How to analytically solve an ODE based on the thermal radiation equation I have to find parameter for $C\text{, }R_0\text{, }\alpha\text{, }T_0\text{, }T_A\text{, }T_L\text{ and }\epsilon \sigma A$ based on pairs of $I,t$ with $T\left(t\right)=T_L$.
Also I know the pair $I, T$, for which $ \frac{\partial T}{\partial t} = 0 $.
All unknown are real and strictly positive.
$$
\frac{\partial T}{\partial t} = \frac{\left( R_0 \left( 1 + \alpha \left( T - T_0 \right) \right) \right) I^2 - \epsilon \sigma A \left( T^4 - T_A^4 \right)}{C} \\
T\left(0\right) = T_A
$$
Further I can assume that:
$$
T_A \le T_0 \\
T_L > T_A \\
R_0 = 200e-6 \\
T_0 \text{ is probably } 23+273.15 \text{ or in the range of } T_A\\
T_A \text{ is probably } 85+273.15 \text{ or } 105+273.15 \\
T_L \ll 1734+273.15
$$
For simplification, I have already substituted complex sub-expressions and factored out $T$:
$$
\frac{\partial T}{\partial t} = D T^4 + E T + F \\
\text{ with: } \\
D = \frac{- \epsilon \sigma A}{C} \\
E = \frac{R_0 \alpha I^2}{C} \\
F = \frac{R_0 I^2 - R_0 \alpha T_0 I^2 + \epsilon \sigma A T_A^4}{C} \\
$$
This can easily be decomposed into:
$$
\frac{1}{D T^4 + E T + F} \partial T = \partial t
$$
Thanks to @mattos I know that this is a Chini equation, for which the Chini invariant is independent of $t$:
$$
\frac{\partial y\left(t\right)}{\partial t} = f\left(t\right) y^n\left(t\right) + g\left(t\right) y\left(t\right) + h\left(t\right) \\
f\left(t\right)=D \\
n=4 \\
y\left(t\right)=T \\
g\left(t\right)=E \\
h\left(t\right)=F \\
C = f^{−n−1}\left(t\right) h^{−2n+1}\left(t\right) \left(f\left(t\right)\frac{\partial h\left(t\right)}{\partial t}−h\left(t\right)\frac{\partial f\left(t\right)}{\partial t}+n f\left(t\right) g\left(t\right) h\left(t\right)\right)^n n^{−n} \\
C = D^{−4−1} F^{−2*4+1} \left(D\frac{\partial F}{\partial t}−F\frac{\partial D}{\partial t}+4 D E F\right)^4 4^{−4} \\
C = D^{−5} F^{−7} \left(0 D−0 F+4 D E F\right)^4 4^{−4} \\
C = D^{−5} F^{−7} 4^4 D^4 E^4 F^4 4^{−4} \\
C = D^{−1} F^{−3} E^4 \\
C = \frac{E^4}{D F^3}
$$
Obviously $C$ is independent of $t$ and $n=4$.
And from there I have no idea how to continue.
Referenced, which do not really fit I found are:

*

*Solve $y' = x^4y+x^4y^4$

*Solve Radiation Total Energy Equation
Side information:

*

*Finally I need the function $ t\left(I, T_A\right)\text{ for which } T\left(t\right)=T_L $.

*The solution should be analytical to allow error calculations.

*The final formula has to be calculated in a low-power embedded system.

*In a first step, we could assume that $ \alpha = 0 $

*The parameters $R_0, \alpha, T_0, T_A, T_L, C \text{ and } \epsilon{}\sigma{}A$ have (later, independent of this question) to be calculated out of pairs of $t\left(I\right)=t_L$ and (one) $\lim_{I \to I_C+}t\left(I\right)=\infty$ with $T_A$ possibly pre-defined and possibly different between both.

P.S.: My math lectures are roughly 20 years ago, please feel free to improve my notation.
 A: You have to solve :
$$\frac{d T}{d t} = D\:T^4 + E\:T + F$$
This is an ODE of separable kind.
$$t=\int \frac{dT}{D\:T^4+E\:T+F}+C_0$$
$C_0$ is a constant.
Firstly, solve the algebraic equation $\quad D\:X^4+E\:X+F=0$ for X. 
This is a quartic equation. The analytic solution is complicated : http://mathworld.wolfram.com/QuarticEquation.html
But if you really want an explicit solution to your problem you cannot avoid to solve it.
Suppose that you does solve it you get the four roots, say $X_1,X_2,X_3,X_4$
More simply, solve numerically the quartic equation and you get the numerical values of  $X_1,X_2,X_3,X_4$.
Secondly, write the fraction on the form 
$$\frac{1}{D\:T^4+E\:T+F}=\frac{C_1}{T-X_1}+\frac{C_2}{T-X_2}+\frac{C_3}{T-X_3}+\frac{C_4}{T-X_4}$$
You have to compute $C_1,C_2,C_3,C_4$ in terms of $D,E,F,X_1,X_2,X_3,X_4$. Very boring job if you want an explicit solution. Much easier with numerical calculus if a numerical solution is sufficient.
At this point $C_1,C_2,C_3,C_4$ are known.
$$t=\int \left(\frac{C_1}{T-X_1}+\frac{C_2}{T-X_2}+\frac{C_3}{T-X_3}+\frac{C_4}{T-X_4}\right)dT+C_0$$
$$t=C_0+\ln\bigg( (T-X_1)^{C_1} (T-X_2)^{C_2} (T-X_3)^{C_3}(T-X_4)^{C_4}\bigg) $$
At this point you got $t(T)$, that is $t$ as a function of $T$ with an unknown parameter $C_0$ in it.
Thirdly, one have to determine $C_0$ according to the initial condition.
Fourthly, the major difficulty arises : Inverting the function $t(T)$ in order to obtain the function $T(t)$. 
Due to the exponents $C_1,C_2,C_3,C_4$ this cannot be analytically done, except for a few cases of particular values of $C_1,C_2,C_3,C_4$, for example $0$ and/or $1$. This is generally never the cases.
Finally we see that $T(t)$ can only be obtained on numerical form. As a consequence, instead of the arduous above calculus, on a practical viewpoint it is advised to directly solve the ODE $\frac{d T}{d t} = D\:T^4 + E\:T + F$ thanks to numerical method. Such methods are implemented in some math-softwares.
As a conclusion, don't expect an analytical solution for $T(t)$. 
An analytical solution is theoretically possible for the inverse function $t(T)$ but the formula would be awfully complicated if we want to write it on a full explicit form (without intermediate variables $X_1,X_2,X_3,X_4,C_1,C_2,C_3,C_4$ which each one is already a complicated formula).
