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: 20


$y=\dfrac{-1+\sin x}{x}$

Work Step by Step

Here, we have $y'+(\dfrac{1}{x}) y=\dfrac{\cos x}{x}$ The integrating factor is: $I=e^{\int (\dfrac{1}{x}) dx}=x$ Now, $x[y'+(\dfrac{1}{x}) y]=x(\dfrac{\cos x}{x})$ This implies that $\int [xy]' =\int \cos x dx $ $\implies (xy)=(\sin x)+c$ Now, apply the initial conditions. Thus, we have $c=-1$ $xy=\sin x-1 \implies y=\dfrac{-1+\sin x}{x}$
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