Determining an equation for the current after the switches close
The initial conditions are \$\small v_C=2.5\:V\$ and \$\small i_L=\frac{5}{400}=12.5\:mA\$.
For \$\small t \ge 0\$, the same current, \$\small i\$, flows through \$\small L\$ and \$\small C\$. If this current is assumed to flow clockwise, KVL gives
$$\small 2.5-\frac{1}{C}\int i\:dt =L\frac{di}{dt}$$
Differentiating and rearranging gives the 2nd order ODE
$$\small \frac{d^2 i}{dt^2}+\frac{i}{LC}=0 $$
Solution is of the form \$\small i=Asin(\omega t+\phi)\$.
Substituting for \$\small i\$ in the ODE
$$\small -A\omega ^2 sin(\omega t+\phi)+\frac{Asin(\omega t+\phi)}{LC}=0$$
Hence \$\small \omega^2 =\frac{1}{LC}\$
Since the \$\small LC\$ circuit is has no resistance, the phase angle between current and voltage is known, hence the value of \$\small A\$ can be found from the initial conditions.