Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - Chapter Review Exercises - Page 526: 6


$$ y=-(\frac{1}{2}\ln(x^2+1)+c)^{-1}.$$

Work Step by Step

By separation of variables, we have $$y^{-2}=\frac{xdx}{x^2+1}$$ then by integration, we get $$ -y^{-1}=\frac{1}{2}\ln(x^2+1)+c .$$ So the general solution is given by $$ y=-(\frac{1}{2}\ln(x^2+1)+c)^{-1}.$$
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