Math / Calculus

Question(b) Let's now consider a modified differential equation given by $\frac{d P}{d t}=\frac{1}{2} P(3-P)$ .
Find a stable equilibrium solution: $P=$ $\square$ Find an unstable equilibrium solution: $P=$ $\square$ If $P(0)$ is positive, describe the long-term behavior of the solution to $\frac{d P}{d t}=\frac{1}{2} P(3-P)$ . A. The value of $P$ approaches a nonzero constant. B. The value of $P$ approaches zero. C. The value of $P$ increases without bound. D. The value of $P$ oscillates and does not approach a limit. E. None of the above
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

What is this asking? We're looking for special solutions to this differential equation, called equilibrium solutions, where the population ( $P$ ) doesn't change over time, and then we want to figure out what happens to the population in the long run if it starts out positive. Watch out! Don't mix up stable and unstable equilibrium solutions!
Stable means the population tends towards that value, while unstable means it tends away from it.

STEP 2

1. Find Equilibrium Solutions
2. Analyze Stability
3. Long-Term Behavior

STEP 3

Equilibrium solutions mean the population isn't changing, so the rate of change is zero.
We set $\frac{dP}{dt}$ to zero and solve for $P$ : $\frac{dP}{dt} = \frac{1}{2}P(3-P) = 0$

STEP 4

This equation is true if either $P = 0$ or $3 - P = 0$ .
So, our equilibrium solutions are $P = \mathbf{0}$ and $P = \mathbf{3}$ .

STEP 5

Let's think about what happens if $P$ is a tiny bit bigger than zero, like $0.1$ .
Then $3 - P$ will be positive (like $2.9$ ), and so will $\frac{1}{2}P(3-P)$ , meaning $\frac{dP}{dt}$ is positive.
If $P$ is slightly negative, then $\frac{dP}{dt}$ is negative.
This means $P$ moves away from $0$ , so $P = 0$ is an unstable equilibrium.

STEP 6

Now, let's think about what happens if $P$ is a tiny bit bigger than $3$ , like $3.1$ .
Then $3 - P$ will be negative (like $-0.1$ ), making $\frac{dP}{dt}$ negative, so $P$ decreases.
If $P$ is slightly smaller than $3$ , like $2.9$ , then $3 - P$ is positive (like $0.1$ ), and $\frac{dP}{dt}$ is positive, so $P$ increases.
This means $P$ moves towards $3$ , so $P = 3$ is a stable equilibrium.

STEP 7

We're told $P(0)$ is positive.
If it starts between $0$ and $3$ , it will grow towards $3$ .
If it starts greater than $3$ , it will shrink towards $3$ .
Either way, as long as it's not exactly zero, the population will approach $3$ in the long run.

STEP 8

Stable equilibrium solution: $P = \mathbf{3}$ Unstable equilibrium solution: $P = \mathbf{0}$ Long-term behavior: A.
The value of $P$ approaches a nonzero constant.

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Studdy Solution

STEP 1

STEP 2

1. Find Equilibrium Solutions
2. Analyze Stability
3. Long-Term Behavior

STEP 3

Equilibrium solutions mean the population isn't changing, so the rate of change is zero.
We set $\frac{dP}{dt}$ to zero and solve for $P$ : $\frac{dP}{dt} = \frac{1}{2}P(3-P) = 0$

STEP 4

This equation is true if either $P = 0$ or $3 - P = 0$ .
So, our equilibrium solutions are $P = \mathbf{0}$ and $P = \mathbf{3}$ .

STEP 5

STEP 6

STEP 7

We're told $P(0)$ is positive.
If it starts between $0$ and $3$ , it will grow towards $3$ .
If it starts greater than $3$ , it will shrink towards $3$ .
Either way, as long as it's not exactly zero, the population will approach $3$ in the long run.

STEP 8

Stable equilibrium solution: $P = \mathbf{3}$ Unstable equilibrium solution: $P = \mathbf{0}$ Long-term behavior: A.
The value of $P$ approaches a nonzero constant.

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Studdy Solution

STEP 1

STEP 2

1. Find Equilibrium Solutions2. Analyze Stability3. Long-Term Behavior

STEP 3

Equilibrium solutions mean the population isn't changing, so the rate of change is zero.We **set** dPdt\frac{dP}{dt}dtdP​ to **zero** and **solve** for PPP: dPdt=12P(3−P)=0 \frac{dP}{dt} = \frac{1}{2}P(3-P) = 0 dtdP​=21​P(3−P)=0

STEP 4

This equation is true if either P=0P = 0P=0 or 3−P=03 - P = 03−P=0.So, our **equilibrium solutions** are P=0P = \mathbf{0}P=0 and P=3P = \mathbf{3}P=3.

STEP 5

STEP 6

STEP 7

We're told P(0)P(0)P(0) is positive.If it starts between 000 and 333, it will grow towards 333.If it starts greater than 333, it will shrink towards 333.Either way, as long as it's not *exactly* zero, the population will approach 333 in the long run.

STEP 8

Stable equilibrium solution: P=3P = \mathbf{3}P=3 Unstable equilibrium solution: P=0P = \mathbf{0}P=0 Long-term behavior: A.The value of PPP approaches a nonzero constant.

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1. Find Equilibrium Solutions
2. Analyze Stability
3. Long-Term Behavior

Equilibrium solutions mean the population isn't changing, so the rate of change is zero.
We set $\frac{dP}{dt}$ to zero and solve for $P$ : $\frac{dP}{dt} = \frac{1}{2}P(3-P) = 0$

This equation is true if either $P = 0$ or $3 - P = 0$ .
So, our equilibrium solutions are $P = \mathbf{0}$ and $P = \mathbf{3}$ .

We're told $P(0)$ is positive.
If it starts between $0$ and $3$ , it will grow towards $3$ .
If it starts greater than $3$ , it will shrink towards $3$ .
Either way, as long as it's not exactly zero, the population will approach $3$ in the long run.

Stable equilibrium solution: $P = \mathbf{3}$ Unstable equilibrium solution: $P = \mathbf{0}$ Long-term behavior: A.
The value of $P$ approaches a nonzero constant.