Answer
See below
Work Step by Step
Given: $\frac{d^2q}{dt^2}+\frac{R}{L}\frac{dq}{dt}+\frac{1}{LC}q=\frac{1}{L}E(t)$
Rewrite as: $\frac{d^2q}{dt^2}+\frac{dq}{dt}+\frac{17}{4}q=\frac{E_0}{4}$
We obtain $A.E=m^2+m+\frac{17}{4}=0\\
\rightarrow 4m^2+4m+17=0\\
\rightarrow (2m+1)^2+16=0\\
\rightarrow 2m+1=\pm 4i\\
\rightarrow m=\frac{-1\pm 4i}{2}$
The complementary function is
$y_c(x)=c_1e^{-\frac{t}{2}}\cos 2t+c_2e^{-\frac{t}{2}}\sin 2t$
Place $q=A_0$
then $\frac{17}{4}A_0=\frac{E_0}{4}$
$A_0=\frac{E_0}{17}$
Hence $q=(c_1\cos 2t +c_2\sin 2t)e^{-\frac{t}{2}}+\frac{E_0}{4}$
Since $t \rightarrow \infty$ then $e^{-\frac{t}{2}} \rightarrow 0$
The current in the circuit $i=\frac{dq}{dt}=-\frac{1}{2}e^{-\frac{t}{2}}(c_1\cos 2t+c_2\sin 2t)+e^{-\frac{t}{2}}(-2c_1\sin2t+2c_2\cos2t)=e^{-\frac{t}{2}}(\frac{3}{2}\cos 2t-\frac{5}{2}\sin 4t)$
