Solving a Differential Equation for Temperature Change with Initial Conditions

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Homework Help Overview

The discussion revolves around solving a differential equation related to temperature change over time, specifically the equation dT/dt + 0.4*T - 6.4 = 0, with an initial temperature of 86 degrees. Participants are attempting to find the temperature at t=2.3.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing various methods to solve the differential equation, including integration techniques and rearranging terms. Some question the correctness of their approaches and results, while others suggest alternative methods and corrections to initial attempts.

Discussion Status

The discussion is ongoing, with participants sharing different interpretations and methods for solving the equation. Some have provided calculations and results, while others have pointed out potential errors in reasoning or methodology. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants have noted a lack of recent experience with differential equations, which may be affecting their understanding and application of the methods discussed. There is also mention of moving the thread to a different section of the forum, indicating a potential misalignment with the original posting guidelines.

TW Cantor
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Homework Statement



dT/dt + 0.4*T - 6.4 = 0

where T is the temperature at time t
initially the temperature is 86 degrees

find the temperature at t=2.3

Homework Equations





The Attempt at a Solution



so i rearrange the equation to get:
dT/dt + 0.4*T = 6.4
then
integral (0.4*T) dT = integral (6.4) dt

so the general solution would be:
0.2*T^2 = 6.4*t + c

since i know that T=86 when t=0 i can work out c which i calculate to be 1479.2

so my final equation is 0.2*T^2 = 6.4*t +1479.2

but when i then put t=2.3 and solve for T my equation gives out a value of around 86.5 degrees so obviously i have gone wrong somewhere? can anyone help me out :-)
 
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First can you tell me is this answer you want when t=2.3,T=52.460 ? :smile:
 
that sounds better than what I've been getting :P how did you get that?
 
TW Cantor said:
that sounds better than what I've been getting :P how did you get that?

∫(dT/dt)dt=∫(6.4-0.4T) dt
T=6.4t-0.4Tt+c
T+0.4Tt=6.4t+c
T(1+0.4t)=6.4t+c

You should have answer!
 
Last edited:
dT/dt+0.4T=6.4 => (T e^(0.4t))dT=6.4 e^(0.4t) => T e^(0.4t)=16 e^(0.4t) + C
=> T=16+Ce^(-0.4t) Substituting in T(0)=86, C=70. So, T=16+70e^(-0.4t). At t=2.3 I got 43.9
 
oops, (T e^(0.4t))dT should be d/dT(T e^(0.4t)) or (T e^(0.4t))'. the answer is still correct though.
 
I was wondering how you were going to solve a differential equation without using Calculus! (I will move this to the "Calculus and beyond" Homework section.)

Your main error was right at the beginning. You have
\frac{dT}{dt}+ 0.4T- 6.4= 0
rewrite it as
\frac{dT}{dt}+ 0.4T= 6.4
and then have
\int 0.4T dT= 6.4dt

How did that sum, dT/dt+ 0.4T, suddenly become a product, 0.4T dT?

Instead, as xiaoB suggest, write it as
\frac{dT}{dt}= -0.4T+ 6.4
and integrate
\frac{dT}{0.4T+ 6.4}= -fdt
 
oh ok, i haven't done differential equations in about a year and that's just the method i thought my teacher taught me haha. thanks for setting me straight though guys :-)
 

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