# Differential equation

## 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

## 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 :-)

First can you tell me is this answer you want when t=2.3,T=52.460 ?

that sounds better than what ive been getting :P how did you get that?

that sounds better than what ive 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

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.

HallsofIvy
Homework Helper
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 havent done differential equations in about a year and thats just the method i thought my teacher taught me haha. thanks for setting me straight though guys :-)