How can I understand that at t<0 the DC source is disconnected?
At time t&lt;0 the switch has been open for a long time, and the circuit looks like figure 7.19.  Note in that figure the source is connected in series with the 2 ohm resistor.  

Since the circuit has been like that for a long time it has reached steady-state.  There's no change in current through the inductor, and since V=L*di/dt (where di/dt is zero) the voltage across the inductor is zero.  The inductor can be replaced by a short for t&lt;0, as shown in 7.20(a).

Once the switch is closed at t=0 currents will now start to change and the inductor's response to that is now important.  However, since you solved the circuit at T just less than zero you know the initial condition for the current in the inductor.  

The effect of the source is removed from the right hand side of the circuit by the switch, which is a short across the source and its 2 ohm resistor. 5A will flow from the source through the 2A resistor, but that has no impact on the circuit to the right of the short, as shown in figure 7.20(b)