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: 15



Work Step by Step

Here, we have $x dy+y dx=-3y^2 dy$ The general solution can be determined when we multiply the equation $x dy+y dx=-3y^2 dy$ with the integrating factor and then integrate on the both sides. we have $\int [xy]' =\int -3y^2 dy$ This implies that $(xy)=-(y^3)+c$ Hence, $xy=-y^3+c$
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