First order seperable ode with IC's

[willmoore21]

Question

According to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air. The differential equation is given byAccording to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air. The differential equation is given by:

dT/dt=-k(T-T_​0)

where T is the temperature of the substance, T₀ is the temperature of air, k is a constant of proportionality and t is the time. Find the general solution of this equation.
If T₀=30°C then find the solution subject to the initial condition that T=100°C at time t=0.
Suppose an experiment is carried to determine the physical constant k. If the substance cools from 100°C to 70°C in 15 minutes then find k. Hence find the time taken for the temperature to fall from 100°C to 40°C

Ok guys, Just so you know where I'm at, I can do seperable ODES quite easily, however, I am useless and wordy questions and I don't know how to set it up with IC's in. I need to bring the T under the dT, but don't know how to seperate it from -k(T-T₀).

If I multiply out I get

dT/dt=-kT+T₀

and I know know how to get it under the dT, can someone explain how I do this? I may need further help with the initial conditions but I need to know how to find the general solution for this type of ode with IC's first.

 

[galactus]

\(\displaystyle \frac{dT}{dt}=-k(T-30)\)

Separate variables:

\(\displaystyle \frac{dT}{T-30}=-kt\)

Now integrate:

\(\displaystyle ln(T-30)=-kt+c\)

Solve for T:

\(\displaystyle T-30=e^{-kt+c}\)

Note that \(\displaystyle e^{-kt+c}=e^{-kt}e^{c}\). Since \(\displaystyle e^{c}\) is a constant we can write it as \(\displaystyle C=e^{c}\):

\(\displaystyle T-30=Ce^{-kt}\)

Now, use the initial conditions to solve for C and k.

\(\displaystyle T(0)=100, \;\ T(15)=70\)

 

[willmoore21]

Thanks

Ok,

T=70exp(-kt)+30 I guess because T=100, and if t=0 then C has to be 70 because exp(0)=1,

This is the particular solution when T0=30.



Then T(15)=70

I am slightly confused,

I think it's

70=Cexp(-15k)+30,

However, I do not know what C would be in this case? exp(-15k) might not equal 1 anymore so it can't be 40?

Or do I just keep it at 70?

To give ln(4/7)/(-15)=k
=0.0373

which means for T=40

40=70exp(-0.0373t)+30
t=52mins?

 

[galactus]

Yep, that's it.

 

[willmoore21]

Thanks galactus, very detailed explanation, have managed to do another similar one following this so hopefully I understand now.