Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises: 3

Answer

$ y $ = $ 3t^{3} + C $

Work Step by Step

$\frac{dy}{dt} $ = $ 9t^{2}$ To get original equation, we have to integrate the aforementioned differential equation. $y=\int dy=\int9t^{2} dt$ = $\frac{9t^{2+1}}{3} + C $ = $ 3t^{3} + C $ Hence, $ y $ = $ 3t^{3} + C $
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