2
$\begingroup$

$$C\frac{dT}{dt}=-\sigma T(t)^{4}+(1-\alpha)Q$$ I need help to solve the above pde where $C,\alpha ,Q$ are constants, I'm really unsure on how to even start to solve it

$\endgroup$
6
  • 1
    $\begingroup$ It looks like an ODE to me. I presume $t(t)$ is a typo? $\endgroup$ – Yuriy S Mar 14 '18 at 18:34
  • $\begingroup$ Yes sorry that is a typo thanks $\endgroup$ – Gibberish Mar 14 '18 at 18:35
  • $\begingroup$ Is $*$ multiplication? If so you should write $\sigma T(t)^4$ instead. If it is not you need to say so. $\endgroup$ – AzJ Mar 14 '18 at 18:36
  • $\begingroup$ It is multiplication not convolutions $\endgroup$ – Gibberish Mar 14 '18 at 18:38
  • $\begingroup$ $\sigma$ is a constant as well? You can separate variables, but the integral is gross $\endgroup$ – operatorerror Mar 14 '18 at 18:48
4
$\begingroup$

As you have written it, your problem is an ODE not a PDE as $T$ is a function of only one variable. As you only have one function and everything else is a constant you can solve using separation of variables.

For example let $p=(1-\alpha)Q$. We can can separate variables as

\begin{align} \frac{C}{-\sigma T^4+p} \frac{d T}{dt}=1 \end{align} The tricky part is integration of the left side. Also see

The second link we be more helpful as the it shows how to express the solution as an integral. I beilive this form is going to be part of the expected answer as this is a very diffcult integral.

$\endgroup$
3
  • $\begingroup$ should $d$ be $C$? $\endgroup$ – Gibberish Mar 14 '18 at 18:56
  • $\begingroup$ yes, my mistake $\endgroup$ – AzJ Mar 14 '18 at 18:57
  • $\begingroup$ Please accept my answer, if you found it satisfactory. $\endgroup$ – AzJ Mar 14 '18 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.