Results 1 to 4 of 4

Math Help - Setting up a Growth Problem (ODE, probably separable or linear)

  1. #1
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1

    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 P= population and t = years: \frac{dP}{dt} = kP \Rightarrow

    \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,

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

    When I compute this I get something ridiculously wrong.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    The population due to net addition of 400 people per year is

    <br />
P=400t<br />

    and

    <br />
\frac{dP}{dt}=400.<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1
    Ah, right, if I catch your drift: \frac{dP}{dt} = \frac{\ln 2}{50}P + 400 \Rightarrow. So if that's right, to find the population in ten years,

    \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

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

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    May be you need not dif. equation.
    The double of population in 50 years is

    <br />
\displaystyle {<br />
P=P_0 \: 2^{{t}/{\tau}}<br />
}<br />

    and net addition of 400 people per year is

    <br />
\displaystyle {<br />
P=400t.<br />
}<br />

    So all population change is

    <br />
\displaystyle {<br />
P=P_0 \: 2^{{t}/{\tau}}+400t<br />
}<br />

    or

    <br />
\displaystyle {<br />
P=40000 \: \; 2^{{t}/{50}}+400t.<br />
}<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Setting up linear equations.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 2nd 2011, 06:46 AM
  2. Setting up a Linear ODE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 24th 2011, 03:59 PM
  3. how to determine separable & linear?
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: November 8th 2009, 02:45 AM
  4. Separable Linear Equations
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 28th 2009, 12:47 PM
  5. Linear and separable equations
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 12th 2009, 06:17 AM

Search Tags


/mathhelpforum @mathhelpforum