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First Order ODE Growth and Decay Modelling

  • Thread starter mjk11
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Homework Statement


A culture of bacteria have a growth rate (as a percent) given by [itex]kb[/itex] per year, constant [itex]k>0[/itex] and [itex]b[/itex] is the number of bacteria. A virus removes bacteria at a rate of [itex]m[/itex] bacteria per year. I am trying to model this information using an ODE, but might be making a mistake.

Homework Equations





The Attempt at a Solution


If the percent growth rate is kb, [itex](dy/y)/dt = kb[/itex] so [itex]dy/dt = kb^2[/itex].
Add this to the removal rate, and [itex]dy/dt = kb^2 - m[/itex].
I am not sure the model is as simple as this, it would mean the units of k are [itex]1/(bacteria*year)[/itex], which sounds odd.
 
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Answers and Replies

  • #2
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I think that the real confusion comes about when we try to determine what the question means by a "growth rate (as a percent)". To be honest with you, I have no idea what that means. The rate (in percent, per year) is kb, which depends on the number of bacteria. This suggests that the "actual" rate is [itex]kb^2[/itex], as you inferred. However, bacteria don't reproduce proportional to the square of their population; they reproduce by binary fission. The rate of reproduction should be proportional to [itex]b[/itex], not [itex]b^2[/itex].

If you're concerned, ask your professor about this problem, and how you should interpret "as a percent".

EDIT: Welcome to the forum, by the way.
 
  • #3
cepheid
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I think that growth rate as a percent refers to fractional growth rate i.e. growth rate as a fraction of the existing population. So if the fractional growth rate is 5%, it means that if the population starts out with 'b' individuals at the beginning of the year, then by the end of a year, the number of additional individuals will be 0.05b. So k=0.05 in this example. So if dy/dt is the fractional growth rate, then I think that what the problem means is that dy/dt = k. But the fractional growth rate is the ratio of the growth rate to the population size. Hence $$\frac{dy}{dt} = \frac{1}{b}\frac{db}{dt} = k$$ OR $$\frac{db}{dt} = kb$$ which gives you the standard exponential growth for populations of this type. I think that this is what the problem is saying, but I could be wrong.
 

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