How to solve the inrush current?
Before the switch closes, the steady state voltage across the inductor is zero, i.e. U_L(0-)=0. Ohm's law tells you that I_R3=0 since it has the same voltage U_L. So I_L(0-) is simply I_q. This is the initial condition.

Your second schematic shows a set up after the switch is closed. One approach would be to simplify the schematic further by translating the voltage source U_q and the 3 resistors to the Thevenin equivalent. If you have done that, you get a simple series voltage-source, resistor and inductor circuit in a loop. The would look exactly like the circuit inside the Example linked, after the switch is closed.

The Example linked has a initial condition of I_L(0)=0. Remember that for this actual circuit, the initial condition is I_L(0)=I_q, use this to solve for the integration constant ("A" in the Example).