Lecture 4. Modeling With First Order Equations1. Interest2. Mixing ProblemsDifferential EquationsLECTURE 4Modeling With First Order EquationsBefore we resume our discussion of how to solve certain types of first order equations, let’s moveto some applications of linear and separable equations. Our goal is to get a rough introduction tothe process of modeling: how do we write a differential equation to model (fairly basic) situationsand what can the solution tell us?We’ve already seen some examples of first order models. In Lecture 1, we saw two examples ofdifferential equations meant to describe physical situations.Example 4.1 (Radioactive Decay).dNdt= −λN(t),where N (t) is the number of atoms of a radioactive isotope and λ > 0 is the decay constant. Thisequation is separable, and it’s easy to see that, if the inital data is N (0) = N0, the solution isN(t) = N0e−λt.So we can see that radioactive decay is exponential. Example 4.2 (Newton’s Law of Cooling). If we immerse a body in an environment withconstant temperature E, then if B(t) is the temperature of the body we havedBdt= κ(E − B),where κ > 0 is a constant related to the material of the body and how it conducts heat. Thisequation is also separable; we solved it with initial condition B(0) = B0to getB(t) = E −E − B0eκt. How do we write down a model given a description of the situation? There are a few differentapproaches that we’ll see in this class:(1) Remember that we can think of the derivative of a function as its rate of change. It’spossible that the description of the problem tells us directly what the rate of change is.Newton’s Law of Cooling was an example of this; the original quote told us that the rateof change of the body’s temperature was proprotional to the difference in temperaturebetween the body and the environment. All we had to do was set the relevant termsequal.(2) There are also going to be cases where we’re not explicitly given the formula for rate ofchange in the problem, but if we can glean how a function should change from its physicaldescription, we can set the derivative equal to that. The basic premise that we’ll be usinghere is derivative = increase - decrease. Next, we’ll see some more in-depth examples of thisprocess. It should be noted that this can only apply to first order equations, since higherorder equations can’t just be interepreted as “the rate of change is equal to something.”1Differential Equations Lecture 4: Modeling With First Order Equations(3) We may just be adapting a known differential equation to a particular situation. Newton’sSecond Law F = ma is a prime example of this. It’s either a first or second order differentialequation (for velocity or position, respectively). If we compile the list of forces acting onthe object in question, we can just plug them in for F to yield the differential equation for aparticular situation. We’ll see this approach later this lecture with regard to falling bodies,but it will also appear later in the course concerning harmonic motion and pendulums.(4) The last possibility is that we are able to determine (via physical principles) two differentexpressions for some quantity, one or both of which involve derivatives; setting them equalyields the desired differential equation. We’ll see this when we discuss partial differentialequations later.It’s important that, when faced with a problem, we begin by determining which of these ap-proaches (or other ones that may be out there) is most applicable before proceeding. This is a skillthat will prove valuable in the future.One other very important point: in general, your differential equation should not depend on theinitial condition. The initial data will tell you the starting point, but the way the system evolves(which is given by the differential equation) ought not to depend on that. Now, let’s look at someparticular examples.1. InterestThese are among the most straightforward problems to write down. Suppose we’ve got a bankaccount (or a mortgage, or some other loan, or so on...the particulars don’t change much about theequation) that gives r% interest per year. If I withdraw a constant w dollars per month, what isthe differential equation modeling my account’s balance?Let’s take our time unit for t to be years, and denote the balance after t years by B(t). B0(t)is the rate of change of my account balance from year to year, so it will be the difference betweenthe amount that’s added to my account and the amount that’s withdrawn. The only income I’mgetting is interest and we know how much I withdraw each year (it’s the monthly withdrawal times12). HenceB0(t) =r100B(t) − 12w.This is a linear equation, so we could solve it using integrating factors to know the account balanceat any time t. Make sure you pay attention to units: we’re not withdrawing w dollars per year,we’re withdrawing 12w.The same setup will work perfectly for something like the following problem.Example 4.3. Bill wants to take out a 25 year loan to buy a house. He knows that he canafford, maximum, monthly payments of $400. If the going interest rate on housing loans is 4%,what is the largest loan Bill can take out that he will be able to pay off in time?Let’s say the amount that Bill owes at time t (it’s convenient here to measure t in years, sowe’ll do that) is B(t). We want B(25) = 0. We also know that the balance will gain 4% interest(i.e., the amount being added to the balance is 0.4B) and he can make payments totalling $4800each year. So the relevant initial value problem should beB0(t) = .04B(t) − 4800 B(25) = 0.This is a linear equation with standard formB0(t) − .04B(t) = −4800,2Differential Equations Lecture 4: Modeling With First Order Equationsso the integrating factor is µ(t) = eR−.04 dt= e−0.4t.e−4100tB(t)0= −4800e−4100te−4100tB(t) = −4800Ze−4100tdt= 120000e−4100t+ cB(t) = 120000 + ce4100tB(25) = 0 = 120000 + ce100100⇒ c = −120000e−1B(t) = 120000 − 120000e4100t−1.We want to know the size of the loan, which is the amount Bill owes to begin with, hence B(0).B(0) = 120000 − 120000e−1= 1200001 − e−1 2. Mixing ProblemsSuppose we have an efficient mixing tank that can instantly completely mix whatever is insideof it. The tank has some liquid inside of it. Contaminant is being added to the tank at someconstant rate and the mixed solution is drained out at a (possibly different) rate. We can
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