University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 7


$xy'=-sin x -y$ and $y(\pi/2)=0$

Work Step by Step

Apply the quotient rule of differentiation: Thus, $y'=\dfrac{-x \sin x -\cos x}{x^2}=(\dfrac{-\sin x}{x}) -(\dfrac{1}{x})\dfrac{\cos x}{x}$ Since, we have $y=\dfrac{\cos x}{x}$ Therefore, $y'=(\dfrac{-\sin x}{x}) -\dfrac{y}{x}$ or, $xy'=-sin x -y$ Apply the initial conditions: $y(\dfrac{\pi}{2})=\dfrac{\cos \dfrac{\pi}{2}}{\dfrac{\pi}{2}}=0$
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.