Voltage of reverse-biased series diodes?
So, a common approximation for the current through a diode is (the Shockley diode equation) at voltage \$V\$ is:

$$ I = \left(e^{\frac{V}{nV_T}}-1\right) I_S$$

with \$V_T\$ being the thermal voltage (temperature-dependent), \$n\$ being a quality factor (device-dependent) and \$I_S\$ being the saturation voltage.

So; since we know that 

$$I_{D1}=I_{D2}$$
we can infer that

$$ \left(e^{\frac{V_1}{n_1V_{T1}}}-1\right) I_{S1}=\left(e^{\frac{V_2}{n_2V_{T2}}}-1\right) I_{S2}$$

Now normally, these diodes will not be at the same temperature, they wouldn't have the same quality factor, and thus, this equation would be underdefined, but with \$V_{T1}=V_T=V_{T2}\$ and \$n_1=n_2=n\$, as you imply in your comment with:


  I think we can suppose that the only difference between the diodes is the Is and everything else is identical! 


we get, also noting that \$V_2 = V_0 -V_1\$:

$$\begin{align}
\left(e^{\frac{V_1}{nV_{T}}}-1\right) I_{S1}&amp;=\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right) I_{S2}\\
%\implies\\
%\frac{e^{\frac{V_1}{nV_{T}}}-1}{e^{\frac{V_0-V_1}{nV_{T}}}-1} &amp;= \frac{I_{S2}}{I_{S1}}
\end{align}$$

Now, this equation is hard to solve analytically, but both sides are really easy to plot (use something like 30 mV for \$nV_T\$). Just find the intersection of these two curves!

Analytically, we can move forward with:
$$\begin{align}
\left(e^{\frac{V_1}{nV_{T}}}-1\right) I_{S1}&amp;=\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right) I_{S2}\\
&amp;=\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right) 50I_{S1}\\
\implies\\
e^{\frac{V_1}{nV_{T}}}-1&amp;=50e^{\frac{V_0-V_1}{nV_{T}}}-50\\
\implies\\
\ln \left(e^{\frac{V_1}{nV_{T}}}-1\right)&amp;=\ln\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right)+\ln 50\\
\end{align}$$