Question

A rabbit population satisfies the logistic equation
dy/dt = 2 × 10^−7 y (10^6 − y)

where t is the time measured in months. The population is suddenly reduced to 40% of its steady state size by myxamatiosis. If the myxamatosis then has no further effect, how large is the population 8 months later? How long will it take for the population to build up again to 90% of its steady state size?

I keep getting a really big number.... HELP!!

Answer 1

General solution of logistic equation:

dy/dt=Ky(L-y) is given by

y(t)=LA/A+e^(-LKt), where A=y_0/L-y_0, y_0=y(0) and y(t)=L is the steady state solution.

Then, y_0 = 0.4L = 0.4x10^6 = 4x10^5 and A=y_0/L-y_0 = 4x10^5/(10^6 - 4x10^5) = 6 2/3.

If t is the time it takes the population to build up to 90%, y(t)=0.9L = 0.9x10^6 = 9x10^5.

So that, 9x10^5 = 6 2/3x10^6/(6.67+e^(-2x10^-7t)

6 2/3+e^(-2x10^-7t)=200/27

t=1500522.96...... this just seems way to high!

Answer 2

this number equates to more than 127,000 years... cant be right!

Answer 3

ok i got 13.. i may have made an error, i'll post my workings

Answer 4

the steady state size is where there is no increase or decrease, hence dy/dt = 0
we get this at either y = 0 or y = 10^6

only the 10^6 value makes sense, so this is our steady state size

\[\frac{dy}{dt} = (2 \times 10^{-7})y(10^6 - y)\]

\[\frac{1}{y(10^6 - y)} \frac{dy}{dt} = (2 \times 10^{-7})\]

integrating:

\[\int\limits{\frac{1}{y(10^6 - y)} \frac{dy}{dt}} \text{ } dt = \int\limits{(2 \times 10^{-7}) } \text{ } dt\]

we need to seperate the LHS into partial fractions:

\[\frac{1}{y(10^6 - y)} = \frac{A}{y} + \frac{B}{10^6 - y}\]

i got \[A =B = 10^{-6}\]

so:

\[\int\limits{\frac{1}{y(10^6 - y)} }dy =\frac{1}{10^6} \int\limits{ \frac{1}{y} + \frac{1}{10^6 - y}}\text{ }dy\]

Answer 5

instead of adding arb. constants im going to do a definite integration, its neater

\[\frac{1}{10^6}\int\limits_{4 \times 10^5}^{9 \times 10^5}{\frac{1}{y} + \frac{1}{10^6 - y} } \text{ } dy = (2 \times 10^{-7})t\]

Answer 6

the limits of y are 40% to 90%

Answer 7

ie 4 x 10^5 to 9 x 10^5

and the limits of t are 0 to t

Answer 8

Thanks for your help with this!
ok, so by working out the 90%, this gives me what the number of fish will be at 90%? or how much time it takes to get to 90%? How do I work out how many fish will be in the pond after 8 months?

Answer 9

\[[lny - \ln(10^6 - y) ]^{9 \times 10^5}_{4 \times 10^5} = 0.2t\]

Answer 10

we first found the value of y where the rate of change is zero, think of it as the upper limit of the population of rabbits. by multiplying this by 0.9, we get 90% of the "steady state size"

Answer 11

oh ok

Answer 12

Then to get 8 month population, I just sub in t=8?

Answer 13

yes, as long as we have y in terms of t

Answer 14

isn't y already in terms of t? I mean, in the initial equation?

Answer 15

ah no the initial equation is a differential equation, you see it has a dy/dx term? we dont know the rate of change at t = 8 , we have to solve the equation

Answer 16

like before, we separate variables, but this time we have y as the upper limit,
getting:

\[\int\limits^y_{4 \times 10^5}\frac{1}{y(10^6 - y)} dy = \int\limits^t_0(2 \times 10^{-7}) dt\]

Answer 17

in fact it would have been better to do this first, and then just sub in values of y and t to answer the questions

Answer 18

oh ok, Im catching on now!... Thanks so much for your help tonight. Much appreciated!

Answer 19

cool, glad to help :)
partial fractions etc etc..

we get:

\[lny - \ln(10^6 - y) - \ln(4 \times 10^5) + \ln(6 \times 10^5) = 0.2t\]

Answer 20

then using that equation, solve for y.... where t=8

Answer 21

yeah :)

Answer 22

you're a champ! Thanks mate. Have a good one!

Answer 23

you to! :D cya

Answer 24

:) cheers