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: 3


$$ y= (\frac{4}{3}t^3+c)^{1/4}.$$

Work Step by Step

By separation of variables, we have $$y^3dy=t^2dt $$ then by integration, we get $$ \frac{1}{4}y^4=\frac{1}{3}t^3+c .$$ So the general solution is given by $$ y= (\frac{4}{3}t^3+c)^{1/4}.$$
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