Breakdown of a Logistic Equation

In summary, the given conversation discusses how to solve the Logistic Equation using various methods including finding equilibrium solutions, separating the equation, and using partial fraction decomposition and integration. However, there are some mistakes in the attempt at a solution and a need to revise the integration process.
  • #1
defaultusername
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Homework Statement


I feel so stuck.
Given the Logistic Equation:
$$\frac{dP}{dt}=kP(1-\frac{P}{A})$$

a.). Find the equilibrium solutions by setting $$\frac{dP}{dt}=0$$ and solving for P.
b.). The equation is separable. Separate it and write the separated form of the equation.
c.). Use partial fraction decomposition and then integrate both sides of the equation to solve for P.

Homework Equations


$$\frac{dP}{dt}=kP(1-\frac{P}{A})$$

The Attempt at a Solution


a.) $$\frac{dP}{dt}=0$$
$$⇒P=0, A$$

b.) $$⇒\frac{1}{P}+(\frac{\frac{1}{A}}{1-\frac{P}{A}})dP=k dt$$
$$⇒(\frac{dP}{1-\frac{P}{A}})=k dt$$

c.) $$⇒(\frac{dP}{1-\frac{P}{A}})=k dt$$
$$⇒P=\frac{B}{P}+\frac{C}{1-\frac{P}{A}}$$

I am not even sure if I am doing everything correctly or not. I need to find a common denominator to solve for B and C but my attempts always end up as a huge mess.
 
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  • #2
defaultusername said:

Homework Statement


I feel so stuck.
Given the Logistic Equation:
$$\frac{dP}{dt}=kP(1-\frac{P}{A})$$

a.). Find the equilibrium solutions by setting $$\frac{dP}{dt}=0$$ and solving for P.
b.). The equation is separable. Separate it and write the separated form of the equation.
c.). Use partial fraction decomposition and then integrate both sides of the equation to solve for P.

Homework Equations


$$\frac{dP}{dt}=kP(1-\frac{P}{A})$$

The Attempt at a Solution


a.) $$\frac{dP}{dt}=0$$
$$⇒P=0, A$$

b.) $$⇒\frac{1}{P}+(\frac{\frac{1}{A}}{1-\frac{P}{A}})dP=k dt$$
I can't tell what you did here (above). Start with the given diff. equation and separate it, using my hint below.
$$\frac{dP}{dt}=kP(1-\frac{P}{A})\\
\Rightarrow \frac{dP}{dt}=\frac k A P(A - P)$$
Now it should be easier to separate.
defaultusername said:
$$⇒(\frac{dP}{1-\frac{P}{A}})=k dt$$

c.) $$⇒(\frac{dP}{1-\frac{P}{A}})=k dt$$
$$⇒P=\frac{B}{P}+\frac{C}{1-\frac{P}{A}}$$

I am not even sure if I am doing everything correctly or not.
What you did is definitely wrong. You should get P as a function of t.
defaultusername said:
I need to find a common denominator to solve for B and C but my attempts always end up as a huge mess.
 
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  • #3
Following where you write b) you have just made a mistake in getting the partial fractions.

In the following lines, you've lost yourself - you drop a term for no reason and integrate incorrectly.

You will need to integrate a LHS with respect to P, and right hand side with respect to t.

Revise the integrals of things like dP/P if you have forgotten.
 
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  • #4
Thanks for your help!
 
  • #5
defaultusername said:
Thanks for your help!

See my sig. :oldsmile: :oldwink:
 
Last edited:

Related to Breakdown of a Logistic Equation

1. What is a logistic equation and how does it work?

A logistic equation is a mathematical model used to describe the growth of a population over time. It takes into account the limiting factors of the environment, such as resources and competition, and predicts the population's growth rate as it approaches a maximum value, known as the carrying capacity.

2. What are the key components of a logistic equation?

The key components of a logistic equation are the initial population size, the carrying capacity, and the growth rate, also known as the intrinsic rate of increase. These factors determine the shape and speed of the population's growth curve.

3. What is the significance of the carrying capacity in a logistic equation?

The carrying capacity is a crucial concept in a logistic equation as it represents the maximum population size that can be sustained by the available resources in the environment. As the population approaches the carrying capacity, the growth rate decreases and eventually reaches a stable equilibrium.

4. Can a logistic equation be applied to real-world populations?

Yes, a logistic equation can be used to model the growth of various populations in the real world, such as animal populations, human populations, and bacterial populations. However, it is important to note that it is a simplified model and may not accurately reflect the complexities of natural populations.

5. Are there any limitations to using a logistic equation?

One limitation of using a logistic equation is that it assumes a constant carrying capacity and growth rate, which may not always hold true in real-world populations. It also does not take into account external factors such as disease outbreaks or natural disasters, which can significantly impact population growth. Additionally, the logistic equation may not accurately predict the growth of populations with fluctuating or unknown carrying capacities.

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