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
$-5=3(0)-0^2+C$
$C=-5$
$y=-t^2+3t-5$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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 - Page 252: 48
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