Question

Suppose a country's population at any time t grows in accordance with the rule dP dt = kP + I where P denotes the population at any time t, k is a positive constant reflecting the natural growth rate of the population, and I is a constant giving the (constant) rate of immigration into the country. If the total population of the country at time t = 0 is P0, find an expression for the population at any time t.

Answer 1

Answer:

$\mathbf{P =\bigg (P_o +\dfrac{ I}{k} \bigg)e^{kt}- \dfrac{I}{k}}$

Step-by-step explanation:

Given that:

A country population at any given time (t) is:

$\dfrac{dP}{dt}= kP+I$

where;

P = population at any time t

k = positive constant

I = constant rate of immigration into the country.

Using the method of separation of the variable;

$\dfrac{dP}{kP+1}= dt$

Taking integration on both sides:

$\int \dfrac{dP}{kP+I}= \int \ dt$

$\dfrac{1}{k} log (kP + I) = t+c_1 \ \ \ here: c_1 = constant \ of \ integration$

$log (kP + I) =k t+kc_1$

By applying the exponential on both sides;

$e^{log (kP + I) }=e^{k t+kc_1 }$

$KP+I = e^{kt} *e^{kc_1}$

Assume $e^{kc_1 }= C$

Then:

$kP + I = Ce^{kt}$

$kP = Ce^{kt}-I$

$P =\dfrac{ Ce^{kt}-I}{k} \ \ \---- Let \ that \ be \ equation \ (1)$

When time t = 0, The Total population of the country is also $P_o$

$P_o = \dfrac{Ce^{0(t)} -I}{k}$

$P_o = \dfrac{Ce^{0} -I}{k}$

$P_o = \dfrac{C-I}{k}$

C - I = kP₀

C =  kP₀ + I

Substituting the value of C back into equation(1), we have:

$P =\dfrac{ (kP_o+1)e^{kt}-I}{k}$

$P =\dfrac{ (kP_o+1)e^{kt}}{k} - \dfrac{I}{k}$

$\mathbf{P =\bigg (P_o +\dfrac{ I}{k} \bigg)e^{kt}- \dfrac{I}{k}}$