dP/dt= P(a-bP) = bP(300 − P)
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.
I know you use partial fractions, but I get caught up by bP part. I didn't know if you carried it all the way to the end or what...
Steve
answered
P(a-bP) = bP(300 − P)
aP - bP^2 = 300bP - bP^2
so, a = 300b
dP/dt = P(300b-bP) = bP(300-P)
Try plugging that in. You already know what the logistic growth function looks like, so that should guide your efforts some.
Explain Bot
answered
The logistic equation is as follows:
dP/dt = bP(300 − P)
To make it a separable equation, we can rewrite it as:
dP/(P(300 − P)) = b dt
To integrate both sides, we can use partial fractions to simplify the left side of the equation:
1/(P(300 − P)) = A/P + B/(300 − P)
Next, we need to find the values of A and B. To do so, we can multiply both sides of the equation by P(300 - P) to eliminate the denominators:
1 = A(300 - P) + B(P)
Expanding the equation:
1 = 300A - AP + BP
Rearranging terms:
1 = (300A) + (-A + B)P
By comparing the coefficients of P on both sides of the equation, we get:
-1 = -A + B (equation 1)
And by comparing the constant terms on both sides, we get:
1 = 300A (equation 2)
Solving equation 2 for A, we find A = 1/300. Substituting this value into equation 1, we can solve for B:
-1 = - (1/300) + B
-1 + 1/300 = B
-300/300 + 1/300 = B
-299/300 = B
Now that we have A and B values, we can rewrite the left side of the equation using partial fractions:
1/(P(300 − P)) = 1/300P - 299/300(300−P)
Now, we can integrate both sides of the equation:
∫ (1/(P(300 − P))) dP = ∫ (1/300P - 299/300(300−P)) dt
Integrating the left side will require the use of logarithmic functions, while integrating the right side is straightforward:
ln|P(300 − P)| = (1/300)ln|P| - (299/300)ln|300 - P| + C
Simplifying the right side:
ln|P(300 − P)| = (1/300)ln|P| - (299/300)ln|300 - P| + C
Finally, we can exponentiate both sides to solve for P:
|P(300 − P)| = e^((1/300)ln|P| - (299/300)ln|300 - P| + C)
Since the exponential function is always positive, we can remove the absolute value signs:
P(300 − P) = e^((1/300)ln|P| - (299/300)ln|300 - P| + C)
This is the solution to the given differential equation in the form of a separable equation. The parameter b remains unknown in the solution.
