Please show all work! Thank you!! 3. It is spring and the heating and cooling sy

Question

Please show all work! Thank you!!

3. It is spring and the heating and cooling systems of a house are turned off. The weather is calm for many days and the external temperature follows a sine wave: ft) 10sin(nt 120 +05. Here the time, t, is measured in hours starting from 9am, which is the time in the morning when the temperature is equal to its average, 65F. The temperature inside the house follows Newton's law of cooling. Observation show that, in the long run, the maximum and minimum temperatures inside the house occur 3 hours after those for outside. If the initial temperature is 65 F, find the formula for the temperature inside the house Use the data on the long run solution to find a value of the constant of cooling Data: From the textbook (front end pages) we find asin bt bcos (bt)) +c ea sin bt dt a b which we write: e sin bt dt sin bt-o)+c, a +b where cos (p) and sin (p) a +b a +b For the long run solution discard all the terms in the general solution containing e since this is very small for sufficiently large t

Explanation / Answer

By Newton's Law of Cooling, the range of change of temperature inside the house is directly proportional to the difference between its current temperature and the temperature outside. Thus, if we let T we have the DE:
T' = k[f(t) - T], for some k > 0.

Notice that if f(t) - T < 0 and k > 0, then T' < 0 as expected. In a similar fashion, f(t) - T > 0 yields T' < 0 as expected.
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Subbing in f(t) = 10sin(t/12) + 65 gives:
T' = k[10sin(t/12) + 65 - T] ==> T' + kT = k[10sin(t/12) + 65].
(note: the equation was re-written in standard form.)

This DE has integration factor e^(kt), so multiplying both sides by e^(kt) yields:
e^(kt)T' + kT*e^(kt) = [e^(kt)T]' = k*e^(kt)[10sin(t/12) + 65].

Integrating both sides:
e^(kt)T = k*e^(kt)[10sin(t/12) + 65] dt = 10k e^(kt)sin(t/12) dt + 65k dt.

The first integral can be computed by integrating by parts twice and then solving for the unknown integral, or we can resort to tables---either way works. The second integral should be trivial to compute. After both of these are integrated, you will have two constants: k and C (or whatever you want to call the constant of integration that occurs when these two integrals are computed). You can find these constants by using the fact that f(t) ranges between 55 and 75 and finding the values of t such that f(t) = 55 and f(t) = 75. Using these two values of t, add 3 to these values of t (since the temperature in the house occurs 3 hours later) and you will get two initial conditions, allowing you to solve for k and C.

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