Dfq Prob, using Logisitc Equation

[Naeem]

Q. Develop the model of the logisitic equation and use it to solve the following. At first there are 100 fruit flies. After one day there are 200 fruit flies.The maximum population is 10,000 fruit flies.

a) Determine the population size P as a function of days t.

We know,

P(0) = 100

P(1) = 200

P max = 10,000

Logisitc equation:

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

Integrating both sides:

Integral ( dP/P(1-P/m) = Integral kdt

Using Partial fractions, we get:

PMax - P^2 = Ce^kt

Therefore,

C = PM - P^2 / e^kt

P(t) = Ce^kt / Pmax - P

Is this correct,

b. How many flies are present after 3 days?

For, this we can find C, using the initial condition: P(0) = 100

To find k, we can use P(1) = 200, and Pmax = 10,000

After finding all this,

We can do, P(3) = Ce^kt / Pmax - P

Any suggestions / ideas, Please help

 

[Naeem]

Please note, that M , in the previous parts, is actually referring to Pmax

 

[M98Ranger]

.

Yes, your approach is correct. To find the value of C, we can use the initial condition P(0) = 100:

C = PM - P^2 / e^kt
C = 10,000 - 100^2 / e^k(0)
C = 10,000 - 10,000 / 1
C = 0

Now, we can substitute the values of C, k, and Pmax into the equation we derived earlier to find the population size after 3 days:

P(3) = Ce^kt / Pmax - P
P(3) = 0 * e^k(3) / 10,000 - 0
P(3) = 0 / 10,000
P(3) = 0

Therefore, after 3 days, there are no fruit flies present. This is because the population reaches its maximum size of 10,000 and then starts to decline due to limited resources and competition. This can also be seen in the graph of the logistic equation, where the population initially grows rapidly, reaches a peak, and then starts to decline.