Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 7

Answer

$xy'=-sin x -y$ and $0$

Work Step by Step

Need to use quotient rule of differentiation. Thus, $y'=\dfrac{-x \sin x -\cos x}{x^2}=(\dfrac{-\sin x}{x}) -(\dfrac{1}{x})\dfrac{\cos x}{x}$ As we are given that $y=\dfrac{\cos x}{x}$ Thus, $y'=(\dfrac{-\sin x}{x}) -\dfrac{y}{x}$ and, $xy'=-sin x -y$ Now, apply the initial conditions. $y(\dfrac{\pi}{2})=\dfrac{\cos \dfrac{\pi}{2}}{\dfrac{\pi}{2}}=0$
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