I didn't read your whole question, which seemed to go out of its way to make a simple thing complicated. As I understand it, you have a capacitor, resistor, and LED all in series, and you want to know how things decay if the capacitor starts out initially charged up.
At first apporoximation, you can consider the LED a voltage source. That means the current will decay just as if the LED wasn't there and the cap was charged up to the LED voltage less than what it really was. This is now a simple R-C systems which follows a basic exponential decay with a time constant of RC, which it seems you already understand. The question of when the LED goes "off" then comes down to at what current you consider the brightness to be low enough to be off. This can vary a lot by the efficiency of the LED, ambient light level, and how obvious "on" is supposed to be. For example, if the cap is initially charged so that the initial current is 20 mA (a common maximum for LEDs) and you consider 1 mA the "off" level, then the on time will be the 95% decay time, which is 3.0 time constants.
As I said, this was the basic first approximation where the LED has a fixed voltage accross it. That will be largely true, but of course its voltage will drop with current somewhat. For practicle purposes, this is a small effect compared to the slop of deciding what current level "off" really is, unless that current is small, like less than a mA.
That article is crap, and you should probably forget you ever saw it. Besides the fact that it's totally wrong, as you've discovered, it's full of half-truths that cultivate an incorrect understanding of how electrical phenomena actually work.
The plate on the capacitor that attaches to the negative terminal of the battery accepts electrons that the battery is producing.
wrong. Batteries don't produce electrons. In fact, nothing in your circuit produces electrons. As far as physicists have been able to demonstrate, charge is never created nor destroyed. A physicist can tell you how to make an electron by assembling more fundamental particles, but unless your circuit includes a particle accelerator or operates inside a star, there won't be any relevant electron creation or destruction in your circuit. Batteries pump electrons. They don't produce them.
In the first few paragraphs, that article uses the word charge in several senses. An experienced engineer can distinguish the senses by context, but the novice is more likely to confuse them. Go read Bill Beaty to immunize yourself against this misconception.
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacity.
Not exactly true. The bulb will go out once the voltage across the capacitor is equal to the battery voltage. When this happens, there can be no voltage across the bulb, and thus no current, thus no light. I don't know what they mean by capacity in this sense, but it seems to me that they are phrasing it this way to avoid explaining how capacitors actually work. You might define a capacitor's "capacity" any number of ways, but they haven't defined it at all. This kind of half-thinking isn't going to help you understand once you figure out that the explanation is incomplete.
A capacitor's storage potential, or capacitance, is measured in units called farads.
If you take "capacity" and "storage potential" as synonymous, as anyone would naturally do, this is wrong and self-contradictory. If this were true, then we could re-write the previous statement as
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacitance
...which is totally bogus. Ordinary capacitors don't change capacitance under normal operation conditions. Again, the problem here is they haven't fully defined the underlying concepts. If capacitance is a capacitor's storage potential, then what is it storing?.
A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt.
Here, they try to define the stuff that the capacitor is storing, but it hardly makes sense. The problem is if you use words like "capacity", this implies there is some sort of "full" or "at capacity". But, there is no concept for an ideal capacitor. If you push 1C of charge through a 1F capacitor, the capacitor will have a voltage of 1V. For 2C, you get 2V, and 1000C gets you 1000V. You can push 1000C through a 0.5F capacitor also, but it will then be at 2000V. An ideal capacitor is never full, and you can push charge through it forever and it will never be "full". That's not a capacity, it's the ratio of charge to voltage, which is evident in the definition of farad (a farad is a coulomb per volt):
$$ F = \frac{C}{V} $$
Real capacitors, incidentally, have a maximum voltage, which if exceeded will damage or destroy the capacitor. So in this sense, a capacitor can be "full" and can have a "capacity". But, this is not how this article is using the word.
I could probably write an entire article on the problems with that article, but hopefully you get the idea. Please wipe your memory of what that article has said, and seek a more sound explanation.
Best Answer
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
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.