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: 3

Answer

$ y = \sqrt {x^2 + 2ln|x| + C}$ or $-\sqrt {x^2 + 2ln|x| + C }$

Work Step by Step

$xyy' = x^2 + 1$ can be rewritten as: $xy (dy/dx) = x^2 + 1$ Now let's separate y on one side and x on the other side: $y dy = ((x^2 + 1)/x )dx$ Now let's integrate both sides: $\int y dy = \int((x^2 + 1)/x )dx$ $y^2/2 = \int(x+ 1/x )dx$ $y^2/2 = x^2/2 + ln|x| + C_1 $ $y^2 = x^2 + 2ln|x| + 2C_1 $ Let's replace $2C_1 = C$ $ y = \sqrt {x^2 + 2ln|x| + C }$ or $-\sqrt {x^2 + 2ln|x| + C} $
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