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U.S. Population Modeling

  1. Mar 11, 2008 #1
    Use P0 = 3.9 (1790 population) as your initial condition to find the particular solution for this differential equation. Note: You may find it easier to solve in terms of the constants a and b. Show all the steps in your solution.

    This is the last step to a multi-part problem. I basically did a scatter plot of the population for each year (x) and the relative growth rate (y) which was found by dividing an approximate growth rate by the US population between 1790 and 2000, did a linear regression and obtained the equation below.

    y = -.0000917x + .0287

    Given (1/P)(dP/dt) = b +aP

    (1/P)(dP/dt) = .0287 - .0000917(P)
    I think I need to integrate and get 1/P(dt) = (.0287 - .0000917(P))/dP

    I don't know if this is set up right to integrate and if it is the fractions are confusing me and I don't know where to start. Any help or suggestions would be appreciated. Thanks.
     
  2. jcsd
  3. Mar 12, 2008 #2
    Any suggestions or advice would be appreciated. Thanks.
     
  4. Mar 12, 2008 #3

    HallsofIvy

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    No, you don't want to get 1/P(dt) = (.0287 - .0000917(P))/dP because you can't integrate "(1/P(dt))" is an unknown function of t (and you surely don't want the differentials in the denominator!). Get all the "P"s over on the left and "t" on the right. Then you can integrate, probably by using "partial fractions" on the left. If it were me, I would leave "a" and "b" in the equation until the final answer- less distracting.
     
  5. Mar 12, 2008 #4
    I thought that didn't look right. Here is what I get:

    (1/P(.0287-.0000917P)dp = dt

    Integrate using partial fractions

    1/.0287 ln|P/.0287-.0000917P| + C = t + C

    ln|P/.0287-.0000917P| = .0287t + C

    If t = 0, P = 3.9 so C = ln 137.6

    Now I think I should take the exponential of each side to get:

    |P/.0287 - .0000917P| = 137.6e^.0287t

    Not sure if this is right thus far and plus since I am solving for P not sure how to isolate P. A little more guidnace would be appreciated. Thanks.
     
  6. Mar 13, 2008 #5
    Just a little more help would be appreciated. Thanks.
     
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