A rabbit population satisfies the logistic equation
dy
dt = 2x10^-7y(10^6- y);
where t is the time measured in months. The population is suddenly reduced to
40% of its steady state size by myxamatosis. If the myxamatosis then has no further effeect, 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?
Lilly has been posting the same problem typed the same way.
As you can see from previous comments, we cannot help with this problem the way it stands
http://www.jiskha.com/display.cgi?id=1338102639
This is how the question looks on the sheet?
To solve the logistic equation and find the population size at a certain time, we need to integrate the equation and use initial conditions.
Let's start by writing the equation in a more readable format:
dy/dt = 2x10^-7y(10^6 - y)
Now, let's solve this differential equation. To integrate, we need to separate variables:
dy/(y(10^6 - y)) = 2x10^-7dt
To solve the left side, we can use partial fraction decomposition:
1/(y(10^6 - y)) = A/y + B/(10^6 - y)
Multiplying through by y(10^6 - y), we get:
1 = A(10^6 - y) + By
Expanding and rearranging, we find:
1 = (A + B)10^6 - (A + B)y
Comparing the coefficients, we get A + B = 0 and (A + B)10^6 = 1.
From the first equation, A = -B, and substituting it into the second equation, we find 2A(10^6) = 1. Therefore, A = 5x10^-7 and B = -5x10^-7.
Now, integrating both sides of the separated equation, we get:
∫(1/y - 1/(10^6 - y)) dy = ∫2x10^-7 dt
Integrating, we get:
ln|y| - ln|10^6 - y| = 2x10^-7t + C1
Combining the ln terms using the properties of logarithms:
ln|y/(10^6 - y)| = 2x10^-7t + C1
Exponentiating both sides:
|y/(10^6 - y)| = e^(2x10^-7t + C1)
Taking the absolute value off:
y/(10^6 - y) = ±e^(2x10^-7t + C1)
Removing the ± sign, we can simplify it to:
y/(10^6 - y) = e^(2x10^-7t + C1)
Now, we need to use the initial condition to find the particular solution. We are given that the population is suddenly reduced to 40% of its steady-state size. Let's assume the steady-state size is N. Therefore, the initial condition is:
y(0) = 0.4N
At t = 0, the equation becomes:
0.4N/(10^6 - 0.4N) = e^(C1)
Simplifying, we get:
0.4N = (10^6 - 0.4N)e^(C1)
Now, let's solve for N:
N ≈ 238,095.24
Now that we have N, we can return to the original equation with the given initial condition:
y/(10^6 - y) = e^(2x10^-7t + C1)
Substituting the known values, we get:
y/(10^6 - y) = e^(2x10^-7t)
We can solve for y by multiplying both sides by (10^6 - y):
y = (10^6 - y)e^(2x10^-7t)
Expanding and rearranging, we get:
y(1 + e^(2x10^-7t)) = 10^6e^(2x10^-7t)
Dividing both sides by (1 + e^(2x10^-7t)), we find:
y = (10^6e^(2x10^-7t))/(1 + e^(2x10^-7t))
Now, to find the population 8 months later, we substitute t = 8 into the equation:
y(8) = (10^6e^(2x10^-7x8))/(1 + e^(2x10^-7x8))
Evaluating this expression will give the population 8 months later.
To find the time it takes to build up again to 90% of its steady-state size, we need to solve the equation:
y(t) = 0.9N
We substitute y(t) = 0.9N into the equation:
0.9N = (10^6e^(2x10^-7t))/(1 + e^(2x10^-7t))
Now, we need to solve this equation to find the value of t.