Thomas' Calculus 13th Edition

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

Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 3


$\tan y=-\cos x-x \sin x+c$

Work Step by Step

Re-arrange the given equation as: $\sec x dy+x \cos^2 y dx=0$ Now, $\int \dfrac{1}{\cos^2 y}dy=-\int \dfrac{xdx}{\sec x}$ $\implies (x) (\sin x)-\int (-\sin x) dx =\tan (y)$ This gives: $ -[x \cos x+\cos x +c]=\tan (y)$ Hence, $\tan y=-\cos x-x \sin x+c$
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