) = 0{\text{ implies that}} \cr & 0 = C{e^0} - 2 \cr & C = 2 \cr & {\text{so the solution of the initial value problem is}} \cr & y = 2{e^{ - 2t}} - 2 \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.