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: 28

Answer

$$u = 9{e^{2x - 2}} - 3$$

Work Step by Step

$$\eqalign{ & \frac{{du}}{{dx}} = 2u + 6 \cr & {\text{separate the variables}} \cr & \frac{{du}}{{2u + 6}} = dx \cr & \frac{{2du}}{{2u + 6}} = 2dx \cr & {\text{integrate both sides}} \cr & \int {\frac{{2du}}{{2u + 6}}} = 2\int {dx} {\text{ }} \cr & {\text{ln}}\left| {2u + 6} \right| = 2x + c \cr & {\text{solve for }}u \cr & {e^{{\text{ln}}\left| {2u + 6} \right|}} = {e^{2x + c}} \cr & 2u + 6 = {e^c}{e^{2x}} \cr & 2u = {e^c}{e^{2x}} - 6 \cr & u = \frac{{{e^c}{e^{2x}}}}{2} - \frac{6}{2} \cr & u = \frac{{{e^c}}}{2}{e^{2x}} - 3 \cr & {\text{set }}\frac{{{e^c}}}{2} = C \cr & u = C{e^{2x}} - 3 \cr & {\text{the initial condition }}u\left( 1 \right) = 6{\text{ implies that}} \cr & 6 = C{e^{2\left( 1 \right)}} - 3 \cr & 6 = C{e^2} - 3 \cr & C = \frac{9}{{{e^2}}} \cr & {\text{so the solution of the initial value problem is}} \cr & u = \frac{9}{{{e^2}}}{e^{2x}} - 3 \cr & u = 9{e^{2x - 2}} - 3 \cr} $$
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