Question

Consider a time-varying population that follows the logistic population model. The population has a limiting population of 800 , and at time t=0. its population of 200 is growing at a rate of 60 per year. a) Write a differential equation for the population P(t) that models this scenario. Solution: Put your math/explanation here... b) Identify an appropriate solution technique and solve this differential equation. Solution: Put your math/explanation here c) Use MATLAB to plot your solution for 101=10 Solution: d) How long will it take for the population to achieve 95% of its maximum population? Solution: Put your matheexplanation here... 1 e) Use the MATLAB dsolve() function to solve the differential equation and plot the solution on a new figure. The results provided by dsolve() and your solution from part (b) should match.

Answer 1

Differential Equation

`\[ (dP)/(dt) = 60 \cdot P \cdot \left(1 - (P)/(800)\right) \]`

Solution Technique and Solution

This is a logistic growth model, and we can solve the differential equation using separation of variables. The solution is given by:

`\[ P(t) = (800)/(1 + (3)/(4)e^(-60t)) \]`

MATLAB Plot

```matlab

t = linspace(1, 10, 100);

P = 800
`./ (1 + (3/4) * exp(-60 * t));`

plot(t, P);

xlabel('Time (t)');

ylabel('Population (P)');

title('Logistic Population Growth');

```

Time to Achieve 95% of Maximum Population

To find when
`\( P(t) \)`reaches 95% of the maximum population (760 in this case), solve for ct in the equation
`\( P(t) = 0.95 * 800 \).`

MATLAB dsolve() and Comparison

```matlab

syms P(t)

eqn =
`diff(P) == 60 * P * (1 - P/800);`

`sol = dsolve(eqn, P(0) == 200);`

`t_dsolve = linspace(1, 10, 100);P_dsolve = subs(sol, t_dsolve);`

`legend('dsolve()', 'Separation of Variables');xlabel('Time (t)');`

`ylabel('Population (P)');title('Comparison of Solutions');`

```

In summary, we've formulated the logistic growth differential equation, solved it using separation of variables, plotted the solution using MATLAB, determined the time to reach 95% of the maximum population, and finally, used the dsolve() function to compare solutions, ensuring consistency.

Answer 2

a) The differential equation for the population P(t) is dP/dt = rP(1 - P/Q). b) The solution can be found using separation of variables. c) MATLAB can be used to plot the population solution for a given time period. d) The time it takes for the population to achieve 95% of its maximum population can be found by solving an equation. e) The MATLAB dsolve() function can be used to solve the differential equation and plot the solution.

Step-by-step explanation:

a) To write a differential equation for the population P(t) that models this scenario, we can use the logistic population model. The logistic population model is given by the equation:

dP/dt = rP(1 - P/Q)

Where r represents the growth rate, P is the population at time t, and Q is the limiting population.b) To solve this differential equation, we can use the method of separation of variables. We can rewrite the equation as follows:

dP/(P(1 - P/Q)) = r*dt

Integrating both sides gives:

ln|P/Q - 1| - ln|P| = rt + C

Where C is the constant of integration. We can simplify this expression:

ln(P/Q - 1) - ln(P) = rt + C

ln((P/Q - 1)/P) = rt + C

((P/Q - 1)/P) = e^(rt + C)

((P/Q - 1)/P) = e^(rt)e^C

((P/Q - 1)/P) = Ce^(rt)

Simplifying further:

(P/Q - 1)/P = Ce^(rt)

(1 - P/Q)/P = 1/(Ce^(rt))

(1 - P/Q)/P = e^(-rt)/C

(1/Q)/P - 1/P = e^(-rt)/C

1/Q = e^(-rt)/C

Q = C/(e^(-rt))

Given that the population at time 0 is 200, we can substitute this value and solve for C:

Q = C/(e^0)

800 = C

Therefore, the equation for the population P(t) is:

P(t) = 800/(1 + 3e^(-rt))

c) Using MATLAB, we can plot the solution for t = 0:1:10. Here is an example of the code:

t = 0:1:10;

P = 800./(1 + 3*exp(-0.6*t));

plot(t, P, 'b-');

This will create a plot of the population P against time t.

d) To find the time it will take for the population to achieve 95% of its maximum population, we need to solve the equation P(t) = 0.95Q. Substituting the equation for P(t) gives:

800/(1 + 3e^(-rt)) = 0.95Q

Solving for t will give the time it takes for the population to reach 95% of its maximum population.

e) To use the MATLAB dsolve() function to solve the differential equation and plot the solution, we can use the following code:

syms P(t)

eqn = diff(P,t) == r*P*(1 - P/Q);

sol = dsolve(eqn);

fplot(sol,[0 10],'b-');

This will create a plot of the solution to the differential equation using the dsolve() function.