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Math Help - separable and exact ODE

  1. #1
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    separable and exact ODE

    I'm unsure of how to set up the differential equation for this problem:

    Given that the earth's population on July 1980 was 4,473,000,000 and that on July 1987 it was 5,055,000,000. Assuming that P satisfies model A. When would the earth's population reach 10 billion? Thanks to all in advance for your help!
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  2. #2
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    Re: separable and exact ODE

    What is model A?
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    Re: separable and exact ODE

    I'm sorry model A is P' = kP where P' is the first derivative.
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    MHF Contributor MarkFL's Avatar
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    Re: separable and exact ODE

    Let t represent time measured in years and let July 1980 coincide with t=0. Let P represent the population at time t in billions. Using the given model, we have the IVP:

    \frac{dP}{dt}=kP where P(0)=4.473

    Separating the variables in the ODE, we have:

    \frac{1}{P}\,dP=k\,dt

    Integrate:

    \int\frac{1}{P}\,dP=k\int\,dt

    \ln(c_1\cdot P)=kt

    t=\frac{\ln(c_1\cdot P)}{k}

    Now, you have two points on the curve, from which you may determine the two constants. Once you have these, then set P=10 to answer the question.
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    Re: separable and exact ODE

    So I'm still a little lost. What is the k value? Is it the rate?

    I found k = 5.055 - 4.473 = .582
    where .582/7 = .0831 so that's the rate of the increase of people per year.

    -plug k into the t equation to solve for c1 which gives me .2235

    put c1 back into t equation and my answer is not correct. The answer is between mid 2020 and 2035 according to the book. Please show me where i'm going wrong.
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    MHF Contributor MarkFL's Avatar
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    Re: separable and exact ODE

    t=\frac{\ln(c_1\cdot P)}{k}

    We have the two points (0,4.473) and (7,5.055) so this gives us the system:

    0=\frac{\ln(c_1\cdot4.473)}{k}

    7=\frac{\ln(c_1\cdot5.055)}{k}

    From the first equation, we find:

    c_1\cdot4.473=1\,\therefore\,c_1=\frac{1000}{4473}

    Putting this into the second equation, we find:

    k=\frac{\ln\left(\frac{5055}{4473} \right)}{7}

    And so we have:

    t=\frac{7\ln\left(\frac{1000P}{4473} \right)}{\ln\left(\frac{5055}{4473} \right)}

    Now, let P=10 and we find:

    t=\frac{7\ln\left(\frac{10000}{4473} \right)}{\ln\left(\frac{5055}{4473} \right)}\approx46.04

    This is the year 2026.
    Thanks from nivek0078
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  7. #7
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    Re: separable and exact ODE

    Cool I was thinking too hard. Thank you again!
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