Another change of variable problem

1. Jul 24, 2005

spoon

I'm also attempting another problem...

A population has a periodic growth rate r(t) = A[1 + sin(t/(2*pi))], but otherwise follows the logistic population model with carrying capacity K. There is no threshold and the initial population is Yo = Y knot = K/2.

a. Modify the basic logistic equation for this population.
b. Use a change of variable z(t) = y/K to find a initial value equation in z(t). Then Solve for z(t).

So far I have:
Using the logistic growth equation: dy/dt = r(1-y/k)y
integral of [dy/((1-y/k)y)] = rt +c
substituting for r:
integral of [dy/((1-y/k)y)] = t*A[1 + sin(t/(2*pi))] + c

I can't really follow how to use the change of variables well so part b is the most confusing part.

2. Jul 25, 2005

qbert

ah,
as a friend of mine likes to say, "Classic mistake."
if r = r(t) you can't integrate it as if it were a constant then
plug in the value. you need:

$$\int \ldots dy = C + \int r(t) dt$$

There shouldn't be anything tricky to a change of variables.
(except, of course, deciding what varibles to use.)
It's a plug-and-chug operation.

if y = k z then
y' = ?

now, just replace every occurence of y with (kz) and every
occurence of y' with ?.

3. Jul 25, 2005

spoon

Thanks for the point about r(t), so I fixed that part, but I'm still confused as how to go about the rest...
If I substitute y with kz and y' with ?, then how do I integrate the following:
?/[(1-z)kz]
Because I could integrate it if the "?" were a "dz" right?

4. Jul 25, 2005

qbert

I guess I was unclear. it's your job to find out what
to put in in place of the ?.

What I thought i had hinted at
was if y = kz then y' = kz', because
k is a constant.

5. Jul 25, 2005

spoon

Sorry, looking back that was pretty clear...
So then it would be the integral of:
dz/(1-z)z = ln(1-z) + ln(z)
Then set this equal to:
C + Integral( r(t))
and just solve for z?

Last edited: Jul 25, 2005