# Differential Equation Help

## Homework Statement

Well, I finally figured out the differential equation from a previous post (thank god) but know I am having some trouble getting the result I want as it has been 6 years since diff eq.

500dx = 20dT + 20T - 9.47(T-15)dx

N/A

## The Attempt at a Solution

Well, first shot I went for seperation of variables which seemed logical to me:

500dx + 9.47(T-15)dx = 20dT + 20T

(500 + 9.47(T-15)dx = 20dT + 20T

dx = (20dT + 20T) / [500 + 9.47(T-15)]

And this seems unrealistic.

Someone kick me and tell me how wrong I am please. I have tried digging though my old books but I don't seem to have my diff eq or calc books anymore.

Dick
Homework Helper
Either you aren't using parentheses right, or that's complete rubbish. 500dx = 20dT + 20T - 9.47(T-15)dx is crazy. Every term in that equation should have 'd' something in it if one does.

Hrmmm.

Well, this was what I worked out with my professor last week and I have been trying to work on it for awhile.

I am not sure if the link to the other post I had worked or not, but basically it is a heat transfer problem. The control volume being from x to x+dx which corresponds to T and T+dT.

In the middle of typing this I think I may have figured out my error....I believe the first should have been T+dT.

500dx = 20(T+dT) - 20T - 9.47(T-15)dx

[500 + 9.47(T-15)]dx = 20 dT

dx = 20 dT / [500 + 9.47(T-15)]

Does that look better?

Continuing on that thought...

Integating both sides...

x = 2.11*ln(T+37.8) + c

x/2.11 = ln (T +37.8) + c

e^(x/2.11) = T + 37.8 + c

T = e^(x/2.11) - 37.8 + c

T(0) = 15 -> c = 51.8

T = e^(x/2.11) + 14

hrmmm...still along way from the books answer of T = 15 + 52.8(1-e^(-x/2.11))

Last edited:
Dick
Homework Helper
Hrmmm.

Well, this was what I worked out with my professor last week and I have been trying to work on it for awhile.

I am not sure if the link to the other post I had worked or not, but basically it is a heat transfer problem. The control volume being from x to x+dx which corresponds to T and T+dT.

In the middle of typing this I think I may have figured out my error....I believe the first should have been T+dT.

500dx = 20(T+dT) - 20T - 9.47(T-15)dx

[500 + 9.47(T-15)]dx = 20 dT

dx = 20 dT / [500 + 9.47(T-15)]

Does that look better?

Looks better to me. At least the 'd's are balanced. I don't know the whole context of your problem, but at least it looks like a differential equation now.

Sir,

I appriciate your help...but I think I am still missing some as you can see from my above answer. I am getting closer but not quite there yet. It may be that I am missing something in my equation still.

Dick
Homework Helper
Sir,

I appriciate your help...but I think I am still missing some as you can see from my above answer. I am getting closer but not quite there yet. It may be that I am missing something in my equation still.

It doesn't look like anything is missing to me, except for actually solving the equation. Now you can integrate both sides, right?

I believe so, I should have replied but I just edited the text. My solution is about 4 posts up. If you have time I would very much appriciate your approval / laughter.

=)

You are very close. I think you may have screwed up your initial integration (mine looks a lot different). This might help...

#### Attachments

• Calculus_Cheat_Sheet_Integrals.pdf
183.5 KB · Views: 102
I like the cheat sheet! I am not quite sure that I follow you though.

When I worked it out (already erased), my integration of both sides resulted in an x that was much different than yours. I used the integral of 1/at+b to get the x in terms of t. I also did u substitution and got the same answer...

...this is also very similar to a first order transient response in electric circuits...

I have tried this problem several different ways and keep coming up with the answer that I originally had. I looked into some FO Transient Response problems and can't seem to get it to fit together.

Maybe the answer in the book is wrong... Have you tried plotting your results or running a simulation to see if your results make sense?

Hrmmm, just plotted it and it doesn't make sense. Back to the drawing board.

500dx = 20(T+dT) - 20T - 9.47(T-15)dx

[500 + 9.47(T-15)]dx = 20 dT

dx = 20 dT / [500 + 9.47(T-15)]

This is wrong. You should have 20(t + dt) in the numerator. Based on the equation you first posted.

500dx = 20dT + 20T - 9.47(T-15)dx

Actually, I think you are going to have to use a transform to solve this.

did you get it?