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: 16


$y=x^{-3}(\sin x+c)$

Work Step by Step

Here, we have $y'+\dfrac{3}{x} y=x^{-3} \cos x$ The integrating factor is: $I=e^{\int (\dfrac{3}{x}) dx}=x^3$ Now, $x^3[y'+\dfrac{3}{x} y]=(x^3)(x^{-3}) (\cos x)$ This implies that $\int [x^3y]' =\int \cos x dx$ Hence, $y=x^{-3}(\sin x+c)$
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