US Population Modeling

[pjallen58]

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.

[blochwave]

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.

[pjallen58]

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.

[blochwave]

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

[pjallen58]

Thanks. After I sent the last reply it clicked that I understood what you ment by solve for P as the variable. Thanks again.

[pjallen58]

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.