# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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When considering the cooling of an egg, ODE (1) is separable because Tout is constant in this instance. Let’s consider what happens when the outside temperature changes with time.

We can still use Newton’s law of cooling, so that if T(t) is the egg’s temperature and Tout(t) is the room’s temperature, then

~77 — k( T0ut( t) — T) dt

(4)

Every ODE text discusses integrating factors and first-order linear ODEs.

Note that ODE (4) is not separable (because Tout varies with time) but it is linear, so we can find its general solution as follows. Rearrange the terms to give the linear ODE in standard form:

“77 + kT = ê Tout{ t) dt

Multiply both sides by ekt, so that

^{^Tt+kT)=kTout(t)^ (5)

Since the left-hand side of ODE (5) is (d/dt)(ektT(t)), it can be rewritten:

jt{e?trI)=kToat(t)eP (6)

Integrating both sides of ODE (6) we have that

ektT = j kToUt(t)ektdt + C

where C is an arbitrary constant. The magic factor fz(t) = ekt that enabled us to do this is called an integrating factor. So ODE (4) has the general solution

kt

T(t) = e~*(J kT0Ut(t)ektdt + Cj

Air Conditioning a Room

47

Time (minutes)

Figure 3.2: Eggs at initial temperatures of 180, 150, 120, and 90°F cool in a room whose temperature oscillates sinusoidally about 70°F for k = 0.03 min-1. Do the initial temperatures matter in the long term?

Finally, letting T(0) = T0 and integrating from 0 to t, we get the solution

T(t) = e~kt^j^ kToUt(s)eksds + T^ (7)

It may be possible to evaluate the integral (7) analytically, but it is easier to use ODE Architect right from the start. See Figure 3.2 for egg temperatures in a room whose temperature oscillates between hot and cold.

? Show that if Tout is a constant, then formula (7) reduces to formula (3).

You may find a computer algebra system or a table of integrals helpful!

? Use equation (7) to find a formula for T(t) if

' 2n(t + 3)

ToUt(t) = 82 - 10sin

24

Use a table of integrals to carry out the integration.

¦ Air Conditioning a Room

Now let’s build a model that describes a room cooled by an air conditioner. Without air conditioning, we can model the change in temperature using ODE (1). When the air conditioner is running, its coils remove heat energy at a rate proportional to the difference between Tr, the room temperature, and the

48

Chapter 3

Newton’s law of cooling (twice)!

Time t is measured in minutes.

The modeling here is more advanced than you have seen up to this point. You may want to just use the equations and skim the modeling.

The step function is one of the engineering functions. You can find them by going to ODE Architect and clicking on Help, Topic Search, and Engineering Functions.

temperature Tac of the coils. So, using Newton’s law of cooling for the temperature change due to both the air outside the room and the air conditioner coils, our model ODE is

³ô

—77 = k{ Tout — Tr) + kac( Tac — Tr) dt

where Tout is the temperature of the outside air and k and kac are the appropriate cooling coefficients. If the unit is turned off, then kac = 0 and this equation reduces to ODE (1).

Let’s assume that the initial temperature of the room is 60°F and the outside temperature is a constant 100°F. The air conditioner operates with a coil temperature of 40°F, switches on when the room reaches 80°F, and switches off at 70°F. Initially, the unit is off and the change in the room temperature is modeled by

dT

— = 0.03(100- Tr), Tr(0) = 60 (8)

dt

where we have taken the cooling coefficient k = 0.03 min-1. As we expect, the temperature in the room will rise as time passes. At some time ton the room’s temperature will reach 80°F and the air conditioner will switch on. If kac = 0.1 min-1, then for t > ton the temperature is modeled by the IVP

(Ô

—^ = 0.03(100- Tr) + 0.1(40- Tr), Tr(ton) = 80 (9)

dt

which is valid until the room cools to 70°F at some time toff. Then for t> toff the room temperature satisfies the IVP (8) but with the new initial condition Tr (toff) = 70. Each time the unit turns on or off the ODE alternates between the two forms given in (8) and (9).

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