[Math] problem about population growth

calculusexponential functionintegrationordinary differential equations

At the beginning of the Gold Rush, the population of Coyote Gulch,Arizona was $365$.From then on ,the population would have grown by a factor of $e$ each year,except for the high rate of "accidental" death, amounting to one victim per day among every 100 citizens.By solving an appropriate differential equation determine, as functions of time:
(a) the actual population of Coyote Gulch $t$ years from the day the Gold Rush began, and
(b) the cumulative number of fatalities.

The question is from Apostol's Calculus I. In other questions,Apostol uses the statement "… increases at a rate proportional to the amount present. …"But this one says "grown by a factor of $e$", I have difficulty in understanding the meaning of "a factor of $e$".Does it mean $y=be^{kt}$ or $y'=ey$ or what?

So I decide to denote "a factor of $e$" by $f_e$ and focus on the ' "accidental" death '.So let $y$ denote the population at present year, each day one victim dies among every 100 citizens .So the population remains in that year is $y(1 -\frac{1}{100})^{365}$. And the number of fatalities in that year is $$y-y(1 -\frac{1}{100})^{365} = y(1- (\frac{99}{100})^{365})$$ .
And as far as I can get $$y'=f_e -y(1- (\frac{99}{100})^{365})$$

The answer of the question is given
a) $365e^{-2.65t}$
b) $365(1-e^{-2.65t})$

Any help is appreciated.

Best Answer

Try to write it out in words.

The rate of change in population is the population we have minus the loss ratio of that population (of course, we could have other factors, but that is what we are working with here), so we have:

$$\dfrac{dP}{dt} = P - \alpha P = P(1 - \alpha)$$

Now how we can we find the loss ratio $\alpha$ of the population per year? The problem tells us that we lose "$1$ victim per day among every $100$ citizens", so for the year, we lose $365 \times \dfrac{1}{100}$ of the population $P$. This gives us $\alpha = \dfrac{365}{100}$.

So we have:

$$\dfrac{dP}{dt} = P\left(1 - \dfrac{365}{100}\right), P(0) = 365$$

Note: $P(0)$ is the initial population which is given as $365$.

This DEQ is separable and gives us:

$$20 \int \dfrac{1}{P}~dP = -53 ~\int dt, P(0) = 365$$

Thus,

$$\large P(t) = 365 ~e^{-\frac{53 t}{20}} = 365~ e^{-2.65~ t}$$

If we want to find the cumulative number of fatalities over time, we write that out in words as: we start out with a population of $365$ and we lose them at a rate of $P(t)$ (which we just calculated), so this is:

$$CP(t) = 365 - P(t) = 365 - 365 ~e^{-\frac{53 t}{20}} = 365\left(1 - e^{-\frac{53 t}{20}}\right) = 365\left(1 - e^{-2.65 ~t}\right)$$

We can plot these curves to verify that they are inverses of each other as:

enter image description here

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