Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises: 9

Answer

$p = e^{t^{3}/3-t+C}-1$

Work Step by Step

$dp/dt = t^2p - p + t^2 -1$ We need to rearrange the equation so that it becomes separable: $dp/dt = p(t^2 - 1) + (t^2 -1)$ $dp/dt = (t^2 - 1)(p + 1)$ $(1/(p+1))dp = (t^2 -1 ) dt$ Now let's integrate both sides: $\int(1/(p+1))dp =\int (t^2 -1 ) dt$ $ln|p+1| = t^3/3 -t + C$ Now to isolate p, we can include "e" on both sides: $e^{ln|p+1|} = e^{t^3/3 -t + C}$ $p+1 = e^{t^3/3 -t + C}$ $p = e^{t^3/3 -t + C}-1$
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