Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises: 1


$ y = -1/(x^3 + C) $

Work Step by Step

$dy/dx = 3x^2y^2$ Let's separate the y on one side and the x on the other side: $ (1/y^2 )dy = 3x^2dx$ Now let's integrate both sides: $\int (1/y^2 )dy = \int 3x^2dx$ $-1/y = x^3 + C$ Isolate y: $ y = -1/(x^3 + C) $
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