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Math Help - Population growth DE

  1. #1
    Super Member craig's Avatar
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    Population growth DE

    \frac{dP}{dt} = B_0e^{-at}P

    We know that P(0) = P_0, and we know that B(t) =B_0e^{-at} where a and B_0 are positive and real constants.

    I need to solve this for P(t) and find the limiting size.

    Here's how far I got, could someone please take a look and see if this seems ok.

    \frac{dP}{dt} = B_0e^{-at}P

    Need to separate the variables.

    \frac{1}{P} dP = B_0e^{-at} dt

    Integrating both sides.

    \ln{P} = \frac{-B_0e^{-at}}{a} + C

    If we now take both sides to the power e

    P = e^{\frac{-B_0e^{-at}}{a} + C} = Ae^{\frac{-B_0e^{-at}}{a}} - **Sorry I'm not sure how to enlarge the latex code**

    Now we know the initial condition, P(0) = P_0, this then gives us:

    P_0 = Ae^{\frac{-B_0}{a}}, so A = P_0e^{\frac{B_0}{a}}

    Giving us P = P_0 \cdot e^{\frac{B_0}{a}} \cdot e^{\frac{-B_0e^{-at}}{a}}.

    The reason I'm asking is that it doesn't look the tidiest of equations.

    Thanks in advance

    Craig
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  2. #2
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    Looks fine to me. You could make it a bit prettier.
    P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}<br />
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  3. #3
    Super Member craig's Avatar
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  4. #4
    Super Member craig's Avatar
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    Just one more little thing I've just found

    I also need to find out the limiting size of the population.

    P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}

    As t \to \infty, then (1 - e^{-at}) \to 1.

    So am I correct in saying that as t \to \infty, then P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)} \to P_0 e^{\frac{B_0}{a}}

    Thanks again
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  5. #5
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    No,
    <br />
\lim_{t \to \infty} (1 - e^{-at}) = 1<br />
,
    but
    <br />
P_{\infty} = \lim_{t \to \infty} P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)} = P_0 e^{\frac{B_0}{a}}<br />
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  6. #6
    Super Member craig's Avatar
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    Quote Originally Posted by lvleph View Post
    No,
    <br />
\lim_{t \to \infty} (1 - e^{-at}) = 1<br />
,
    but
    <br />
P_{\infty} = \lim_{t \to \infty} P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)} = P_0 e^{\frac{B_0}{a}}<br />
    Sorry there was a typo in my work above, was meant to be \lim_{t \to \infty} (1 - e^{-at}) = 1.

    Thanks for the reply again
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