I did'nt understand why the solution says that V (0) = -30 V.

Also, I did not understand why the capacitor's voltage (Vc) is -30 V.

Isn't the voltage for t <0 (when the capacitor is loaded) the same as the voltage source?

The exercise asked for the circuit below i(t) for t>0 . The switch opens at t=0 after a long time closed

My attempt:

For t<0: V(0)= 30 V and i(0)= 15 A

For t>0:

α= R/(2L)= 2/(2*0,5)= 2

omega = 1/sqrt (LC) = 1/sqrt(0.5*0.25) = 2.82942712475

Under-damped:

di/dt= -2(15 cos(2t)+ B sin(2t))e^-2t + (-2*15 sin (2t) + 2 Bcos(2t))e^2t

di(0)/dt = -30+ 2B = -(1/L) [R*i(o)+ Vc(0)] = -2 [30+30] = -120

-30+2B=-120

B= -45

i(t)= 15 cos (2t) + (-45 sin (2t))e^-2t A

So, the answer isn't the same as the book's solution.