Population modeled by function

Question

A population is modeled by a function P that satisfies the differential equation dP/dt = k(1200-P), where

t is time in years and k is the constant of proportionality. When t = 0, the population is 300, P(0) = 300.

(a) Find P(t) in terms of t and k.

(b) If P(4) = 600, find the value of k.

(c) Find lim P(t ) as t approaches positive infinity

Answer 1

(a) we separate variables and integrate: ∫dP/(1200 - P) = ∫kdt;

ln(P - 1200) = -kt + lnC ; P - 1200 = Ce^-kt; P(t) = 1200 + Ce^-kt; P(0) = 1200 + C = 300. C = - 900;

P(t) = 1200 - 900e^-kt.

(b) P(4) = 1200 - 900e^-4k = 600; e^-4k = 2/3; -4k = ln(2/3); k = -ln(2/3)/4 = 0.1013663

(c) P(t) = 1200 - 9000e^-0.1013663t

lim P(t) = 1200 - 900·0) = 1200

t→∞

Answer 2

To solve differential equations one of the fundamental strategies you can use is separability. By separating the independent and dependent variables we can then find anti-derivatives and solve the equation. Here we move the (1200-P) to the other side and get dP/(1200-P) = k dt. This implies that log(-1 / ( 1200-P)) = kt + c where c is a constant. We can then proceed as normal to solve for P:

-1/(1200-P) = exp(kt+c1) = c2 * exp(kt)

c3*exp(-kt) = 1200-P

P = 1200 - c3 * exp(-kt)

This satisfies the original equation because dP/dt = k*c3*exp(-kt). This solves part a).

To solve b) simply plug in 0 and get 300 = 1200 - c3*exp(-t*0) = 1200 - c3, so c3 = 900. Thus P = -900*exp(-kt) + 1200

For b) we plug in 4 for t to get 600 = -900*exp(-k*4) + 1200 which becomes exp(-4k)= 2/3 so k = -1/4 * log(2/3)

For large x the function exp(-k*x) goes to zero if k is positive but diverges to infinity if k is negative. In this case k is positive because 2/3 < 1, therefore it's behavior at large x is to go to zero. Thus lim P(t) for large t is 1200.