Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 8



Work Step by Step

Need to separate the variables to determine the differential equation and then integrate the obtained equation on the both sides. $\int \dfrac{1}{y^2-1}dy=\int \dfrac{dx}{x}$ This implies that $\dfrac{\ln [{\dfrac{y-1}{y+1}}]}{2}=\ln |x|+c'$ or, $\ln [{\dfrac{y-1}{y+1}}]=2 \ln |x|+\ln c'$ Hence, $\dfrac{y-1}{y+1}=cx^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.