University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 2


a) $\dfrac{1}{x^2}=y^2$ b) $\dfrac{1}{(x+3)^2}=y^2$ c) $\dfrac{1}{(x+C)^2}=y^2$

Work Step by Step

a) $y^2=\dfrac{1}{x^2}$ Thus, $y'=-[(-1) x^{-2}]=\dfrac{1}{x^2}=y^2$ b) $y^2=\dfrac{1}{(x+3)^2}$ Thus, $y'=-[(-1) (x+3)^{-2}]=\dfrac{1}{(x+3)^2}=y^2$ c) $y^2=\dfrac{1}{(x+C)^2}$ Thus, $y'=-[(-1) (x+C)^{-2}]=\dfrac{1}{(x+C)^2}=y^2$
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