HELP! i have no idea how to do this!
A rabbit population satisfies the logistic equation dy/dx=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 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?
thank you!
To solve this problem, we will need to integrate the differential equation and use some initial conditions.
Let's start by integrating the logistic equation. The given equation is:
dy/dt = 2 * 10^7 * y * (10^6 - y)
To integrate this equation, we need to separate the variables:
(dy / y * (10^6 - y)) = 2 * 10^7 * dx
Now, let's simplify this expression:
(dy / (y * (10^6 - y))) = 2 * 10^7 * dx
Now, we can integrate both sides of the equation:
∫ (dy / (y * (10^6 - y))) = ∫ (2 * 10^7) dx
To integrate the left side, we can use partial fractions. We can express the fraction as:
1 / (y * (10^6 - y)) = A / (y) + B / (10^6 - y)
To find the values of A and B, we can multiply both sides by the denominator and then substitute y = 0 and y = 10^6:
1 = A * (10^6 - y) + B * (y)
Substituting y = 0 gives:
1 = A * 10^6
So, A = 1 / 10^6
Substituting y = 10^6 gives:
1 = B * 10^6
So, B = 1 / 10^6
Now we can rewrite the integral as:
∫ (1 / y) * (1 / 10^6) + (1 / (10^6 - y)) dy = 2 * 10^7 * ∫ dx
The integral on the left side is now straightforward to solve:
(1 / 10^6) * ln(y) - (1 / 10^6) * ln(10^6 - y) = 2 * 10^7 * x + C1
Now, let's solve for y as a function of x:
(1 / 10^6) * ln(y) - (1 / 10^6) * ln(10^6 - y) = 2 * 10^7 * x + C1
Multiply both sides by 10^6 to cancel out the denominators:
ln(y) - ln(10^6 - y) = 2 * 10^13 * x + C2
Use the property of logarithms to combine the terms:
ln(y / (10^6 - y)) = 2 * 10^13 * x + C2
Now, exponentiate both sides to eliminate the natural logarithm:
y / (10^6 - y) = e^(2 * 10^13 * x + C2)
Rearrange the equation to solve for y:
y = (10^6 - y) * e^(2 * 10^13 * x + C2)
Now, let's apply the initial condition that the population is suddenly reduced to 40% of its steady-state size by myxomatosis. At t = 0, y = 0.4 * y_ss, where y_ss is the steady-state size.
Substituting these values into the equation:
0.4 * y_ss = (10^6 - 0.4 * y_ss) * e^(2 * 10^13 * 0 + C2)
Now, solve for C2:
C2 = ln(0.4 * y_ss / (10^6 - 0.4 * y_ss))
Substitute the value of C2 back into the equation:
y = (10^6 - y) * e^(2 * 10^13 * x + ln(0.4 * y_ss / (10^6 - 0.4 * y_ss)))
To find the population 8 months later (t = 8 months), substitute x = 8 into the equation:
y = (10^6 - y) * e^(2 * 10^13 * 8 + ln(0.4 * y_ss / (10^6 - 0.4 * y_ss)))
Solve this equation to find the population 8 months later.
To find the time it takes for the population to build up again to 90% of its steady state size, we need to find the value of t when y = 0.9 * y_ss.
Substitute this value into the equation and solve for t.
I hope this helps!