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: 9


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

Work Step by Step

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