$$C\frac{dT}{dt}=-\sigma T(t)^{4}+(1-\alpha)Q$$ 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

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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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