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: 23

Answer

$$y = C{e^{ - 2x}} - 2$$

Work Step by Step

$$\eqalign{ & y'\left( x \right) = - 2y - 4 \cr & \frac{{dy}}{{dx}} = - 2y - 4 \cr & {\text{separate the variables}} \cr & \frac{{dy}}{{ - 2y - 4}} = dx \cr & {\text{integrate both sides}} \cr & \int {\frac{{ - 2dy}}{{ - 2y - 4}}} = - 2\int {dx} {\text{ }} \cr & {\text{ln}}\left| { - 2y - 4} \right| = - 2x + c \cr & {\text{solve for }}y \cr & {e^{{\text{ln}}\left| { - 2y - 4} \right|}} = {e^{ - 2x + c}} \cr & - 2y - 4 = {e^c}{e^{ - 2x}} \cr & - 2y = {e^c}{e^{ - 2x}} + 4 \cr & y = \frac{{{e^c}{e^{ - 2x}}}}{{ - 2}} + \frac{4}{{ - 2}} \cr & {\text{set }}\frac{{{e^c}}}{{ - 2}} = C \cr & y = C{e^{ - 2x}} - 2 \cr} $$
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