\[i_C+i_L=0\]
\[C\frac{dv_{OUT}}{dt}+\frac{1}{L}\int v_{out}=0\]
\[\frac{d^2v_{OUT}}{dt^2}+\frac{1}{LC}v_{OUT}=0\]
First, notice that the nodal equation is a second order homogeneous equation. Therefore, we will not need to find a particular solution.
As usual with problems like these, guess that the solution is $e^{st}$.
Plugging in $v_{OUT=}e^{st}$ into the nodal equation yields:
\[e^{st}\left ( s^2 + \frac{1}{LC} \right )\]
Solving the above equation for s gives:
\[s=\pm \sqrt{\frac{-1}{LC}}\]
\[s=\pm j\sqrt{\frac{1}{LC}}\]
So..
\[v_{out}=e^{j\sqrt{\frac{1}{LC}}t}\]
and
\[v_{out}=e^{-j\sqrt{\frac{1}{LC}}t}\]
There are two properties of homogeneous differential equations that are useful. A constant times the solution is still a solution, and the sum of solutions is still a solution.
Using those two properties we can write:
\[v_{out}=Ae^{j\sqrt{\frac{1}{LC}}t}+Ae^{-j\sqrt{\frac{1}{LC}}t}\]
Any good math student will recognize that is just,
\[v_{out}=A'cos \left [t\sqrt{\frac{1}{LC}} \right ]\]
At $t=0$, $v_{OUT}=v_{IN}$.
Finally we get,
\[v_{out}=v_{IN}cos \left [t\sqrt{\frac{1}{LC}} \right ]\]