Assume that the number of fish P(t) after t years grows according to the logistic equation

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...

well, apparently

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.

To solve the given differential equation, we can first rewrite it as a separable equation.

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.