Question

Hi all,

The temperature of a building obeys the differential equation \[x'(t) = K(T-x(t))\], where T = 20 when the room is heating up, and 9 when cooling down. Starting from 8am, the heat cycles on for three hours, and off for 0.5 hours until 6pm.

The temperature of the building is 13 at 8am, and 19 at 11am (after the first 3-hour heating period).

(a) Find the temperature at 12 noon.

My attempts so far:

\[x' = K(20-x) \: ; \: x(0) = 13\] for the first heating period, which becomes
\[x' + Kx = 20K\] This is not separable as far as I can see, so the integrating factor is (cont'

Answer 1

\[\mu = e^{\int K dt}\], so we get \[\frac{d}{dt}(e^{Kt} x) = 20Ke^{Kt}\] so \[e^{Kt}x = 20e^{Kt} + c_1\]Using the initial condition, \[13e^{Kt} = 20e^{Kt}+c_1\]Similarly for the cooling down period, \[e^{Kt}x = 9e^{Kt} + c_1\], and \[19e^{Kt} = 9e^{Kt} + c_1\], trying to solve for \(c_1\) gives gibberish, so... what do I do?

Answer 2

Some progress, I forgot to replace t with 0 when filling in the initial condition. When I do that for the first equation I get \[c_1=0\], so then \[e^{Kt}x=20e^{Kt}\]and \[x=20\]As x is a function of time, there's clearly something wrong somewhere in here.