Calculus (3rd Edition)

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

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 34



Work Step by Step

By separation of variables, we have $$y^2dy=x^{-3}dx$$ then by integration, we get $$\frac{1}{3} y^3=-\frac{1}{2}x^{-2}+c\Longrightarrow y^3=-\frac{3}{2}x^{-2}+A.$$ Now, since $y(1)=0$, then $A= \frac{3}{2}$. So the general solution is given by $$y^3=-\frac{3}{2}(x^{-2}-1)$$
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