Question

The population of a suburb grows at a rate proportional to the population. Suppose the population doubles in size from 3000 to 6000 in a 6- month period and continues at the current rate of growth. Find the particular solution to the differential equation with the initial condition P(0)=3000.

Answer 1

Answer:

$P(t) = 3000e^{0.1155t}$

Step-by-step explanation:

The growth of the population can be modeled by the following differential equation:

$\frac{dP}{dt} = rP$

Where r is the growth rate, P is the population, and t is the time measures in months.

I am going to solve the above differential equation with the separation of variables method.

$\frac{dP}{P} = rdt$

Integrating both sides:

$ln P = rt + P(0)$

Where P(0) is the initial condition

We need to isolate P in this equation, so we do this

$e^{ln P} = e^{rt + P(0)}$

So

$P(t) = P(0)e^{rt}$

The problem states that P(0)=3000, so:

$P(t) = 3000e^{rt}$

The problem wants us to find the value of r:

It states that the population doubles in size from 3000 to 6000 in a 6- month period, meaning that P(6) = 6000. So

$6000 = 3000e^{6r}$

$e^{6r} = 2$

To isolate 6r, we apply ln to both sides.

$ln (e^{6r}) = ln 2$

$6r = 0.693$

$r = \frac{0.693}{6}$

r = 0.1155

The particular solution to the differential equation with the initial condition P(0)=3000 is:

$P(t) = 3000e^{0.1155t}$