Q 3.3-15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Stefan’s law of radiation states that the rate of change of temperature of a body at T degrees Kelvin in a medium at M degrees Kelvin is proportional to ${\mathit{M}}^{\mathbf{4}}\mathbf{-}{\mathit{T}}^{\mathbf{4}}$. That is $\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\left(M^4-T^4\right)\mathbf{,}$where k is a positive constant. Solve this equation using separation of variables. Explain why Newton’s law and Stefan’s law are nearly the same when T is close to M and M is constant.  [Hint: Factor ${\mathit{M}}^{\mathbf{4}}\mathbf{-}{\mathit{T}}^{\mathbf{4}}$]

Step-by-Step Solution

Answer

The results are $\left(\mathbf{T}\mathbf{-}\mathbf{M}\right)\mathbf{=}\mathbf{C}\left(\mathbf{T}\mathbf{+}\mathbf{M}\right){\mathbf{e}}^{\mathbf{2}\mathbf{arctan}\frac{\mathbf{T}}{\mathbf{M}}\mathbf{-}\mathbf{4}{\mathbf{kM}}^{\mathbf{3}}\mathbf{t}}$ and When is T close to M then

$\frac{\mathbf{1}}{{\mathbf{M}}^{\mathbf{4}}\mathbf{-}{\mathbf{T}}^{\mathbf{4}}}\mathbf{=}\mathbf{4}{\mathbf{M}}^{\mathbf{3}}\mathbf{(}\mathbf{M}\mathbf{-}\mathbf{T}\mathbf{)}$

$$\begin{gathered} \frac{\mathrm{dT}}{\mathrm{dt}}=4{\mathrm{M}}^3\mathrm{k}(\mathrm{M}-\mathrm{T}) \\ ={\mathrm{k}}_1(\mathrm{M}-\mathrm{T}) \end{gathered}$$

Step1: Important concept.

According to Stefan’s law,

$\frac{dT}{dt}=k\left(M^4-T^4\right),$

 

Step 2: Analyzing the given statement

According to Stefan’s law,

$\frac{dT}{dt}=k\left(M^4-T^4\right),$ 

Where, k is a positive constant, T degrees Kelvin is the temperature of the body degrees and M degrees Kelvin is the change in the temperature in a medium.

We have to solve this equation and to explain why Newton’s law and Stefan’s law are nearly the same when T is close to M and M is constant.

Step3: Factorizing M 4 - T 4 in the given differential equation s eparating variables

$$\begin{gathered} \frac{dT}{dt}=k\left(M^4-T^4\right) \\ \frac{dT}{\left(T^4-M^4\right)}=-kdt \\ \frac{1}{T^4-M^4}=\frac{1}{\left(T^2-M^2\right)\left(T^2+M^2\right)} \end{gathered}$$

Now using the partial fractions

Now,

$$\begin{gathered} \int \left(\frac{1}{2M^2\left(T^2-M^2\right)}-\frac{1}{2M^2\left(T^2+M^2\right)}\right)dT=-\int 2k{M}^{2}dt \\ \int \frac{dT}{2M\left(T-M\right)}-\int \frac{dT}{2M\left(T+M\right)}-\int \left(\frac{dT}{\left(T^2+M^2\right)}\right)=-2k{M}^{2}t+C \\ \frac{1}{2M}\ln \left|T-M\right|-\frac{1}{2M}\ln \left|T+M\right|-\frac{1}{M}\arctan \frac{T}{M}=-2k{M}^{2}t+C \\ \ln \left|\frac{T-M}{T+M}\right|=2\arctan \frac{T}{M}-4k{M}^{3}t+C \\ \left|\frac{T-M}{T+M}\right|=C{e}^{2\arctan \frac{T}{M}-4k{M}^{3}t} \\ \left(T-M\right)=C\left(T+M\right)e^{2\arctan \frac{T}{M}-4k{M}^{3}t} \end{gathered}$$

When is T close to M then 

$$\begin{gathered} \frac{1}{M^4-T^4}=4M^3\left(M-T\right) \\ \frac{dT}{dt}=4M^{3}k\left(M-T\right) \\ =k_1\left(M-T\right) \end{gathered}$$

Therefore, the results are $\left(\mathbf{T}\mathbf{-}\mathbf{M}\right)\mathbf{=}\mathbf{C}\left(\mathbf{T}\mathbf{+}\mathbf{M}\right){\mathbf{e}}^{\mathbf{2}\mathbf{arctan}\frac{\mathbf{T}}{\mathbf{M}}\mathbf{-}\mathbf{4}{\mathbf{kM}}^{\mathbf{3}}\mathbf{t}}$ and When T is close to M then

$\frac{\mathbf{1}}{{\mathbf{M}}^{\mathbf{4}}\mathbf{-}{\mathbf{T}}^{\mathbf{4}}}\mathbf{=}\mathbf{4}{\mathbf{M}}^{\mathbf{3}}\left(\mathbf{M}\mathbf{-}\mathbf{T}\right)$

$$\begin{gathered} \frac{dT}{dt}=4M^{3}k\left(M-T\right) \\ =k_1\left(M-T\right) \end{gathered}$$