Results 1 to 6 of 6

Thread: Population growth DE

  1. #1
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1

    Population growth DE

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

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

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

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

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

    Need to separate the variables.

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

    Integrating both sides.

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

    If we now take both sides to the power $\displaystyle e$

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

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

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

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

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

    Thanks in advance

    Craig
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Looks fine to me. You could make it a bit prettier.
    $\displaystyle P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}
    $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Just one more little thing I've just found

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

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

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

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

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    No,
    $\displaystyle
    \lim_{t \to \infty} (1 - e^{-at}) = 1
    $,
    but
    $\displaystyle
    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}}
    $
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by lvleph View Post
    No,
    $\displaystyle
    \lim_{t \to \infty} (1 - e^{-at}) = 1
    $,
    but
    $\displaystyle
    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}}
    $
    Sorry there was a typo in my work above, was meant to be $\displaystyle \lim_{t \to \infty} (1 - e^{-at}) = 1$.

    Thanks for the reply again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Population Growth
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 19th 2009, 02:01 PM
  2. Population growth with non-constant growth factor
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Feb 3rd 2009, 04:46 PM
  3. Population Growth
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 25th 2009, 01:09 PM
  4. Population Growth
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Aug 14th 2008, 12:38 PM
  5. Population Growth
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: Sep 26th 2007, 12:59 PM

Search Tags


/mathhelpforum @mathhelpforum