logistic model: suppose growth is according to logistic mode

[math101]

Suppose that a population grows according to the logistic model.
k=.01 and carrying capacity is 60,000.

Solve the equation with the initial condition P(0) = 1,000.

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

dP/dt = .01P(1-P/60,000)

For P(0)= 1,000 do i just substitute P as 1,000?

dP/dt = .01(1,000) (1-(1,000/60,000))

dp/dt = 10(59/60)

dP/dt = 59/6 = 9.83

I wasn't sure if i was suppost to put P as 1,000 and solve, or something else.

 

[skeeter]

uhhh ... there's a bit more to this problem than you seem to realize. Solving the equation means using the technique of partial fractions to integrate the given differential equation to find P in terms of t.

to avoid confusion between the letter k's, let k = growth rate 0.01 and M = carrying capacity of 60000.

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

dP/dt = (k/M)P(M - P)

separate variables ...

dP/[P(M - P)] = (k/M)dt

using the method of partial fractions, 1/[P(M - P)] = (1/M)[1/P + 1/(M - P)]

(1/M)[1/P + 1/(M - P)]dP = (k/M)dt

[1/P - -1/(M - P)]dP = k dt

ln(P) - ln(M - P) = kt + C

ln[P/(M - P)] = kt + C

P/(M - P) = Ce^kt

when t = 0, P = M/60 ...

(M/60)/(M - M/60) = C ... C = 1/59

P/(M - P) = (1/59)e^kt

59P/(M - P) = e^kt

59P = Me^kt - Pe^kt

59P + Pe^kt = Me^kt

P(59 + e^kt) = Me^kt

P = Me^kt/(59 + e^kt)

in the standard form of a logistic equation ...

P = M/(1 + 59e^-kt) = 60000/(1 + 59e^-0.01t)

 

[math101]

Thank you for the explanation. I didn't relize that i needed to find the logistic equation from the differential equation.

So, i can use:

60,000/(1+59e^(-.01(t)))

to find a population after t hours.

so after 10 hours:

60,000/(1+59e^(-.01(10))) = 1,103