I am having trouble figuring out where to start with this one.
Biologists stocked a lake with 40 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 300.
(a) Assume that the number of fish P(t) after t years grows according to the logistic equation
dP/dt = P(a − bP) = bP(a/b − P)
Here, according to Section 2.15 of the course notes, a/b is the carrying ca-pacity of the lake. Find the solution of this DE (with the given initial value at t = 0) as a separable equation. Your solution formula will contain b as an unknown parameter.
The logistic function is
y = a/(1+be^(-kx))
y(∞) = 300
==> a = 300
y(0) = 40
==> 300/(1+b) = 40
b = 15/2
The k just affects how slowly the population approaches the final value. See
http://www.wolframalpha.com/input/?i=plot+y%3D300%2F(1%2B7.5e%5E-x),+y%3D300%2F(1%2B7.5e%5E(-2x))+for+-5+%3C%3D+x+%3C%3D+5
To find the solution of the differential equation, we need to solve it as a separable equation. The given logistic equation is:
dP/dt = P(a - bP)
To solve this equation as a separable equation, we need to separate the variables and integrate both sides.
First, we rewrite the equation in a more convenient form:
dP/(P(a - bP)) = dt
Next, we separate the variables by moving the P terms to one side and the t terms to the other side:
dP/(P(a - bP)) = dt
1/(P(a - bP)) * dP = dt
Now, we can integrate both sides. On the left side, we use partial fractions to integrate:
∫(1/(P(a - bP))) dP = ∫dt
To find the partial fractions, we need to decompose the P terms in the denominator:
1/(P(a - bP)) = A/P + B/(a - bP)
Multiplying both sides by P(a - bP), we get:
1 = A(a - bP) + BP
Expanding the right side:
1 = Aa - AbP + BP
Now, we match coefficients of the same powers of P:
0P: 0 = -Ab + B
1P: 0 = A
From the equation -Ab + B = 0, we can solve for A:
B = Ab
Substituting the value of B into the equation 0 = A, we get:
0 = A(a - bP)
Since a - bP is not zero (given in the question statement), then A must be zero:
A = 0
Therefore:
1/(P(a - bP)) = B/(a - bP)
Going back to the integral equation:
∫(1/(P(a - bP))) dP = ∫dt
We can now integrate both sides:
∫(B/(a - bP)) dP = ∫dt
B∫(1/(a - bP)) dP = ∫dt
Evaluating the integrals:
B * (-1/b) * ln|a - bP| = t + C
Multiplying both sides by -b:
B * ln|a - bP| = -bt - bC
Taking the exponent of both sides:
|a - bP|^B = e^(-bt - bC)
Since ln|a - bP| could be positive or negative, we remove the absolute value sign and consider two cases:
Case 1: a - bP > 0
a - bP = e^(-bt - bC)
Case 2: a - bP < 0
-(a - bP) = e^(-bt - bC)
Now, we can solve for P in each case:
Case 1: a - bP = e^(-bt - bC)
a - bP = Ce^(-bt)
Solving for P:
P = (a - Ce^(-bt))/b
Case 2: -(a - bP) = e^(-bt - bC)
a - bP = -Ce^(-bt)
Solving for P:
P = (a + Ce^(-bt))/b
Therefore, the solution of the differential equation (dP/dt = P(a - bP)) in separable form with the given initial value at t = 0 is:
P(t) = (a - Ce^(-bt))/b for a - bP > 0
P(t) = (a + Ce^(-bt))/b for a - bP < 0
Where C is a constant determined by the initial condition.