- #1

shan

- 57

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The population P(t) of a city suburb is governed by the initial value problem

P' = P((10^-2) - (10^-6)P) P(0) = 2000

Time measured in months.

a) What is the limiting value of the population at t approaches infinity?

b) How long before the population is one-half of this limiting value?

a) I used the formula for the general solution

birth-rate/death-rate = 10^-2/10^-6 = 10^4

But I'm not sure if that is what I was supposed to do...

b) Solving the DE

P' = P((10^-2) - (10^-6)P)

[tex]\int \frac{dP}{P(10^{-2}-10^{-6})} = \int dt[/tex]

Using partial fractions, the left can be seperated

[tex]\int \frac{10^2}{P} + \frac{10^{-4}}{10^{-2}-10^{-6}P} = \int dt[/tex]

[tex]10^2 * ln(P-ln(10^{-2}-10^{-6}P) = t + c[/tex]

Using the initial value P(0) = 2000, C = 250,000

Half of the limiting value = (10^4)/2 = 5000

Substituting into the equation, t = 139 months (3sf)