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DiffyQ Word problem (growth rate)

  1. Dec 4, 2011 #1
    I have been trying to figure this problem out for more then 90 minutes and I think I'm past the point where I'm being productive:

    A population P satisfies the differential equation:

    [tex]P^'(t)=10^{-5}P(t)(15000-P(t))[/tex]

    For what value P(0) of the initial population is the initial growth rate P'(0) greatest.

    I end up with [tex]\frac{10^5}{P(15000-P)}dP=dt[/tex] which I have no clue how to integrate (I spent most of the 90 minutes trying to figure that out). Before I went on trying to integrate this I want to make sure I didn't set it up wrong. I thought about partial fractions but have no clue how to separate the denominator.

    This is the first homework assignment I have ever had for DE and I apologize for posting two threads in two days, I'm trying my best to figure this stuff out.
     
  2. jcsd
  3. Dec 4, 2011 #2

    cepheid

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    I don't even think you need to solve this DE in order to answer the question. You're considering t = 0 only, so forget that these are functions, forget the time dependence. You have some variable P'(0) that depends on some other variable P(0). So if you consider P(0) to be an independent variable x, then the dependent variable y = P'(0) is given by the expression y = 10^(-5) * x * (15,000 - x). All you have to do is find the value of x for which y is a maximum. It seems like that's all the question is asking.
     
  4. Dec 4, 2011 #3
    That makes sense but even with the equation y = 10^(-5) * x * (15,000 - x) I have absolutely no clue how to go about finding the value of x where y is maximum..
     
  5. Dec 4, 2011 #4

    cepheid

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    Well now that's just straight up differential calculus. The function y(x) has extrema (maxima or minima) at points where its derivative goes to 0.
     
  6. Dec 4, 2011 #5
    *face palm*

    Thanks.
     
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