University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 16 - Section 16.3 - Applications - Exercises - Page 16-21: 5

Answer

$x^2+y^2=K$

Work Step by Step

Write the differential equation as follows: $-\dfrac{dx}{dy}=\dfrac{y}{x}$ or, $ -x \ dx =y \ dy$ $-\int x dx=\int y dy$ or, $\dfrac{y^2}{2}+C= -\dfrac{x^2}{2}$ or, $ \dfrac{-x^2-y^2}{2}=C$ or $ x^2+y^2=-2C$ or, $x^2+y^2=K$
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