Second-Order RLC Circuit
It looks to me like the voltage source at \$t=0\$ is just supplying some charge to the capacitor, and the differential equation would just be the decaying transient response of the RLRC circuit with an initial capacitor voltage of 8V (v_c(0) = 8V).

So yes, at \$t -&gt; ∞\$, \$i(t) = 0\$; there would be no voltage drop across any component, there would be no current flowing through any component.  

By the definition of a capacitor:
$$ i_c(t) = C\frac{\mathrm{d}V_c(t)}{\mathrm{d}t}, $$
Since the voltage across C is the same as the voltage across R and RL:
$$ Vc(t) = R{i(t)} + L\frac{\mathrm{d}i(t)}{\mathrm{d}t} = Ri_r(t), $$
And since \$i_r(t) + i(t) = i_c(t) = \frac{v_c(t)}{R} + i(t)\$.

It has been a while since I have solved differential equations, but I think this simplifies to:
$$ i_c(t) = C\left[R\frac{\mathrm{d}i(t)}{\mathrm{d}t} + L\frac{\mathrm{d^2}i(t)}{\mathrm{d}t^2}\right]= {i(t)} + \frac{L}{R}\frac{\mathrm{d}i(t)}{\mathrm{d}t} + i(t) $$

$$ i_c(t) = 2\frac{\mathrm{d}i(t)}{\mathrm{d}t} + 2\frac{\mathrm{d^2}i(t)}{\mathrm{d}t^2} = {i(t)} + \frac{\mathrm{d}i(t)}{\mathrm{d}t} + i(t) $$

$$ \frac{\mathrm{d^2}i(t)}{\mathrm{d}t^2} + 0.5\frac{\mathrm{d}i(t)}{\mathrm{d}t} - i(t) = 0 $$