Solving a Differential Equation for Temperature Change with Initial Conditions

In summary, the conversation involves solving a differential equation for the temperature at a given time, given an initial temperature. The problem is approached by rearranging the equation and taking the integral, but an error is made in the process. The correct approach involves writing the equation as a product and then integrating to find the solution. Ultimately, the correct temperature at t=2.3 is found to be 52.460 degrees.
  • #1
TW Cantor
54
1

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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  • #2
First can you tell me is this answer you want when t=2.3,T=52.460 ? :smile:
 
  • #3
that sounds better than what I've been getting :P how did you get that?
 
  • #4
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:
  • #5
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
 
  • #6
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.
 
  • #7
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
[tex]\frac{dT}{dt}+ 0.4T- 6.4= 0[/tex]
rewrite it as
[tex]\frac{dT}{dt}+ 0.4T= 6.4[/tex]
and then have
[tex]\int 0.4T dT= 6.4dt[/tex]

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

Instead, as xiaoB suggest, write it as
[tex]\frac{dT}{dt}= -0.4T+ 6.4[/tex]
and integrate
[tex]\frac{dT}{0.4T+ 6.4}= -fdt[/tex]
 
  • #8
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 :-)
 

What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It involves the use of variables and their rates of change to model real-world phenomena.

What is the purpose of solving a differential equation?

The purpose of solving a differential equation is to find a function or a set of functions that satisfy the equation. This allows us to understand and predict the behavior of complex systems in various fields such as physics, engineering, and economics.

What are the two types of differential equations?

The two types of differential equations are ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs involve only one independent variable, while PDEs involve multiple independent variables.

How do you solve a differential equation?

The method for solving a differential equation depends on its type and complexity. Some common techniques include separation of variables, substitution, and using integrating factors. In some cases, numerical methods may also be used to approximate the solution.

What are some real-world applications of differential equations?

Differential equations are used in many fields to model and analyze various phenomena. Some common applications include predicting population growth, modeling the spread of diseases, predicting weather patterns, and designing electrical circuits.

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