Modeling with differential equations

Section 1.27 Modeling with differential equations

Procedure 1.27.1. Modeling with first-order differential equations.

Many applications present information about a quantity in a form that can be modeled by a first-order differential equation. Here is an outline of the steps to take in these settings.

Explicitly identify the quantity \(Q\) under consideration as a function of some other quantity \(x\text{,}\) and give a name to this function: \(Q=f(x)\text{.}\)
Translate the given information about \(Q\) into a first-order differential equation:

\begin{equation} f'(x)=F(x, f(x)) \text{.}\tag{1.27.1} \end{equation}

This is often the trickiest step!
- Look for phrases that indicate rate of change of \(Q\) and translate these into equalities involving \(f'\text{.}\)
- When there is a combination of components to the rate of change, a diagram may be useful.
- Translate phrases like “\(Q_1\) is proportional to \(Q_2\)” as \(Q_1=k\cdot Q_2\text{,}\) where \(k\) is the (possibly undetermined) constant of proportionality.
Ask yourself whether the differential equation (1.27.1) is linear or separable. If so, use the appropriate technique to solve it in as general a form as you can.
- If the equation is both separable and linear, you can use either method. It is often quicker to use the separation of variables technique.
- If the differential equation is linear, make sure to bring it into standard form before using the integrating factor method.
Once you have solved (1.27.1) you will have a formula for \(Q=f(x)\) that includes some undetermined constants. If possible, use any additional information given about \(Q\) to determine these undetermined constants.

Example 1.27.2. Mixing problem.

A large tank in a pickle factory initially contains 50 liters of brine in which 20 kg of salt is dissolved. The mixture is kept uniform by stirring. Brine containing 0.2 kg of dissolved salt per liter enters the tank at a rate of 10 liters per minute. At the same time the mixture from the tank leaves at a rate of 6 liters per minute. How much salt is in the tank after 30 minutes.

Solution.

First we make explicit the function we wish to determine: let \(A=A(t)\) be the amount of salt (in mg) in the tank after \(t\) minutes.

Next we translate the given information into a differential equation for \(A=A(t)\text{.}\) To do so, we will have to translate statements about the concentration of salt into statements about the amount of salt. To this end, we let \(V=V(t)\) be the volume of the tank after \(t\) minutes. Since fluid enters the tank at a rate of \(10\) liters per minute and exits the tank at a rate of \(6\) liters per minute, we see that the rate of change of \(V(t)\) is \(4\) liters per second: i.e., \(V'(t)=4\text{.}\) It follows that \(V(t)=4t+C\text{.}\) Since the initial volume is 50 liters, we have \(C=50\text{.}\) Thus \(V(t)=4t+50\text{.}\)

We can now write down a differential equation for \(A(t)\text{:}\)

\begin{align*} A'(t)=(\text{rate of change of } A) \amp = (\text{rate of inflow of salt})-(\text{rate of outflow of salt}) \\ \amp = \underset{\text{inflow rate}}{\underbrace{10}}\cdot \underset{\text{brine concent.}}{\underbrace{0.2}} - \underset{\text{outflow rate}}{\underbrace{6}}\cdot \underset{\text{tank concent.}}{\underbrace{\frac{A(t)}{V(t)}}} \\ \amp = 2-\frac{6}{4t+50}A(t)\text{.} \end{align*}

Next we identify that the differential equation \(A'(t)=2-\frac{6}{4t+50}A(t)\) is linear. Following Procedure 1.26.3 we first bring the differential equation into standard form:

\begin{equation*} A'(t)+\underset{p(t)}{\underbrace{\frac{6}{4t+50}}}A(t)=\underset{q(t)}{\underbrace{2}}\text{.} \end{equation*}

Since \(\frac{3}{2}\ln(4t+50)\) is an antiderivative of \(p(x)\text{,}\) we set

\begin{equation*} v(t)=e^{\frac{3}{2}\ln(4t+50)}=(4t+50)^{3/2}\text{.} \end{equation*}

Now compute

\begin{equation*} \int v(t)q(t)\, dt=\int 3(4t+50)^{3/2}\, dt=\underset{G(t)}{\underbrace{\frac{1}{5}(4t+50)^{5/2}}}+C\text{,} \end{equation*}

and conclude

\begin{equation*} A(t)=\frac{G(t)}{v(t)}+\frac{C}{v(t)}=\frac{1}{5}(4t+50)+\frac{C}{(4t+50)^{3/2}}\text{.} \end{equation*}

Lastly, since initially there is 20 mg of salt in the tank, we have \(A(0)=20\text{,}\) which implies

\begin{equation*} 20=10+C/50^{3/2}\text{,} \end{equation*}

or \(C=10\cdot 50^{3/2}\text{.}\) Thus we have

\begin{equation*} A(t)=\frac{1}{5}(4t+50)+\frac{10\cdot 50^{3/2}}{(4t+50)^{3/2}}\text{,} \end{equation*}

from which it follows that the amount of salt after 30 minutes is

\begin{equation*} A(30)=\frac{1}{5}(170)+\frac{10\cdot 50^{3/2}}{170^{3/2}}\text{.} \end{equation*}

Example 1.27.3. Trout population.

Left to their own devices, the population of trout in the Straits of Dudley increases at a rate of \(25\%\) of the present population. A recent oil spill, however, has resulted in an ongoing depletion of the population: specifically, \(t\) months after the spill the trout are dying off at a rate of \(2700t\, e^{-t/2}\) fish per month.

Derive a differential equation describing trout population in the Strait of Dudley, considered as a function of the time \(t\) (in months) since the oil spill. You may assume the oil spill has not affected the trout population’s breeding habits.
Derive a formula for the population of trout \(t\) months after the oil spill, assuming that at the moment of the spill there were six thousand fish in the Strait of Dudley.

Solution.

First, we explicitly identify the function in question: let \(P=f(t)\) be the population of trout \(t\) months after the spill. Now translate the given information into a differential equation for \(f\text{:}\) we have

\begin{align*} f'(t)=(\text{rate of change of } P) \amp = (\text{rate of breeding})- (\text{rate of depletion}) \\ \amp = 0.25\underset{\text{current pop.}}{\underbrace{f(t)}}-2700t\, e^{-t/2}\text{.} \end{align*}

The differential equation \(f'(t)=\frac{1}{4}f(t)-2700t\, e^{-t/2}\) is linear. Following Procedure 1.26.3, we first bring it into standard form

\begin{equation*} f'(t)\underset{p(t)}{\underbrace{-\frac{1}{4}}}f(t)=\underset{q(t)}{\underbrace{-2700t\, e^{-t/2}}}\text{.} \end{equation*}

Skipping some details, we find that the integrating factor is \(v(t)=e^{-t/4}\) and

\begin{equation*} \int v(t)q(t)\, dt= \underset{G(t)}{\underbrace{1200 e^{-3t/4}(3t+4)}}+C\text{.} \end{equation*}

The integral computation was done using integration by parts (followed by some algebra). We conclude that

\begin{equation*} P=f(t)=\frac{G(t)}{v(t)}+\frac{C}{v(t)}=1200e^{-t/2}(3t+4)+Ce^{t/4}\text{.} \end{equation*}

Since the population is 6000 at the time of the spill, we have \(f(0)=6000\text{,}\) which implies \(6000=4800+C\text{,}\) or \(C=1200\text{.}\) We conclude that

\begin{equation*} P=f(t)=1200e^{-t/2}(3t+4)+1200e^{t/4}\text{.} \end{equation*}

As \(t\to\infty\) the first term \(1200e^{-t/2}(3t+4)\) tends toward 0, making \(f(t)\) behave like asymptotically like \(1200e^{t/4}\text{.}\) So the long-term consequence of the oil spill is that the population, which would have grown as \(6000e^{t/4}\) in the absence of the oil spill, grows instead like \(1200e^{t/4}\text{.}\)

Example 1.27.4. Spreading rumor.

A false rumor is being spread within a certain university community. Let \(q=f(t)\) be the percentage (from 0 to 100) of the community that has heard the rumor \(t\) days after the rumor is introduced. Assume that percentage of the population that has heard the rumor grows at a rate proportional to the product of the current percentage of those in the know with the current percentage of those not in the know.

Write down a differential equation for \(q=f(t)\text{.}\) Use \(k\) as your constant of proportionality.
Derive a formula for \(q=f(t)\) assuming that at time \(t=0\) five percent of the population has heard the rumor. Your answer will be expressed in terms of \(t\) and \(k\text{.}\)
Investigate the long-term behavior of the rumor’s spread. Is there a time when the entire population has heard the rumor?

Solution.

We have

\begin{align*} f'(t)=(\text{rate of change of } q) \amp= k \underset{\text{per. in know}}{\underbrace{f(t)}} \underset{\text{per. not in know}}{\underbrace{(100-f(t))}} \text{.} \end{align*}

The differential equation \(\frac{dq}{dt}=k\, q (100-q)\) is separable (and not linear). We solve using the separation of variables technique:

\begin{align*} \frac{dq}{q(100-q)} \amp = k\, dt\\ \int \frac{dq}{q(100-q)} \amp = \int k\, dt\\ \frac{1}{100}\int \frac{1}{q}-\frac{1}{100-q}\, dq \amp =\int kt\, dt \amp (\text{part. frac. decomp.})\\ \frac{1}{100}(\ln q-\ln(100-q)\amp = kt+C \amp (q\geq 0, 100-q\geq 0)\\ \ln\left( \frac{q}{100-q}\right) \amp=100kt+100C=rt+100C \amp (r=100k) \\ \frac{q}{100-q} \amp = e^{rt}e^{100C}=De^{rt} \amp (D=e^{100C})\\ q \amp = 100De^{rt}-De^{rt}q\\ q(1+De^{rt}) \amp = 100De^{rt}\\ q \amp = \frac{100De^{rt}}{1+De^{rt}}=\frac{100D}{D+e^{-rt}}\text{.} \end{align*}

Since initially \(5\%\) of the population knows the rumor, we have \(f(0)=5\text{,}\) and thus

\begin{equation*} \frac{100D}{D+1}=5\text{.} \end{equation*}

This implies \(D=\frac{5}{95}=\frac{1}{19}\text{,}\) and thus

\begin{equation*} q=f(t)=\frac{100}{19}\frac{e^{rt}}{\frac{1}{19}+e^{-rt}}=\frac{100}{1+19e^{-rt}}\text{,} \end{equation*}

where \(r=100k > 0\text{.}\)

This is called a logistic growth function. Since \(e^{-rt}\) is decreasing, we see that \(q=f(t)\) is increasing. Furthermore, we have

\begin{equation*} \lim_{t\to\infty}f(t)=\lim_{t\to\infty}\frac{100}{1+19e^{-rt}}=\frac{100}{1+0}=100\text{.} \end{equation*}

Thus as time passes the percentage of the population that have heard the rumor increases and approaches 100 percent asymptotically.

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