Differential equation--Population growth

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Daniel_Feldman

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Problem:

P(t) is the population of a country, measured in millions of people.
P(t) satisfies the differential equation dP/dt=kP(185-P) with k constant.

In 1940, population was 150 million and was then growing at a rate of 1 million ppl/year.

I need to predict the population for 1999.

Solution attempt:

dP/dt when P=150 is 1

k(150)(35)=1....so k=1.905E-4

Separation of variables

1/[P(185-P)]=kt

Integrating both sides (using partial fractions) gives

(1/185)ln(abs(185P-P^2))=kt+C

P(0)=150 (setting 1940 as year 0)

(1/185)ln(5250)=C

(1/185)ln(abs(185P-P^2))=(1.905E-4)t+(1/185)ln(5250)

Multiplying through by 185 gives

ln(abs(185P-P^2))=0.03524t+ln(5250)

so

185P-P^2=+/- 5250 e^(0.03524t)

in 1999, t=59

So I get

185P-P^2=+/- 41994

P^2-185P+/- 41994=0

Using quadratic formula

[185 +/ sqrt(185^2 +/- 4(1)(41994))]/2

So the +/- inside the radical must be a +, and I get

P=(185 +/- 449.6)/2

This +/- must also be a +, so the answer should be

P=317 million people.

The computer is telling me this is wrong. Any ideas??
 
I found my mistake (in the partial fractions)....should be lnabs((P/(185-P)))


However, I am still not getting the correct answer. Can anyone do it and see what they get?


(I am now getting 180 million)
 
Daniel_Feldman said:
P(t) is the population of a country, measured in millions of people. P(t) satisfies the differential equation dP/dt=kP(185-P) with k constant. In 1940, population was 150 million and was then growing at a rate of 1 million ppl/year.
Click to expand...
A few steps seem to be missing in your work...?

This is what I get:

. . . . .\(\displaystyle \frac{dP}{dt}\, =\, kP(185\, -\, P)\)

. . . . .\(\displaystyle \frac{dP}{P(185\, -\, P)}\, =\, k \, dt\)

. . . . .\(\displaystyle \frac{1}{P(185\, -\, P)}\, =\, \frac{A}{P}\, +\, \frac{B}{185\, -\, P}\)

. . . . .\(\displaystyle 1\, =\, A(185\, -\, P) + B(P)\)

. . . . .\(\displaystyle 1\, +\, 0P\, =\, 185A\, +\, (B\, -\, A)P\)

. . . . .\(\displaystyle 1\, =\, 185A\)

. . . . .\(\displaystyle A\, =\, \frac{1}{185}\)

. . . . .\(\displaystyle B\, -\, A\, =\, 0\)

. . . . .\(\displaystyle B\, =\, A\, =\, \frac{1}{185}\)

So the integration becomes:

. . . . .\(\displaystyle \frac{1}{185}\,\int\, \frac{1}{P}\, dP\, +\, \frac{1}{185}\, \int\, \frac{1}{185\, -\, P}\, dP\, =\, \int \, k\, dt\)

I get:

. . . . .\(\displaystyle \frac{1}{185} \, ln \, \begin{array}{|c|} \frac{P}{185\, -\, P} \end{array} \, + \, C\, =\, kt\)

I'm not sure if any of that lines up with what you did...? :oops:

Eliz.
 
Appreciate the response...

Yeah, sorry I skipped around a bit with steps (it's an ubfortunate habit).

My big error in the original post was that I treated the integral of 1/(185-P) as ln(abs(185-P)) rather than -ln(abs(185-P)).

This eventually gave me ln(abs(P(185-P)) instead of ln(abs(P/(185-P))).

However, I fixed this and got the same answer as you, above (with the C on the other side, but still, minor point).

Despite this, I am still getting the wrong answer for P in 1999. (This is at t=59 if 1940 is 0, right?)

I might be doing something wrong on my calculator...which is why I was hoping that someone could punch in the numbers and see if they get a different answer. I still get 180 million.
 
I am at the point

ln(abs(P/(185-P)))=0.035t+1.455

I need P in 1999 (t=59). I keep getting 180 million but it's wrong.
 
Daniel_Feldman said:
I keep getting 180 million but it's wrong.
Click to expand...
It is unfortunate that you don't show your work. This makes finding errors rather difficult.... :oops:

stapel said:
. . . . .\(\displaystyle \frac{1}{185} \, ln \, \begin{array}{|c|} \frac{P}{185\, -\, P} \end{array} \, + \, C\, =\, kt\)
Click to expand...
Assuming "t = 0" indicates the year 1940 and that we are counting in millions (so "10", for instance, would indicate "ten million"), you have:

. . . . .\(\displaystyle \frac{dP}{dt}\, =\, 1\)

. . . . .\(\displaystyle P\, =\, 150\)

Then, for 1940:

. . . . .\(\displaystyle 150k(185\, -\, 150)\, =\, 1\)

. . . . .\(\displaystyle 150k(35)\, =\, 1\)

. . . . .\(\displaystyle k\, =\, \frac{1}{5250}\)

Returning to the integral, we have:

. . . . .\(\displaystyle \frac{1}{185} \, ln \, \begin{array}{|c|} \frac{P}{185\, -\, P} \end{array} \, + \, C\, =\, kt\)

. . . . .\(\displaystyle ln \, \begin{array}{|c|} \frac{P}{185\, -\, P} \end{array} \, =\, 185kt\, +\, D\)

Raising both sides as powers on "e", we get:

. . . . .\(\displaystyle \begin{array}{|c|} \frac{P}{185\, -\, P} \end{array}\, =\, e^{185kt\, +\, D}\)

. . . . .\(\displaystyle \begin{array}{|c|} \frac{P}{185\, -\, P} \end{array}\, =\, Fe^{185kt}\)

Since we know that, by 1999, the population most likely would have been more than 185 million, we can take off the absolute-value bars, and get:

. . . . .\(\displaystyle \frac{P}{185\, -\, P} \, =\, Fe^{185kt}\)

Solve for P in terms of t, we get:

. . . . .\(\displaystyle P\, =\, 185Fe^{185kt}\, -\, FPe^{185kt}\)

. . . . .\(\displaystyle P\, +\, FPe^{185kt}\, =\, 185Fe^{185kt}\)

. . . . .\(\displaystyle P\, =\, \frac{185Fe^{185kt}}{1\, +\, Fe^{185kt}}\)

Use the known value for k and the data point (0, 150) to solve for the value of F. Then evaluate at t = 59. If you continue to get an invalid answer, please SHOW YOUR WORK.

Thank you! :D

Eliz.
 
P/(185-P)=Fe^(185kt)

At (0,150)

150/(185-150)=Fe^0=F

F=150/35=4.2857

In 1999

P/(185-P)=4.2857e^(185*(1/5250)*59)

P/(185-P)=34.37221

(185-P)/P=1/34.37221=0.02918

(185/P)-1=0.02918

185/P=1+0.02918=1.02918

P=185/1.02918=179.75

Still getting around 180....
 
Hey Dan, I'm getting 185 million exactly. Is that the solution?.

I am supposing k=1/5250, then we eventually get:

\(\displaystyle P=\frac{5550e^{\frac{37t}{30}}}{30e^{\frac{37t}{30}}+7}\)

Setting t=59, we get P=185


EDIT: DUH, I wrote down 1/150 instead of 1/5250.
 
I made an error previously, Dan. I keep getting what you get.

\(\displaystyle P=\frac{5550e^{\frac{37t}{1050}}}{30e^{\frac{37t}{1050}}+7}\)

When t=59, we get P=179.755077875.

I am sticking with that. Remember, computers can be mighty temperamental about how things are entered in.
Is it wanting rounding to a whole number?. Within a hundredth?. Tenth?.
 
Hi again.

Thanks for all the help.

The stupid computer system wanted 3 decimal places despite the units being in millions.

It only gives a set number of tries, hence my concern.

Again, I appreciate the help.

-Daniel-
 
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