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: 20

Answer

$$y\left( x \right) = \frac{3}{2}\sin 2x - \frac{2}{3}\cos 3x + 8$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = 3\cos 2x + 2\sin 3x \cr & {\text{separate the variables}} \cr & dy = \left( {3\cos 2x + 2\sin 3x} \right)dx \cr & {\text{integrate both sides }} \cr & \int {dy} = \int {\left( {3\cos 2x + 2\sin 3x} \right)dt} \cr & y\left( x \right) = \frac{3}{2}\sin 2x - \frac{2}{3}\cos 3x + C \cr & {\text{the initial condition }}y\left( {\pi /2} \right) = 8{\text{ implies that}} \cr & 8 = \frac{3}{2}\sin 2\left( {\pi /2} \right) - \frac{2}{3}\cos 3\left( {\pi /2} \right) + C \cr & 8 = \frac{3}{2}\sin \left( \pi \right) - \frac{2}{3}\cos \left( {3\pi /2} \right) + C \cr & 8 = C \cr & {\text{so the solution of the initial value problem is}} \cr & y\left( x \right) = \frac{3}{2}\sin 2x - \frac{2}{3}\cos 3x + 8 \cr} $$
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