# Homework Help: Verifying and ploting D.E. solution

1. Feb 7, 2009

### morrobay

1. The problem statement, all variables and given/known data

A cup of coffee cools according to;

dT/dt = .17(36-T)
At t=0 T= 85 deg centigrade
Two questions : how to verify the solution below and to plot a continuous curve of T/t

2. Relevant equations

the solution is T(t) = 36 + Ce^-.17t, = 36+49e^-.17t

b]3. The attempt at a solution[/b]

I solved for two (t)'s numerically ;
T= 74 , 1.29 deg=e^.17t , t= 1.5 s
T= 80 ,.108=.17t . t=.64 s
So could someone explain how to integrate this and get the curve?
1. The problem statement, all variables and given/known data

2. Feb 7, 2009

### slider142

This equation is separable. Manipulate it algebraically so that you get the form f(T) dT/dt = g(t), then integrate both sides dt: $$\int f(T) \frac{dT}{dt} dt = \int g(t) dt$$. Note that either f or g may be a constant function, and that some people are more comfortable writing $\frac{dT}{dt} dt$ as just dT.

3. Feb 8, 2009

### morrobay

Thanks.
Just for my edification would you or someone else complete the numerical integration on this problem. ( it is not a homework problem )

4. Feb 8, 2009

### HallsofIvy

Why do you keep insisting on numerical solutions? It is easy to integrate both sides of
$$\int\frac{dT}{36- T}= \int .17 dt$$
(On the left use the substitution u= 36- T.)

Of course, to verify that $T= 36+49e^{-.17t}$ is the solution to the problem you don't even need to integrate. $$dT/dt= (-.17)(49)e^{-.17t}$$ while $$36- .17T= (-.17)(49)e^{-.17t}$$ so this T clearly satifies the equation and, of course, taking t= 0 gives T(0)= 36+ 49= 85.

Last edited by a moderator: Feb 8, 2009
5. Feb 8, 2009

### morrobay

Thanks that is what I was looking for

6. Feb 9, 2009

### morrobay

referring to : 36-.17T = (-.17)(49)e^-.17t
It looks likes the left side above should be (36-T)(.17) since at (t)=1.5 T=74
and the right side ,dT/dt= 6.45

Last edited: Feb 9, 2009