Laplace: iL(t) after switch opens, 4 V

Initial conditions (t = 0⁻, S closed, steady state)

Capacitor open, inductor short. The inductor branch carries no steady current (it ends at the open capacitor), so R₁ and R₂ form a divider from the 4 V source:

iL(0⁻) = 0
vc(0⁻) = 4 × R2/(R1 + R2) = 4 × 2/4 = 2 V

For t > 0 (S open)

The source and R₁ are disconnected. What remains is a source-free series loop of R₂, L and C (R₂ is the shunt resistor, which forms a loop with L and C):

R = 2 Ω, L = 0.5 H, C = 1 F, i_L(0) = 0, v_c(0) = 2 V.

Let i_L be the current from the R₂ node through L into C. KVL (the capacitor is a source of 2/s that drives current backwards):

(R + sL + 1/(sC)) IL = −vc(0)/s
(2 + 0.5 s + 1/s) IL = −2/s
IL(s) = −4 / (s² + 4s + 2)

Inverse transform

Roots of s² + 4s + 2: s = −2 ± √2, i.e. −0.586 and −3.414 (overdamped).

IL(s) = −4 / [(s + 0.586)(s + 3.414)]
      = −1.414/(s + 0.586) + 1.414/(s + 3.414)

i_L(t) = −1.414 (e^(−0.586t) − e^(−3.414t)) A, t > 0

The negative sign means the actual current flows backwards (from the capacitor through L and R₂), as the 2 V capacitor discharges. Its peak magnitude is about 0.81 A at t ≈ 0.62 s. i_L(0) = 0 ✓ and i_L′(0) = −4 A/s = −v_c(0)/L ✓. At t = 1 s the formula gives −0.741 A, the same as a numerical solution of the loop equations.