Breakdown of a Logistic Equation

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SUMMARY

The discussion focuses on solving the Logistic Equation $$\frac{dP}{dt}=kP(1-\frac{P}{A})$$. Participants outline the steps to find equilibrium solutions, separate the equation, and apply partial fraction decomposition for integration. The equilibrium solutions identified are $$P=0$$ and $$P=A$$. The correct approach to separate the equation involves rewriting it as $$\frac{dP}{dt}=\frac{k}{A}P(A-P)$$, which simplifies the separation process.

PREREQUISITES
  • Understanding of differential equations, specifically separable equations.
  • Familiarity with equilibrium solutions in mathematical modeling.
  • Knowledge of partial fraction decomposition techniques.
  • Ability to perform integration with respect to different variables.
NEXT STEPS
  • Study the method of solving separable differential equations in depth.
  • Learn about equilibrium points in logistic growth models.
  • Review techniques for partial fraction decomposition in calculus.
  • Practice integration techniques, particularly with respect to different variables.
USEFUL FOR

Students studying differential equations, mathematicians interested in logistic growth models, and educators teaching calculus and differential equations.

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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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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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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Likes defaultusername
Thanks for your help!
 
defaultusername said:
Thanks for your help!

See my sig. :oldsmile: :oldwink:
 
Last edited:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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