Answer
$y=k(50t - \frac{t^2}{2})+C$
Work Step by Step
$$\frac{dy}{dt} = k(50-t)$$
$$\int dy = \int k(50-t^1) dt$$
$$y=k(50t - \frac{t^{1+1}}{1+1})+C$$
$$y=k(50t - \frac{t^2}{2})+C$$
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: 22
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