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Homework Help: US Population Modeling

  1. Feb 28, 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 relative growth rate by dividing an approximate growth rate by the US population between 1790 and 2000, did a linear regression and obtained the equation below.

    y = -.0001x + .0338 and (1/P)(dP/dt) = ax + b

    I do not exactly know what they are asking to be done. Do they want me to solve for P'(t)? If so, how does the 1/P on the left side of the equation effect the right side? Any help would be appreciated. Thanks.
  2. jcsd
  3. Feb 28, 2008 #2
    P'? They want you to solve for P, it's a differential equation. Remember that the x-axis is time, and then it's separation of variables.
  4. Feb 28, 2008 #3
    Thanks for the reply. I already have all the P's. If you could give a little more information to clear things up it would be appreciated. Thanks.
  5. Feb 28, 2008 #4
    You typed (1/P)(dP/dt) = ax + b

    Where a is -0.0001 and b is .0338 and x should be time

    So you haven't found the function P(t) yet, but there is the equation. After you separate variables you'd have

    dP/P=(a*t+b)*dt, then you need to integrate and use your initial condition
  6. Feb 28, 2008 #5
    Thanks. After I sent the last reply it clicked that I understood what you ment by solve for P as the variable. Thanks again.
  7. Feb 28, 2008 #6
    Here is what I calculated:

    Use P0 = 3.9 as your initial condition to find the particular solution for this differential equation.

    (1/P)(dP/dt) = b + at

    y = -.0001t + .0338

    dP/P = (.0338 - .0001t)dt

    ln P = .0338t - (.0001t^2/2) + C

    At t = 0, P0 = 3.9 so then C = ln 3.9

    ln P = .0338t - (.0001t^2/2) + ln 3.9

    Take the exponential of both sides,

    P = 3.9 e^.0338t-(.0001t^2/2)

    This formula seems to work well for the first 100 years but gets out of control after that so not sure if I have everything right. Let me know. Thanks.
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