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Thread: Setting up a Growth Problem (ODE, probably separable or linear)

  1. #1
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    Setting up a Growth Problem (ODE, probably separable or linear)

    I'm doubtful whether I've set up the equations correctly just because the computations are getting ugly:

    By natural increase a city, the population of which is 40,000, will double in 50 years. There is also a net addition of 400 people per year because of people moving to and from the city.

    I figger to solve for the natural increase function, where $\displaystyle P=$ population and $\displaystyle t$ = years: $\displaystyle \frac{dP}{dt} = kP \Rightarrow$

    $\displaystyle \displaystyle \int_{40,000}^{80,000}\frac{dP}{P} = k\int_{0}^{50}dt = 50k = \ln 2 \Rightarrow k = \frac{\ln 2}{50}$

    Now taking into consideration the additional 400 people per year,

    $\displaystyle \frac{dP}{dt} = \frac{\ln 2}{50}P + 400t$

    When I compute this I get something ridiculously wrong.
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  2. #2
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    The population due to net addition of 400 people per year is

    $\displaystyle
    P=400t
    $

    and

    $\displaystyle
    \frac{dP}{dt}=400.
    $
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  3. #3
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    Ah, right, if I catch your drift: $\displaystyle \frac{dP}{dt} = \frac{\ln 2}{50}P + 400 \Rightarrow$. So if that's right, to find the population in ten years,

    $\displaystyle \displaystyle \int_{40,000}^{x}\frac{dP}{\frac{\ln 2}{50}P + 400} = \int_{0}^{10}dt = 10 = \frac{50}{\ln 2}\ln \Big( \frac{\ln 2}{50}P + 400} \Big) \Rightarrow$

    $\displaystyle \frac{\ln 2}{5} = \ln \Big( \frac{\ln 2}{50}P + 400} \Big) \Rightarrow$

    $\displaystyle 2^{\frac{1}{5}} = \frac{\ln 2}{50}P + 400 \Rightarrow ...$

    I'll stop here because this computation gets me something egregiously wrong and so I'm guessing I've done something wrong by now.
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  4. #4
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    May be you need not dif. equation.
    The double of population in 50 years is

    $\displaystyle
    \displaystyle {
    P=P_0 \: 2^{{t}/{\tau}}
    }
    $

    and net addition of 400 people per year is

    $\displaystyle
    \displaystyle {
    P=400t.
    }
    $

    So all population change is

    $\displaystyle
    \displaystyle {
    P=P_0 \: 2^{{t}/{\tau}}+400t
    }
    $

    or

    $\displaystyle
    \displaystyle {
    P=40000 \: \; 2^{{t}/{50}}+400t.
    }
    $
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