Classical: capacitor voltage, switch opened, 15 V and 60 V

Assumptions: the 15 V source (positive terminal up), the 1 H inductor, the 3 Ω resistor and the 5 Ω resistor are in series along the top wire; the 10 μF capacitor is connected from the 3 Ω/5 Ω junction to ground. At the right end, the switch connects a 60 V source (positive terminal up) to the 5 Ω resistor. Switch closed for a long time, then opened at t = 0.

Before t = 0 (dc steady state, switch closed)

The inductor is short and the capacitor is open. Net source voltage in the loop 15 V − 60 V with total resistance 8 Ω:

i_L(0⁻) = (15 − 60)/(3 + 5) = −5.625 A
v_C(0⁻) = 15 − 3(−5.625) = 31.875 V

After t = 0 (switch open)

The 5 Ω resistor and the 60 V source are disconnected. The remaining circuit is a series RLC circuit: 15 V, L = 1 H, R = 3 Ω, C = 10 μF.

L C d²v_C/dt² + R C dv_C/dt + v_C = 15
10⁻⁵ v″ + 3×10⁻⁵ v′ + v = 15

Complementary solution

s² + 3s + 10⁵ = 0
s = −1.5 ± j316.22      (underdamped: α = 1.5, ω_d = 316.22 rad/s)
v_c = e^(−1.5t)(A cos 316.22t + B sin 316.22t)

Particular solution: v_p = 15 V (the capacitor charges to 15 V finally).

Constants

v_C(0⁺) = 31.875  →  15 + A = 31.875  →  A = 16.875
dv_C/dt (0⁺) = i_L(0⁺)/C = −5.625/10⁻⁵ = −562 500 V/s
−1.5A + 316.22 B = −562 500
B = (−562 500 + 25.31)/316.22 = −1778.7

Result

v_C(t) = 15 + e^(−1.5t)(16.875 cos 316.22t − 1778.7 sin 316.22t) V,   t > 0

This was verified by numerical integration of the circuit equations (differences below 10⁻⁹ relative).

Answer: v_C(t) = 15 + e^(−1.5t)[16.875 cos(316.2t) − 1778.7 sin(316.2t)] V.