Question

The population model given in (1) in Section 1.3 dP/dt \propto P or dP dt = kP (1)

fails to take death into consideration; the growth rate equals the birth rate. In another model of a changing population of a community it is assumed that the rate at which the population changes is a net rate that is, the difference between the rate of births and the rate of deaths in the community. Determine a model for the population P(t) if both the birth rate and the death rate are proportional to the population present at time t > 0. (Assume the constants of proportionality for the birth and death rates are k1 and k2 respectively. Use P for P(t).)

dP dt = __.

Answer 1

Answer:

$\frac{dP}{dt}=k_1 P -k_2 P= P(k_1 -k_2)$

Step-by-step explanation:

For this case we know that the birth rate is given by $b$ and the death rate is given by $d$.

We also know that these rates are proportional to the population size, so then we have this:

$b \propto P(t)  $

$d \propto P(t)$

And in order to have expression with the sign= we have the proportional constants given $k_1$ for b and $k_2$ for d, so then we convert the system of equations on this:

$b = k_1 P(t)$

$d = k_2 P(t)$

And then the change in the population respect to the time would be calculated on this way:

$\frac{dP}{dt} = b-d$

And if we replace what we found we got:

$\frac{dP}{dt}=k_1 P -k_2 P= P(k_1 -k_2)$

And we can solve the differential equation reordering the terms like this:

$\frac{dP}{P}= (k_1 -k_2) dt$

And if we integrate both sides we got:

$ln |P| = (k_1 -k_2) t +C$

Using exponentials we got:

$P(t) = e^{(k_1 -k_2)t} *e^c$

And we can rewrite this expression like this:

$P(t) = P_o e^{(k_1 -k_2)t}$ where $e^c = P_o$