Population Modeling using DE's

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SUMMARY

The discussion focuses on solving a differential equation (DE) related to population modeling, specifically using the equation dP/dt = kP. The user establishes that if the population doubles in 5 years, the constant k can be calculated as k = ln(2)/5. By applying this value, the user can derive the time required for the population to triple and quadruple using the equations 3P0 = P0e^(kt) and 4P0 = P0e^(kt). The key takeaway is that the initial population P0 cancels out, simplifying the calculations significantly.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear DEs.
  • Familiarity with exponential functions and natural logarithms.
  • Basic knowledge of population growth models and their mathematical representations.
  • Ability to manipulate algebraic equations and solve for unknowns.
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  • Explore advanced topics in differential equations, such as systems of DEs and stability analysis.
  • Learn about logistic growth models and their applications in population dynamics.
  • Study the implications of varying growth rates and their effects on population projections.
  • Investigate numerical methods for solving DEs when analytical solutions are not feasible.
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Students, researchers, and professionals in mathematics, biology, and environmental science who are interested in modeling population dynamics and understanding the underlying mathematical principles.

prace
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Hi,

I was wondering if anyone out here on a Friday night could help me understand population modeling. Here is what I have as a problem (this is pretty simple because my goal here is to understand the thinking behind the madness :rolleyes: )

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The population of a certain community is known to increase at a rate proportional to the number of people present at any time. If the population has doubled in 5 years, how long will it take to triple, to quadruple?
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So I understand that the rate = dy/dt and proportional translates to "something" = "some constant" times "something", or (dy/dt)=Ky, or in this case, dP/dt = kP.

Solving this DE I get the equation P(t)= P(initial)e^(kt). So all I am told is that the population doubles in 5 years. So what can I do? I can't assume an arbitrary number as the initial population can I? So I have one equation with three unknowns, P(initial), P(t), and k.

Any insight would be great. Thanks!
 
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you know that [tex]\frac{dP}{dt} = kP[/tex]. The solution is:

[tex]P(t) = P_{0}e^{kt}[/tex]

You also know that [tex]2P_{0}= P_{0}e^{5k}[/tex]. What can you do from here? Solve for k:

[tex]e^{5k} = 2[/tex]

[tex]k = \frac{\ln 2}{5}[/tex].
Then solve the equations for t:

[tex]3P_{0} = P_{0}e^{(\frac{\ln 2}{5})t}[/tex]
[tex]4P_{0} = P_{0}e^{(\frac{\ln 2}{5})t}[/tex]
 
Last edited:
Oh man... So here the [tex]P_{0}[/tex]'s will cancel and all I have to do is take the natural logs of both sides to get t? man... I think I was making it a lot harder than it really was. Thank you for taking the time to show me this.
 

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