How to Solve the Logistic Equation for Elk Population Growth?

[JRangel42]

Homework Statement

This a problem that I didn't get completely right after a test, so I wouldn't mind figuring out what were my errors.

A state game commission releases 40 elk into a game refuge. After 5 years, the elk population is 104. The commission believes that the environment can support no more than 4000 elk.

a) find the specific solution to this differential equation; be sure to find values for all constants.

Homework Equations

dP/dt = kP(1- P/4000) 40≤ P ≤ 4000 A = (K - P initial)/K P = K/(1 + Ae^-kt)

The Attempt at a Solution

A = (K - P initial)/K
A = (4000 - 40)/4000
A = .99

P = K/(1 + Ae^-kt)
104 = 4000/(1 + .99e^-k5)
104(1 + .99e^-k5) = 4000
104 + 102.96e^-k5 = 4000
102.96e^-k5 = 3896
e^-k5 = 37.84
-5k = ln (37.84)
k = -1/5 ln (34.84)

 

[Mark44]

Homework Statement

This a problem that I didn't get completely right after a test, so I wouldn't mind figuring out what were my errors.

A state game commission releases 40 elk into a game refuge. After 5 years, the elk population is 104. The commission believes that the environment can support no more than 4000 elk.

a) find the specific solution to this differential equation; be sure to find values for all constants.

Homework Equations

dP/dt = kP(1- P/4000) 40≤ P ≤ 4000 A = (K - P initial)/K P = K/(1 + Ae^-kt)

The Attempt at a Solution

A = (K - P initial)/K
A = (4000 - 40)/4000
A = .99

P = K/(1 + Ae^-kt)
104 = 4000/(1 + .99e^-k5)
104(1 + .99e^-k5) = 4000
104 + 102.96e^-k5 = 4000
102.96e^-k5 = 3896
e^-k5 = 37.84
-5k = ln (37.84)
k = -1/5 ln (34.84)

What's your question?

 

[JRangel42]

I needed to find k in the equation.

 

[Mark44]

You found one. Does it work in your differential equation?

IOW, does your solution give P(0) = 40 and P(5) = 104?