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 - Page 251: 4

Answer

$ y $ = $ 5t + C $

Work Step by Step

$\frac{dy}{dt} $ = $ 5 $ To get the original equation, we have to integrate the aforementioned differential equation. $y=\int dy=\int5 dt$ = $\int5t^{0} dt$ = $\frac{5t^{0+1}}{1} + C $ = $ 5t^{1} + C $ = $ 5t + C $ Hence, $ y $ = $ 5t + C $
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