Let P(t) be the population of a city after t years. In this city the death rate is alpha times the population size, the birth rate is beta times the population size and the immigration out of city is with rate m.

(a) Write a di fferential equation for the population. What is the general solution?

(b) If the initial population is 1,000,000 people, the birth rate divided by the population size is .02 bigger than the death rate divided by the population size, and 100,000 people leave the city per year. What will be the population be after 2 years?

(c) Under this model and the values given in part b), will the population expand or decline?

(d) If the 100,000 people leave the city per year, then for what value of the di erence between and does the population stay constant?

Im not sure how to approach this problem.

What do they mean by the general solution?

for the differential equation im thinking that P(t) is just a function of the initial population times alpha plus the initial population times beta plus m. does that sound right?

once i get that I think i should be able to figure out the rest.

Any help is appreciated.

  • $\begingroup$ You are looking for a differential equation, i.e. some equation for the function $P(t)$ with derivatives. They give you a hint - rate of change of $P$ wrt $t$ is birthrate - deathrate, so what can you say about $P(t)$? $\endgroup$ – gt6989b Apr 16 '13 at 4:02
  • $\begingroup$ General solution to an equation is the one that solves the equation with drops all terms without $P(t)$ in it in some form; i.e. $P''(t) + P(t) = \sin t$ becomes $P''(t) + P(t) = 0$. Such equations are called homogeneous and solutions are sometimes called that as well. $\endgroup$ – gt6989b Apr 16 '13 at 4:03

We find an expression for $\frac{dP}{dt}$, the rate of change of the population at time $t$. There are three items that contribute: births, deaths, and people leaving. births at time $t$ are occurring at the rate $\beta P$. Deaths are occurring at the rate $\alpha P$. We conclude that $$\frac{dP}{dt}=\beta P -\alpha P-m=(\beta-\alpha)P-m.$$ We are asked to find the general solution of this equation. I don't know the style you use for such things.

There are several possible approaches. One is to note that our DE is separable, and can be rewritten as $$\frac{dP}{(\beta-\alpha)P-m}=dt.$$ Integrate. We get $$\frac{1}{\beta-\alpha}\ln\left|(\beta-\alpha)P-m\right|=t+C.$$ Multiply both sides by $\beta-\alpha$, and take the exponential. We get $$(\beta-\alpha)P-m=D\ast e^{(\beta-\alpha)t} $$ for some constant $D$. Now we can solve explicitly for $P$. The general solution is $$P(t)=\frac{m}{\beta-\alpha}+Ee^{(\beta-\alpha)t},$$ where $E$ is an arbitrary constant.

The above is the general solution of the DE that you were asked for.

I would slightly prefer to make instead the change of variable $y=(\beta-\alpha)P-m$. Then $\frac{dy}{dt}=(\beta-\alpha)\frac{dP}{dt}=(\beta-\alpha)y$.

The equation $\frac{dy}{dt}=(\beta-\alpha)y$ is the familiar differential equation of exponential growth (if $\beta\gt \alpha$). We can write down the general solution from memory, and then use that to get an expression for $P(t)$.

The above is what you explicitly asked about. Now take over. If you have difficulties, please leave a message.

  • $\begingroup$ 1. Good answer! 2. There is a little cleanup (maybe just adding "\" in front of "beta"...). 3. You're almost #1 in rep! What a feat :) $\endgroup$ – The Chaz 2.0 Apr 16 '13 at 4:48
  • $\begingroup$ @TheChaz2.0: Fixed $\beta$. Very restricted answer, OP may need more. So, you are the new model, the latest thing. Or is $2.0$ no longer latest? $\endgroup$ – André Nicolas Apr 16 '13 at 5:19
  • $\begingroup$ Hah! Well, occasionally my words were not in line with who I want to be (or am), so I decided a reboot was in order. $\endgroup$ – The Chaz 2.0 Apr 16 '13 at 5:39
  • $\begingroup$ And while this answer might leave much to be desired from the student's perspective, I see high pedagogical value. $\endgroup$ – The Chaz 2.0 Apr 16 '13 at 5:40
  • $\begingroup$ $E$ is related to the initial state $P(0)$, $E = P(0) - \frac{m}{\beta -\alpha}$. $\endgroup$ – alain Jan 16 '18 at 13:04

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