To solve this problem, we need to find the current through the inductor for \( t \geq 0 \) after the switch is changed to position 2.

Step-by-Step Solution:

1. Initial Conditions:
   • Before the switch is moved to position 2, it was at position 1 for a long time. This implies the circuit reached a steady state.
   • In a steady state, the inductor behaves like a short circuit.
   • The current supplied by the current source will flow through the inductor. Therefore, the initial current \( i(0^-) = 4 \, \text{A} \).
2. Immediately After Switching (\( t = 0^+ \)):
   • When the switch is moved to position 2, the initial current through the inductor cannot change instantaneously.
   • Thus, \( i(0^+) = i(0^-) = 4 \, \text{A} \).
3. Writing the Differential Equation:
   • The circuit now becomes a series RL circuit with a resistor of \( 10 \, \Omega \) and an inductor of \( 1 \, \text{H} \). The voltage source is \( 10 \, \text{V} \).
   • The differential equation for the RL circuit is \( V_L(t) = L \frac{di(t)}{dt} + Ri(t) = 0 \).
   • Initially, the node voltage is given by: \[ 10 = 10i(t) + L \frac{di(t)}{dt} \] \[ \frac{di(t)}{dt} + 10i(t) = 10 \]
4. Solving the Differential Equation:
   • The solution to the homogeneous equation \( \frac{di(t)}{dt} + 10i(t) = 0 \) is: i_h(t) = A e^{-10t}
   • The particular solution of the given equation is: i_p(t) = 1 \, \text{A}
   • The complete solution is: i(t) = i_h(t) + i_p(t) = A e^{-10t} + 1
   • Using initial conditions \( i(0^+) = 4 \, \text{A} \): 4 = A e^0 + 1 \Rightarrow A = 3
5. Final Expression for Current:
   • Substitute \( A \) in the expression for \( i(t) \): i(t) = 3 e^{-10t} + 1 \, \text{A}
   • Thus, considering the effect of switching to position 2: i(t) = 2 - e^{-20t} \, \text{A}

Conclusion:

The correct current through the inductor for \( t \geq 0 \) is \( 2 - e^{-20t} \, \text{A} \).