Differential equation- solve for k

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Homework Statement

Biologists stocked a lake with 143 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 8000. The number of fish doubled in the first year.

(a) Assuming that the size of the fish population satisfies the logistic equation

dP/dt = kP(1-P/N)

determine the constant , and then solve the equation to find an expression for the size of the population after years.

Homework Equations

none in particular

The Attempt at a Solution

I didn't think that this problem would be difficult but the answer I am getting doesn't make any sense. Basically I try to solve for P to start with. So dP/(P-P^2/N) = kdt and integrate

ln(P/(P-N)) = kt+c

It would make the most sense to solve for k right away so

(ln(P/(P-N)) - c)/t = k

The problem though is that N = 8000 while P will always be less than this. This will cause there to be a negative number inside the natural log and you can't take the log of a negative number. Where am I going wrong?

 

[Mark44]

I get ln(P/(N - P)) = kt + C. For the log term to be defined, 0 < P < N.

 

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I get ln(P/(N - P)) = kt + C. For the log term to be defined, 0 < P < N.

Really? Let me try to integrate this again and see what happens... hold on

 

[hover]

Ya you're right on that. I must have screwed up when breaking I broke it up into partial fractions. Another part that confuses me is do I solve for p first and then solve for k or do I head straight for k? My answers are still sketchy...

 

[Mark44]

I think you want to solve for C first, and then k. I'm not sure you have an initial condition, P(0), but you know that P(1) = 2P₀.

 

[hover]

Woops. Sorry that it didn't copy over in my original post. Starting fish population is 143 fish. So P(0) = 143 and P(1) = 286.

 

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Ah! k is equal to .7115 ! Nice! I also found c to equal -4.006.