Logistic model/initial value problem

[ineedhelpnow]

http://www.math.upenn.edu/~zhentao/math114/sol2.pdf

for number two (click the link), can someone explain how the step where they solved for $e^{-k}$

 

[MarkFL]

The number of fish tripled in the first year, hence:

\(\displaystyle P(1)=1200=\frac{10000}{1+24e^{-k}}\)

\(\displaystyle 3=\frac{25}{1+24e^{-k}}\)

\(\displaystyle 3\left(1+24e^{-k}\right)=25\)

\(\displaystyle 3+72e^{-k}=25\)

\(\displaystyle 72e^{-k}=22\)

\(\displaystyle 36e^{-k}=11\)

\(\displaystyle e^{-k}=\frac{11}{36}\)

 

[ineedhelpnow]

1200? that's because the number tripled right?

 

[MarkFL]

1200? that's because the number tripled right?

Yes, the initial population was 400 so after one year there would be 1200 if it tripled during that time. :D

 

[blue_raver22]

is done

Sure, I would be happy to explain that step.

In this problem, we are given a logistic model, which is a type of differential equation that describes the growth or decay of a population over time. The equation is given by:

dP/dt = kP(1-P)

where P represents the population and k is a constant.

To solve this equation, we need to find a function P(t) that satisfies the equation. This is known as an initial value problem because we are given an initial value for P, which we can use to find the solution.

In the link provided, the initial value is given as P(0) = 0.5, which means that at time t=0, the population is 0.5.

To solve for P(t), we first need to separate the variables by dividing both sides of the equation by P(1-P):

1/(P(1-P)) dP/dt = k

Next, we can use the fact that the derivative of ln(x) is 1/x to rewrite the left side of the equation as:

d/dt ln(P(1-P)) = k

Integrating both sides with respect to t, we get:

ln(P(1-P)) = kt + C

where C is a constant of integration.

Now, we can use the initial value P(0) = 0.5 to solve for C. Plugging in t=0 and P(0)=0.5, we get:

ln(0.5(1-0.5)) = C

ln(0.25) = C

Solving for C, we get:

C = ln(0.25)

Now, we can substitute this value back into our equation:

ln(P(1-P)) = kt + ln(0.25)

Using the properties of logarithms, we can rewrite this as:

ln(P(1-P)/0.25) = kt

Next, we can use the fact that e^ln(x) = x to rewrite the left side of the equation as:

P(1-P)/0.25 = e^kt

Finally, we can solve for P by multiplying both sides by 0.25 and taking the square root:

P = 0.5 * √(e^kt)

This is the solution to our initial value problem. To solve for e^-k, we can simply take the reciprocal of both