Let me solve this question from begining.

By KVL

E=L(di/dt)+(1/C)_{[ 0]}^{[t ]} tdt

diff once again

(d²i/dt²)=-i/(LC)

Sol of this problem is of the form

i=Asin(wt)+Bcos(wt) where w^2 = 1/(LC)

at t=0, i=0, therefore B=0

At t=0 di/dt = E/L

Aw=E/L So A = Esqrt(C/L)

 

a) At t= t_o=pi/(2w)

Q =₀^to Asin(wt)dt =-(A/w)[cos(wt_o)-cos(0)]

cos(wt_o) = cos(pi/2)=0

Q=A/w =CE

 

b) i(t₀) =Asin(pi/2) = A = Esqrt(C/L)

 

c) After t=t₀ lets define a new time such that closing of switch 2 is t=0

differential equation will become

(d²i/dt²)=-2i/(LC)

which will have a solution of form

i=Csin(w't)+Dcos(w't) where w'^2=2/(LC)

At t=0 i=Esqrt(C/L)

From last part it can be seen that di/dt=0 at this point

S0 C=0 and D= Esqrt(C/L)

So maximum current will remain same as earlier