Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.9 Introduction to Differential Equations - 7.9 Exercises: 31

Answer

$$y = \pm \sqrt {2{t^3} + C} $$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dt}} = \frac{{3{t^2}}}{y} \cr & {\text{separate}} \cr & ydy = 3{t^2}dt \cr & {\text{integrate both sides}} \cr & \int {ydy} = \int {3{t^2}} dt \cr & \frac{{{y^2}}}{2} = {t^3} + c \cr & {y^2} = 2{t^3} + 2c \cr & {\text{solve for }}y \cr & y = \pm \sqrt {2{t^3} + 2c} \cr & {\text{set }}C = 2c \cr & y = \pm \sqrt {2{t^3} + C} \cr} $$
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