SCENARIO: Suppose that our air conditioner fails at time t0 = 0 one midnight, and we cannot afford to have it repaired until payday at the end of the month. We therefore want to investigate the resulting indoor temperatures that we must endure for the next several days.

Exercise 1: Using the integration factor algorithm we’ve learned for solving first order linear differential equations, and working by hand, solve equation (4.5.6) with the initial condition u(0) = u0 (the indoor temperature at the time of the failure of the air conditioner). You should get the solution u(t) = a0 + c0e
−kt + c1 cos ωt + d1 sin ωt, (4.5.7) p4
where



c0 = u0 − a0 −
k
2a1 − kωb1
k
2 + ω2
c1 =
k
2a1 − kωb1
k
2 + ω2
, d1 =
kωa1 + k
2
b1
k
2 + ω2
with ω = π/12.
Exercise 2: Redo the computation for Exercise 1 using SageMath.
Exercise 3: Redo the computation for Exercise 2 using SageMath and the following
data:
a0 = 80, a1 = −5, b1 = −5
√
3, ω = π/12, and k = 0.2
to get the following solution
u(t) = 80 + e
−t/5
(u0 − 82.3351) + (2.3351) cos πt
12
− (5.6036) sin πt
12
. (4.5.8) p5
Exercise 4: Plot the slope field and solution curves of equation (4.5.6) with the
following initial values u(0) = 65, 70, 75, 80, 85, 90. If your code runs successfully, you
should get the following figure.
68
Applied Calculus IV – AMS 361 September 23, 2020
This figure shows that whatever the initial temperature u0 is, the indoor temperature
“settles down” within about 18 hours to a periodic daily oscillation.
Exercise 5: Observe that the “damped” exponential term in equation (4.5.7) approaches zero as t → ∞, leaving the long-term “steady periodic” solution
usp(t) = 80 + (2.3351) cos πt
12
− (5.6036) sin πt
12
. (4.5.9) p6
Consequently, the long-term indoor temperatures oscillate every 24 hours around the
same average temperature 80◦F as the average outdoor temperature. Verify that
equation (4.5.9) can be rewritten as
u(t) = 80 − (6.0707) cos 
πt
12
− 1.9656
= 80 − (6.0707) cos π
12
(t − 7.5082)
(4.5.10) p7
Exercise 6: Verify that the indoor temperature varies between a minimum of about
74◦F and a maximum of about 86◦F.
Exercise 7: Plot the Ambient temperature, the indoor temperature, and the steadyperiodic indoor temperature on the same screen using SageMath. You should get the
following graph
Finally, comparison of equations (4.5.4) and (4.5.10) indicates that the indoor temperature lags behind the outdoor temperature by about 7.508243.5 hours, as illustrated in
the above figure. Thus the temperature inside the house continues to rise until about
7:30 P.M. each evening, so the hottest part of the day inside is early evening rather
than late afternoon (as outside).
Extra exercise: For a personal problem to investigate, carry out a similar analysis
using average July daily maximum/minimum figures for your own locale and a value
of k appropriate to your own home. You might also consider a winter day instead
69
Applied Calculus IV – AMS 361 September 23, 2020
of a summer day. (What is the winter-summer difference for the indoor temperature
problem?) You may wish to explore the use of available technology both to solve
the differential equation and to graph its solution for the indoor temperature in comparison with the outdoor temperature.

Lecture – AMS 361

Suppose that our air conditioner fails at time t0 = 0 one midnight, and we cannot afford to have it repaired until payday at the end of the month.