How can the differential equation for a sphere in a room be solved?

[atta_bo-y]

Hey guys,

I have a smallish problem with a differential equation.

Set up:
There is a sphere in a room; and it's being illuminated with intensity I from one side.
The room temperature (T_{environment}) is set to be constant.

[ latex ]
\Delta Q = Q_{in} - Q_{out}\\
= I\Delta t A - P_{out}\Delta t + P_{in}\Delta t\\
= I\Delta t A - \sigma A T^{4} \Delta t + \sigma A T^{4}_{environment} \Delta t\\
\Delta Q = cm\Delta T\\
\frac{\Delta T}{\Delta t} = I\Delta t \pi r^2 - \sigma 4 \pi r^2 T^{4} \Delta t + \sigma 4 \pi r^2 T^{4}_{environment} \Delta t
[ /latex ]

Hence we have a differential equation in the form of

!
[ latex ]
\frac{dT(t)}{dt} = a - b*T^4(t)
[ /latex ]
!

I have tried different methods... But none of them worked... (And wolframalpha can only solve for t :-( )

Thanks for your consideration ;-)
atta_bo-y

 

[dextercioby]

Well, first don't put the spaces in the <latex> bracket. Then have you tried to separate variables and then a partial fraction decomposition ?>

 

[AlephZero]

http://integrals.wolfram.com/index.jsp?expr=1/(a+-+bx^4)&random=false

 

[atta_bo-y]

$\frac{dT(t)}{dt} = a - b*T^4(t)$

Any tips for how to get T(t)??

$t+c &= \frac{-\log(bT(t)- a) + \log(a + bT(t)) + 2\arctan\left(\frac{bT(t)}{a}\right)}{4a^3b}$
$(t+c)4a^3b &= \log\left(\frac{a+bT(t)}{-a + bT(t)}\right) + 2\arctan\left(\frac{bT(t)}{a}\right)$
$(t+c)2a^3b = \arctan\left(\tan\left(\frac{1}{2}\log\left(\frac{a+bT(t)}{-a + bT(t)}\right)\right)\right) + \arctan\left(\frac{bT(t)}{a}\right)$
$(t+c)2a^3b = \arctan\left(\frac{\tan\left(\frac{1}{2}\log\left(\frac{a+bT(t)}{-a + bT(t)}\right)\right) + \frac{bT(t)}{a}}{1-\tan\left(\frac{1}{2}\log\left(\frac{a+bT(t)}{-a + bT(t)}\right)\right)\frac{bT(t)}{a}}\right)$

furthermore I can set

$\frac{a+bT(t)}{-a+bT(t)} = 1 + \frac{2a}{bT(t) - a}$

Now this looks like a sisyphean direction to me... Please correct me if I'm wrong...

Thanks

 

[Bill_K]

Usually at this point you define a new special function named after yourself and put a box around it.

 

[0xDEADBEEF]

Easy ;)

$$T(t)=\left(\frac{a}{b}\right)^{\frac{1}{4}}$$

 

[coelho]

Well... if the differential equation youre trying to solve is:

$$\frac{dT(t)}{dt}=a-bT^4$$

Then, there you go: first, separate the variables:

$$\frac{dT}{a-bT^4}=dt$$

and integrate each side:

$$\int \frac{1}{a-bT^4}dT=t+C$$
(Equation 1)

The T integral is somewhat lenghty to calculate. For making the calculations a bit better-looking, write:

$$a=e^4$$ and $$b=f^4$$.

Then decompose the integrand as:

$$\frac{1}{a-bT^4} = \frac{1}{e^4-f^4 T^4} = \frac{1}{2e^2} \left[\frac{1}{e^2+f^2 T^2} + \frac{1}{e^2-f^2 T^2} \right]$$

So the integral becomes:

$$\int \frac{1}{a-bT^4}dT = \frac{1}{2e^2} \left[ \int \frac{1}{e^2+f^2 T^2} dT + \int \frac{1}{e^2-f^2 T^2} dT \right]$$

The first integral can be calculated using the substituition

$$\frac{fT}{e}=tan(x)$$

Which after some calculations leads to:

$$\int \frac{1}{e^2+f^2 T^2} dT = \frac{1}{ef} arctan \left( \frac{fT}{e} \right)$$

The second integral can be decomposed as:

$$\int \frac{1}{e^2-f^2 T^2} dT = \frac{1}{2e} \int \left[ \frac{1}{e-fT} + \frac{1}{e+fT} \right] dT$$

And integrating it we get:

$$\int \frac{1}{e^2-f^2 T^2} dT = \frac{1}{2ef} ln \left[ \frac{e+fT}{e-fT} \right]$$

So, the final result is:

$$\int \frac{1}{a-bT^4}dT =\frac{1}{2e^3 f}\left[ arctan\left(\frac{fT}{e}\right) + \frac{1}{2} ln\left(\frac{e+fT}{e-fT}\right)\right]$$

As shown in the equation called Equation 1, this is equal to t+C (C being an arbitrary constant), so the solution for the differential equation is:

$$\frac{1}{2e^3 f}\left[ arctan\left(\frac{fT}{e}\right) + \frac{1}{2} ln\left(\frac{e+fT}{e-fT}\right)\right] = t+C$$

So you see you can write t as a function of T(t), but i think there is no way to find a way to write T(t) as a function of t. Anyway, its the best we can do... i hope it be enough.

BTW, in the above equation you can remove the e's and f's by writing them in function of the original parameters a and b as:

$$e=\sqrt[4]{a}$$ and $$f=\sqrt[4]{b}$$,

but the equation you will get is even uglier:

$$\frac{1}{2 \sqrt[4]{a^3 b} } \left[ arctan\left(\sqrt[4]{ \frac{b}{a} } T\right) + \frac{1}{2} ln\left(\frac{ \sqrt[4]{a}+\sqrt[4]{b}T }{ \sqrt[4]{a}-\sqrt[4]{b}T }\right)\right] = t+C$$