Answer
$P=k(25t-\frac{1}{2}t^2) +C$
Work Step by Step
Separate Variables then integrate.
$\frac{dP}{dt}= k(25-t)$
$\int dP = \int k(25-t)dt$
$ P= k(25t- \frac{1}{2} t^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 - 6.2 Exercises - Page 412: 12
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