Answer
$y=-\frac{k}{2t^2}+C$
Work Step by Step
$$\frac{dy}{dt} = \frac{k}{t^3}$$
$$\int dy = \int kt^{-3}dt$$
$$y=\frac{kt^{-3+1}}{-3+1}+C$$
$$y=-\frac{k}{2t^2}+C$$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 6 - Differential Equations - Review Exercises - Page 431: 21
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