Extreme Newton's Law of Cooling D.E.

[bdh2991]

Homework Statement

after a very unpleasant valentine's Day, a dead body was found in a downtown warehouse that had no heating or air conditioning. it was February in Florida and we know that the daily temperature in the warehouse fluctuates according to the function T(t)= 63-12sin(∏t/12), where t=0 corresponds to midnight on any given day. The body was discovered at 1:30 am on Feb. 15 and its temperature was 73 degrees F. Two hours later the temperature was 68 degrees F. What was the time of death of this body?

Homework Equations

dT/dt = k (T - T(m))

[exp(at)(-BcosBt + asinBt)]/a^2+b^2


where k is the proportionality constant T is temperature and T(m) is the medium of the environment surrounding the object.

The Attempt at a Solution

i set up my differential as dT/dt - kT - 63k = 12ksin(∏t/12)

after doing the integrating factor method and using the equation to solve the integration i got some huge formula for T

then i rescaled the time so i could solve for C and k ... for C i got

C= 73+ (144∏k)/(144k^2+∏^2)

after that i tried to solve for k but it honestly looks impossible and I'm not sure how to do it
if someone could help me out or at least check my work so far i would greatly appreciate it...this problem seems close to impossible

 

[sid9221]

Not a 100% sure as I've never solved one with a separate function for ambient air temperature but maybe you need to sub T(t)=63-12sin(∏t/12) into Newtons formula rather than equal to it.

I think the formula should be dT/dt = k [T - {63-12sin(∏t/12)}] where T is the initial temperature given by T(0) = 63.

So dT/dt = k(12sin(∏t/12))

and solve that ?

 

[HallsofIvy]

You should not separate the "63k" and "[itex]12ksin(\pi t/12)[/itex]":
[tex]dT/dt- kT= 63k+ 12ksin(\pi t/12)[/tex]

The associated homogeneous equation, dT/dt-kT= 0, has solution [itex]T= Ae^{kt}[/itex]. Now, look for a solution of the form [itex]B+ Csin(\pi t/12)+ Dcos(\pi t/12)[/itex]. Put that into the equation and solve for B, C, and D. Then put the entire solution into y(1.5)= 73 and y(3.5)= 68 to solve for A and k.

 

[bdh2991]

You should not separate the "63k" and "[itex]12ksin(\pi t/12)[/itex]":
[tex]dT/dt- kT= 63k+ 12ksin(\pi t/12)[/tex]

The associated homogeneous equation, dT/dt-kT= 0, has solution [itex]T= Ae^{kt}[/itex]. Now, look for a solution of the form [itex]B+ Csin(\pi t/12)+ Dcos(\pi t/12)[/itex]. Put that into the equation and solve for B, C, and D. Then put the entire solution into y(1.5)= 73 and y(3.5)= 68 to solve for A and k.

ok i see, I'm going to redo it that way, no wonder i was getting nowhere with the way i did it, thanks!