Calculate the turn-off delay of an LED in parallel with a capacitor
You have your differential equation wrong.

In your equation (1) you state \$U_c(t) = U_c(\infty) - \left( U_{c}(\infty) - U_c(0)\right) e^{-\frac{t}{\tau}}\$.  This is the solution to a first-order linear differential equation.  It's probably one you found for a simple RC circuit.  Then you try to plug that equation into another differential equation, and realize you fouled it all up.

The correct relation is 


\$i_d = -i_c = -C \frac{dv_c}{dt}\$, 
\$i_d = I_{ss} \left( e^\frac{v_c}{n V_T} - 1\right)\$


Now you can do a simple algebraic substitution and get 
$$ C \frac{dv_c}{dt} = -I_{ss} \left( e^\frac{v_c}{n V_T} - 1\right) $$
or
$$ \frac{dv_c}{dt} = -\frac{I_{ss}}{C} \left( e^\frac{v_c}{n V_T} - 1\right) $$

This is a \$1^{st}\$-order nonlinear differential equation.  If you've forgotten your diff-eqs (or haven't studied them yet), it's not easily solvable (this is why we have simulation tools, after all).

We can persevere, however.  It's separable, by doing some algebraic shenanigans:

$$ -\frac{C}{I_{ss}}\frac{dv_c}{e^\frac{v_c}{n V_T} - 1} = dt $$

Integrate both sides:

$$ -\frac{C}{I_{ss}}\int{\frac{dv_c}{e^\frac{v_c}{n V_T} - 1}} = \int dt $$

Now, I see that \$-1\$ in the denominator, and I know two things: it's going to be trouble, and \$I_{ss}\$ is so small that \$e^\frac{v_c}{n V_T} \gg 1\$.  So I'm just going to approximate it away:

$$ -\frac{C}{I_{ss}}\int{\frac{dv_c}{e^\frac{v_c}{n V_T}}} = \int dt $$

This is easy to solve.  I integrate both sides:

$$ -\frac{C}{I_{ss}}\left(-n V_T e^{-\frac{v_c}{n V_T}} \right) = t + t_C $$

Do a bit of algebra, and take the log of both sides:

$$ -\frac{v_c}{n V_T} = \ln\left(\frac{I_{ss}}{n V_T C}(t + t_C)\right) $$

And, finally, after skipping enough steps that I am not going to get 100% on my final exam:

$$ v_c = -{n V_T}\ln\left(\frac{I_{ss}}{n V_T C}(t + t_C)\right) $$

This will hold true-ish until \$v_c\$ drops low enough that \$e^\frac{v_c}{n V_T} \gg 1\$ no longer holds -- but that's going to be long after the LED is no longer emitting light, and the diode equation isn't holding up anyway.  You just need to guess at \$n\$, find \$I_{ss}\$ from the LED current vs. voltage, and \$t_C\$ from the initial voltage.  And then do it all over again when the temperature changes, or you change LEDs.