Question

Suppose that your air conditioner fails on Sunday at midnight (t0 = 0),
and you cannot afford to have it repaired until payday at the end of
the month. Assume that the outside temperature varies according to
the function

A(t) = 80 5 cos(π/12)-5√3sin(π/12t)

and that your inside temperature, u(t) obeys Newton’s law of cooling
and is governed by the differential equation
du/dt= -0.2(u-A(t))

(a) If your indoor temperature when the air conditioner failed was
70◦F , determine the dynamics of temperature inside your apart-
ment over time. i.e. find a particular solution to the initial value
problem.
(b) What will the temperature inside the apartment be, 24 hours after
the break down?
(c) In the long run (t → ∞), what is the maximum and minimum
temperature you can anticipate inside your apartment?

Answer 1

+0.24- 0- 2 Ad) Inderatin facker fr tn Sazdt IF Q O.2 1 Ganral Selution af o.24 ACH. O.2+ 161 o.2t "dt: Integradiny y pats, - O.2t O. 2t O 2-4 I, 12 12 d. 2. 0.2 0.2* ).20.24t 2 12 o.2 Cos SIT 12 2 d.2t dt 2 O. 2 (2 O,2 = Se0.2Cot 25 TT Sh 25 T Q /2 12- 1+ 25 +44 12 144 SntCan t+25 2n 0.2t tr /2 12 A12

Go012 Ca 1r O. 2 +SSint 0.24 dt25T2 144 12 pinitauly 60R0.2t 25T214 O.24 in玉t-ST Cor ア大 'dt= 12S 12 12 xubstitut this in(i) Ur

6002t 2ST2+44 932+53 Ca +Bo-12) Sin 4c 0.2+ ung nttu'nl Careli tean, y0)-70 F ni (1).(70)=60 2S12+144 193245 )0+C (1932 +513 ) 7o- 25t2 +144 C

Value af Cin -0.2t (1932+ SJ Ca T(80ST- 125) S Ce 12 2S-12144 132+53) 25r2144 42+ 4 25T2+144 L 1932+53CoT++0Sr-12 Sn to 2

uiny iv (24)= 252+14 (1232 + $) Cor 2r +(80Str-12) Sin2 bo11232 55) 2Sm244 -4. 8 t(70- 60(1932+ 5 60 (1932+53T) 2 ST2144 TO 25サー4149 - 300.84+(70-300.84) (eソ 300. 84 I.9 H(4)298. 94 F Lim -0.2+ or t , uing 4 60 2Sn2t194 -S5)Coot+(8oS-1252) Sinなt /2 12 (120.42) Sinナ (30o.84) Caと 12 oo .8 4+(120.42)2 C.20 (300, 8 324,04 F 4 324.o4 F