I am trying to derive an equation for (RLC model) and i am stuck at one point how he get the final equation (12) second part?

Capacitor discharging in RLC series circuit. Equation (8) general solution for the discharging current,

i(t)=e^(-αt) (A cos⁡(ωt)+B sin⁡(ωt)) (8)

Now we will find the constant A and B using the initial condition assuming the capacitor voltage VDC and current IDC. The initial conditions are taken just before the fault occurs.

At t=0

i(0)=I_DC=A

Then the nominal voltage at the initial condition is equal to to the nominal voltage.

Vc(0)=V_DC

To find B take the derivative of the equation (8) we get

(di(t))/dt=e^(-αt) (-α(A cos⁡(ωt)+B sin⁡(ωt) )+ω(B cos⁡(ωt)-A sin⁡(ωt)) (9)

At t=0

(di(0))/dt=-αA+ωB

The initial rate of change of current is equal to

(di(0))/dt=(V_L (0))/L=-αA+ωB

Using

A=I_DC

B=((V_L (0))/L+αI_DC)/ω

Since the RLC is a series circuit and initial conditions are the same thus the current is also the same across R, L, and C.

(V_L (0))/L=I_L=I_DC

B=(I_DC+αA)/ω (10)

Putting the values of A and B in the equation (8) we get

i(t)=e^(-αt) (I_DC cos⁡(ωt)+(I_DC+αA)/ω sin⁡(ωt)) (11)

For the capacitor

1/C i(0)=V_c (0) and for initial condition i(0)=I_DC so 1/C I_DC=V_DC

“A” becomes equal to

A=I_DC=CV_DC

Expressing the equation in terms of the A and B from equation (10)

I_DC=Bω-Aα

So putting in equation (10)

i(t)=[(Bω-αA) cos⁡(ωt)-(Aω+Bα) sin⁡(ωt)]e^(-αt) (12)

[Aquatris]

This is kinda not related to Matlab though :)