Differential Equations

• Aug 6th 2009, 07:43 AM
Tangera
Differential Equations
Hello!

Q: A population of animals live on an isolated island such that at any time, the number of births per unit time is proportional to the population and the number of deaths per unit time is proportional to the square of the population. If the population at time t is x and the population is changing at the greatest rate when x = 1, show that

$\frac{dx}{dt} = kx (2 - x)$

where k is positive constant.

(I have done the showing step, but I don't know how to solve the following parts.)

i) Find the general solution of the differential equation in terms of k.
ii) Show that there is a limit to the size of the population.

Pleas help! Thank you!
• Aug 6th 2009, 08:15 AM
skeeter
Quote:

Originally Posted by Tangera
Hello!

Q: A population of animals live on an isolated island such that at any time, the number of births per unit time is proportional to the population and the number of deaths per unit time is proportional to the square of the population. If the population at time t is x and the population is changing at the greatest rate when x = 1, show that

$\frac{dx}{dt} = kx (2 - x)$

where k is positive constant.

(I have done the showing step, but I don't know how to solve the following parts.)

i) Find the general solution of the differential equation in terms of k.
ii) Show that there is a limit to the size of the population.

Pleas help! Thank you!

http://www.math.neu.edu/~gilmore/U343su05files/logistic.pdf
• Aug 6th 2009, 11:34 PM
chisigma
In order to simplify the problem we write the equation as...

$\frac{dx}{dt}= - k\cdot x\cdot (x-2)$ , $k>0$ (1)

Before the 'attack' some useful considerations...

a) the (1) has two 'stationary solutions' : $x=0$ and $x=2$

b) for $x>2$ is $x'(t)<0$, for $0 is $x'(t)>0$. That means that if we set for $x(0)=x_{0}$ it is...

$\lim_{t \rightarrow \infty} x(t)= 2$ , $x_{0}>0$

c) $x'(t)$ is independent from t so that if $x_{0}(t)$ is solution od (1), $x_{0}(t-t_{0})$ with $t_{0}>0$ arbitrary is also solution of (1)

The (1) can be solved with standard sepation of variable and the general solution is...

$x= \frac{2}{1-2\cdot c\cdot e^{-kt}}$ (2)

... where the constant c is determined by the 'initial condition'. Some problem exist however when is $x_{0}=0$

Kind regards

$\chi$ $\sigma$