Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises: 48

Answer

$y=-t^2+3t-5$

Work Step by Step

$\frac{dy}{dt}=3-2t;$ $y(0)=-5$ $dy=(3-2t)dt$ $\int dy=\int(3-2t)dt$ $y=3t-t^2+C$ Substitute in initial condition $-5=3(0)-0^2+C$ $C=-5$ $y=-t^2+3t-5$
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