How to analize an ideal diode circuit with sinusoidal sources and storage elements (inductors and capacitors)?
This is the reason simulators were created, because even with the diode, only, it gets complicated.

Consider the case of a diode as a half-wave rectifier, driving an RL load (i.e. your picture without D2). For a mathematical analysis, the diode should be ideal. This would mean that for the 1st half of the period the diode is shorted out, and the circuit disconnected for the 2nd half. But since there is a reactive element in there, the current will not stop when the input voltage would drop to zero. Then, the inductor current will tend to flow the other way, forward-biasing the diode.

But first, let's analyze the circuit without the diode: a simple series RL. The equations would be:

$$L\frac{\mathrm{d}i(t)}{\mathrm{d}t}+R\,i(t)=0$$

with the solution:

$$i(t)=i(0)e^{-{L\over R}t}\tag{1}$$

To solve for \$i(t)\$:

$$\begin{align}
Z&amp;=\sqrt{R^2+\omega^2L^2} \\
L\frac{\mathrm{d}i(t)}{\mathrm{d}t}+R\,i(t)&amp;=V\sin(\omega t) \\
i_{\mathrm{steady}}(t)&amp;={V\over Z}\sin(\omega t-\phi)\tag{2} \\
\phi&amp;=\arctan{{\omega L\over R}} \\
i(0)&amp;={V\over Z}\sin(\phi)
\end{align}$$

So the expression of the total current would be \$(1)\$ plus \$(2)\$:

$$i(t)={V\over Z}\left[\sin(\omega t-\phi)+\sin{\phi}e^{-{R\over L}t}\right]\tag{3}$$

Plotting, side by side, the calculated current next to a SPICE simulation, they would agree:



If the diode is to be considered in the circuit, then the previous equation would only be valid for the first half of the period, plus the part where the diode is forward-biased by the induced voltage. At that point, the current is zero until the 2nd period starts, when the cycle continues. Shown below are the waveforms for the simple RL and for the RL+D:



So, until now, even with the diode, things look like they could be solved analytically fairly easy. But if you insert the D2 from your picture then the things complicate. Until now, there was only a part when the diode was on, and when it was off, and these were two states which could be separated and "stiched" to give the desired waveform. Now, with two diodes, there are 4 states, made by D1 on/off and D2 on/off. In each of these there are different things happening, each influencing the next:




in the 1st half of the period, the current rises according to expression \$(3)\$.
then, the induced voltage forward biases D2, not D1, which conducts until the next period, but it starts by continuing where D1 ended.
D1 again starts to conduct, but this time there are no more zero initial conditions, which means that the waveforms for the two previous states determine the values of the initial conditions for each of the next part.
etc


What remains, though, is the transient solution, \$(1)\$, which is visible in the slowly rising average value of the sum of the two currents, and the solution for the current through D1 (I(R1), green), but only for the first half of the period -- it can be seen that the waveforms coincide in that timespan.

The current through D2 can also be derived (similar to \$(3)\$) and calculated, but, as mentioned, the initial conditions are always changing, until steady-state is reached. So each next half-period has distinct solutions with initial conditions based on the previous half-periods.

At this point, even if I could derive the other formula (the same way, but the voltage has a displacement), I'd rather not because I'm hoping you can see why the simulators are used at this point. And, if you think the simulator decomposes the circuit into formulas, I'd have to dissappoint you, because it does nothing but compose a matrix of voltages, currents, resistances, conductances, what have you (simulator specific), then solves this numerically. It has no idea about complex operators, phasors, and the likes. It simply crunches numbers until a convergence is reached, at which point it declares the result satisfactory.