As you noted, this has to be a boost converter. But operating in continuous mode.
Qualitatively, I'd establish (-) terminal of \$E\$ as "ground reference" and then tend to assume the steady state process is something like this: The switch closes and applies voltage \$E\$ across inductor \$L\$ causing the current in \$L\$ to rise from \$I_L{min}\$ to \$I_L{max}\$. It also pulls \$C\$ to ground, causing \$C\$ to charge rapidly up through \$D_1\$ to voltage E. The switch then opens and forces inductor \$L\$ to reverse its voltage and to start declining its current from \$I_L{max}\$ down to \$I_L{min}\$. Capacitor \$C\$ now sits on top of \$E\$, plus \$V_L\$, making the output voltage developed on \$C_o\$ equal to \$2\cdot E + V_L\$. Current in this phase proceeds through \$D_2\$ to reach \$C_o\$.
Without more information, I can assume that in the continuous steady state it is true that \$I_L{min}=I_L{max}\$. Therefore, there is no change in current in \$L\$ and therefore no voltage developed across it. So the output voltage is twice the input voltage, in steady state.
There are issues, such as the fact that \$C\$ will be charged infinitely fast in the first phase and will discharge itself also infinitely fast in the second phase. And I've no idea the value of \$R\$, which will remove charge on \$C_o\$ that must be made up for by dumped charge from \$C\$ in each cycle. But this is all ideal stuff. So... I guess that's okay.