# Math Help - Population growth model

1. ## Population growth model

The question

The simple population growth model, $\frac{dy}{dt} = ky$ (because of limitation on resources, pollution, ...) is unsatisfactory over a 'long' period. We might look at

$\frac{dy}{dt} = k(y)y$ ...........(1)

(which of course ignores seasonal and other variations with time). Mathematically one of the simplest assumptions we can make is that k(y) decreases linearly as y increases. In this case (1) may be written in the form

$\frac{dy}{dt} = k(1 - \frac{y}{K})y$, ...........(2)
where k and K are constants.

a) i) Equation (2) has two constant (stationary) solutions. What are they?

I'm not sure how to calculate this. Do I have to solve the ODE? Or is there an obvious method? Thanks.

2. Hmm, I think I get it now. I just need find values of 'y' to make the RHS zero?

3. Originally Posted by Glitch

$\frac{dy}{dt} = k(1 - \frac{y}{K})y$, ...........(2)
where k and K are constants.

a) i) Equation (2) has two constant (stationary) solutions. What are they?

I'm not sure how to calculate this. Do I have to solve the ODE?
This is worth solving, separate, decompose with partial fractions and integrate.

Then look at solutions for $\pm K$

4. Ok, I tried to do that, and I got this:

$\frac{1}{y(1-\frac{y}{K})}dy = k dt$
$\frac{A}{y} + \frac{B}{(1 - \frac{y}{K})} = \frac{1}{y(1-\frac{y}{K})}}$

So we get:
A = 1
B = $\frac{1}{K}$

$\int{\frac{1}{y} + \frac{1}{K(1-\frac{y}{K})}$ = kt + C

$ln(y) + \int{\frac{1}{K - y}} = kt + C$

$ln(\frac{y}{K - y}) = kt + C$

Take the exponential of each side:

$\frac{y}{K - y} = e^{kt + C}$

$y = e^{kt + C}(K - y)$

Is this right so far? Thanks.

5. However, to answer the question asked, "i) Equation (2) has two constant (stationary) solutions. What are they?", a "stationary solution" is a constant solution, dy/dt= 0 so, yes, just set the right side of the equation equal to 0: solve $k\left(1- \frac{y}{K}\right)y= 0$.