View Full Version : How to solve : dT/dt = k(T-Tm)... T(0) = To... k,Tm,&To are constants
savage44
Feb 24, 2011, 05:00 AM
How to solve : dT/dt = k(T-Tm)... T(0) = To... k,Tm,&To are constants
I would like to try to solve this using Maple software, but not sure what the steps are. If not with Maple just showing the steps would be a great help.
Thanks
Unknown008
Feb 24, 2011, 07:34 AM
Is that what you are asking for?
\frac{dT}{dt} = k(T - T_m)
Then, to solve this, divide by (T - Tm) and multiply by dt, on both sides, adding the integration sign:
\int \frac{1}{T - T_m} \ dT = k\int\ dt
When t = 0, T = To
When t = t, T = T.
Hence:
\int\ ^T_{T_o} \frac{1}{T - T_m} \ dT = k\int\ ^t_0 \ dt
Can you complete it?
savage44
Feb 24, 2011, 09:21 AM
I'm not sure how to go about the limits(?)... T -> To & t -> 0
Unknown008
Feb 24, 2011, 09:45 AM
I'm not sure what you are asking here... sorry :(
When you evaluate the integral, you get:
\ln (T - T_m) - \ln(T_o - T_m) = k(t - 0)
And then you can simplify.
jcaron2
Feb 24, 2011, 01:58 PM
How to solve : dT/dt = k(T-Tm)....T(0) = To....k,Tm,&To are constants
I would like to try to solve this using Maple software, but not sure what the steps are. If not with Maple just showing the steps would be a great help.
Thanks
First the manual way:
T'=k(T-T_m)
T' - kT = -kT_m
This is a simple linear first-order differential equation. Now we need to find an integrating factor u(t) that satisfies
-k \cdot u(t) = u'(t)
The answer to that, by inspection is simply
u(t) = e^{-kt}
Multiplying your entire equation by u(t), we then get
u(t) \cdot T' - u(t) \cdot kT = -kT_m \cdot u(t)
T'e^{-kt} - kTe^{-kt} = -kT_m e^{-kt}
Now, notice that the entire left side is the derivative of T*u(t):
T'e^{-kt} - kTe^{-kt} = (Te^{-kt})'
so
(Te^{-kt})' = -kT_m e^{-kt}
Now we can integrate both sides with respect to t:
\int (Te^{-kt})' dt= \int -kT_m e^{-kt}dT
Te^{-kt}= T_me^{-kt}+C
T= T_m+Ce^{kt},
where C is a constant. This is where your initial conditions come in (to find the value of C).
T(0)=T_0
so
T(0)= T_m+Ce^{0 \cdot t}=T_0
T_m+C=T_0
C=T_0-T_m
Plugging that back into the solution for T(t), we finally get:
T(t)= T_m+(T_0-T_m)e^{kt}
Tadaaah! There's your final solution.
-------------------------------------------------
Now, to do it with Maple I think you'd do the following:
DE1 := diff(T(t), t) = k*(T(t) - Tm);
dsolve ( { DE1, T(0) = T0 }, T(t) );
Does it give you the same answer? My differential equations are almost as rusty as my Maple skills. :)
jcaron2
Feb 24, 2011, 02:09 PM
Is that what you are asking for?
\frac{dT}{dt} = k(T - T_m)
Then, to solve this, divide by (T - Tm) and multiply by dt, on both sides, adding the integration sign:
\int \frac{1}{T - T_m} \ dT = k\int\ dt
When t = 0, T = To
When t = t, T = T.
Hence:
\int\ ^T_{T_o} \frac{1}{T - T_m} \ dT = k\int\ ^t_0 \ dt
Can you complete it?
In this case it's significantly more complicated than that because T is a function, not a variable. The goal here is not to find the value of T, but rather to find the function T(t) which satisfies the differential equation under the condition that T(0) = To.
If you plug in my answer, T(t)=Tm + (To-Tm)exp(kt), you'll find that it does indeed satisfy the differential equation in general, as well as the specified initial condition that T(0) = To.
There's a pretty good primer on solving this type of first-order ODE here: Pauls Online Notes : Differential Equations - Linear Equations (http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx)
Unknown008
Feb 25, 2011, 06:32 AM
Um... okay. I see what you mean now. I never did those types of differential equations >.<
Thanks for the link! I'll see if I can understand it :)
Unknown008
Feb 25, 2011, 07:12 AM
Hm.. I just read the first example in the site you gave using what I did here and I got the answer, and I then tried that problem, I got the same answer as yours :)
I guess that only when the variables are 'unseparable' that I'll need to use the method of the site you gave me.
jcaron2
Feb 25, 2011, 07:16 AM
Thanks for the link! I'll see if I can understand it :)
I have complete faith that you'll be an first-order ODE expert by the end of the day. :)
jcaron2
Feb 25, 2011, 07:26 AM
Hm.. I just read the first example in the site you gave using what I did here and I got the answer, and I then tried that problem, I got the same answer as yours :)
I guess that only when the variables are 'unseparable' that I'll need to use the method of the site you gave me.
I just went back and did this problem using your technique too, and you're absolutely right.
Sorry I ever doubted you. :o
So now the OP has two different techniques for solving the DE (three if you include Maple :)). Yours is more intuitive.
Unknown008
Feb 25, 2011, 08:01 AM
I'm starting to get it. I did up to example 5. Example 6 is next! :p
jcaron2
Feb 25, 2011, 08:16 AM
Paul's site is pretty awesome, isn't it? He's got everything on there from basic algebra up through partial differential equations.
Like I said, I'm guessing by the end of the day you'll know more than he does. :)