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 525: 4


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

Work Step by Step

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