University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 16 - Practice Exercises - Page 16-34: 7

Answer

$y=\dfrac{c(x-1)}{x}$

Work Step by Step

In order to to solve the given differential equation we will have to separate the variables and then integrate. Here, we have $\int \dfrac{dy}{y}=\int \dfrac{dx}{x(x-1)}$ or, $\ln |y|=\ln (x-1)-\ln |x|+\ln c$ Then, we have $\ln |y|=\ln \dfrac{c(x-1)}{x}$ Hence, $y=\dfrac{c(x-1)}{x}$
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