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Linear algebra  System of differential equations 
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#1
Nov2112, 09:00 PM

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I'm a bit confused on how to do this problem, here is what I have.
Part a) I must set up the set of linear differential equations with the initial values. Using the balance law gives y_{0}'=0.2+0.10.20.1=0 The other 2 net rate of change would be equal to 0 as well. But...I don't think I did that right. 


#2
Nov2112, 10:14 PM

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I would think the flows would be proportional to the temperature differences. If so, the numbers shown on the diagram are coefficients for that.
You will need to invent variables for the three temperatures as functions of time, yes? (It's not clear to me which box is which in the diagram. Hope it is to you.) 


#3
Nov2212, 07:11 AM

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For example, "0" is losing heat to "1" at 0.2 times its temperature and is losing heat to "2" at 0.1 its temperature. It is losing heat at a total of [itex]0.3y_0[/itex]. But it is gaining heat from "1" at [itex]0.2y_1[/itex] and from "2" at [itex]0.1y_2[/itex]. The equation is [itex]y_0'= 0.2y_1+ 0.1y_2 0.3y_0[/itex] and similarly for the others.



#4
Nov2212, 03:39 PM

P: 418

Linear algebra  System of differential equations
Oh thank you guys, I see what I need to do.
Yeah, I was also confused what area each box was referring to. I'm assuming 1 represents the 1st floor and 2 the second. That leaves 0 as the outside, but I could be wrong. 


#5
Nov2812, 12:36 AM

P: 418

Revisiting this problem.
Would the initial conditions be y
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0
y_{1}(0)=70 y_{2}(0)=60 ? 


#6
Nov2812, 04:23 AM

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Sure. What ODEs do you get?



#7
Nov2812, 01:31 PM

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I got
Y0'=0.3y0+0.2y1+0.1y2 Y1'=0.2y00.7y1+0.5y2 Y2'=0.1y0+0.5y10.6y2 Y0(0)=0 Y1(0)=70 Y2(0)=60 The eigenvslues are really ugly 


#8
Nov2812, 03:03 PM

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#9
Nov2812, 06:06 PM

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Hmm I don't think we're supposed to think too Much about thermodynamics for this problem.
Do those equations and ICs look right? 


#10
Nov2812, 06:46 PM

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#11
Nov3012, 01:39 AM

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Hmmmm....something is not right. Part c of the problem asks to compute the time required for each floor to reach 32 degrees fahrenheit.
This does not seem possible. I attached the solutions (solved using wolframalpha). X(t) represents the solution to the outside, y(t) represents 1st floor and z(t) the 2nd. I verified these solutions are correct and satisfy the ICs. The x(t) equatoin was able to be solved for t. It takes t=1.9 (idk units) to reach 32 degrees. However, look @ the y(t) and z(t) equations. The y(t) cannot be solved for y(t)=32. All the terms are positive and each exponential can't be negative. z(t) can be solved for t, but that gives you a negative time.. Hmm, I don't understand this or maybe this is the answer to the question? That x=32 when t=1.9 y=32 is not possible z=32 requires a negative time, and thus, not possible? 


#12
Nov3012, 02:31 PM

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By treating each of the three regions as a closed space of the same specific heat, it was inevitable that the limiting temperature for all three would be the average of the three temperatures you started with. It follows from that that the spaces that started above that average might never make it below that average (and at least one of them would not).
I'll say it again: it is completely obvious that the outside will stay at 0. Put that in and you will get sensible answers. 


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