Equilibrium points of a circuit with a diode
When you added \$I+II\$ you dropped the 2 \$i_L\$'s:
$$
I=5V-u_C-i_L=0\\
II=h(u_d)-i_L=0\\
I+II=5V-u_c+h(u_d)-2i_L=0
$$
If you are only interested in the steady-state solution, you can simplify the circuit a lot. In the steady-state, the inductor will behave like a short circuit and the capacitor will behave like an open circuit.