Classical: capacitor voltage, switch opened, 25 V and 60 V

Assumptions: the 25 V source (+ up), 4 H inductor, 3 Ω and 15 Ω are in series along the top wire; the 3 F capacitor is a shunt from the 3 Ω/15 Ω junction to ground; at the right end a 60 V source (+ up) is connected through the switch to the 15 Ω resistor. The switch was closed for a long time and opens at t = 0.

Before t = 0 (dc steady state)

Inductor short, capacitor open; the 25 V and 60 V sources oppose each other through 3 + 15 = 18 Ω:

i_L(0⁻) = (25 − 60)/18 = −1.944 A
v_C(0⁻) = 25 − 3(−1.944) = 30.83 V

After t = 0 (switch open)

The 15 Ω and 60 V source are disconnected. The circuit becomes series RLC: 25 V, 4 H, 3 Ω, 3 F.

LC d²v/dt² + RC dv/dt + v = 25
12 v″ + 9 v′ + v = 25
s² + 0.75 s + 0.0833 = 0
s = [−0.75 ± √(0.5625 − 0.3333)]/2 = −0.1356, −0.6144

Real, distinct roots (overdamped).

Particular and complementary solution

v_p = 25 V, and

v_C(t) = 25 + K₁ e^(−0.1356t) + K₂ e^(−0.6144t)

Constants

v_C(0⁺) = 30.83:   K₁ + K₂ = 5.833
dv_C/dt(0⁺) = i_L(0⁺)/C = −1.944/3 = −0.648 V/s:
                   −0.1356 K₁ − 0.6144 K₂ = −0.648

Solving: K₁ = 6.132, K₂ = −0.2989.

Result

v_C(t) = 25 + 6.132 e^(−0.1356t) − 0.2989 e^(−0.6144t) V,   t > 0

Numerical integration of the circuit confirms this (for example v_C(5 s) = 28.098 V).

Answer: v_C(t) = 25 + 6.132e^(−0.1356t) − 0.299e^(−0.6144t) V.