James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

614 Chapter 9 Differential Equations
M 2 P
ln Z
P
M 2 P
Z
P
Z − 2kt 2 C
Z − e 2kt2C − e 2C e 2kt
6
M 2 P
P
− Ae 2kt
where A − 6e 2C . Solving Equation 6 for P, we get
M
P 2 1 − Ae2kt ?
P
M − 1
1 1 Ae 2kt
so
P −
M
1 1 Ae 2kt
We find the value of A by putting t − 0 in Equation 6. If t − 0, then P − P 0 (the initial
population), so
M 2 P 0
P 0
− Ae 0 − A
Thus the solution to the logistic equation is
7
Pstd −
M
where A − M 2 P 0
1 1 Ae 2kt P 0
Using the expression for Pstd in Equation 7, we see that
which is to be expected.
lim Pstd − M
t l `
Example 2 Write the solution of the initial-value problem
dP
dt
− 0.08PS1 2
1000D P Ps0d − 100
and use it to find the population sizes Ps40d and Ps80d. At what time does the population
reach 900?
SOLUTION The differential equation is a logistic equation with k − 0.08, carrying
capacity M − 1000, and initial population P 0 − 100. So Equation 7 gives the population
at time t as
Pstd −
1000
1000 2 100
20.08t
where A −
1 1 Ae 100
− 9
Thus
Pstd −
1000
1 1 9e 20.08t
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