Worrisome result from pdetoode method in solving moving boundary condition problem

Question

The problem I am trying to solve is the following :

$$(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+1)\psi(x,t)=0$$

$$\psi (0,t)=\psi(L(t),t)=0$$

$$\psi(x,0)=\sqrt{\frac{1}{c\sqrt{1+\pi^2/4}}} sin(\pi \frac{x}{2})$$

$$\frac{d\psi (x,t)}{dt}\Bigr|_{t=0}=0$$

$$L(t)=2+sin(\omega t)$$

I have used the pdetoode method as was used here to solve this problem and here is my code (some codes might be redundant as I copied and modified the code from the link above) :

ydum = x/L[t];
eqn1 = 1/c^2 D[\[Psi][ydum, t], {t, 2}] - 
      D[\[Psi][ydum, t], {x, 2}] + (m^2 c^2)/h^2 \[Psi][ydum, t] == 
     0 /. \[Psi][ydum, t] -> \[Psi][y, t] /. x -> y L[t] // Expand
m = 1;
c = 1;
h = 1;
[Omega] = 1;
L[t_] := 2 + Sin[[Omega] t];
bc = {[Psi][0, t] == 0, [Psi][1, t] == 0};
ic = {[Psi][y, 0] == 
    Sqrt[2 (m c)/(c Sqrt[m^2 c^2 + [Pi]^2 h^2/L[0]^2] L[0])]
      Sin[ [Pi] y], D[[Psi][y, t], t] == 0 /. t -> 0};
points = 25;
xdifforder = 4;
{yL, yR} = domain = {0, 1};
grid = Array[# &, points, domain];
pto = pdetoode[[Psi][y, t], t, grid, xdifforder];
remove = #[[2 ;; -2]] &;
{ode, odebc, odeic} = MapAt[remove, pto /@ {eqn1, bc, ic}, 1];
soll = NDSolveValue[{ode, odebc, odeic[[2]], odeic[[1]]}, [Psi] /@ 
    grid, {t, 0, 50}];
sol1 = rebuild[#, grid, -1] &@soll
Manipulate[
 Plot[sol1[y, t] /. y -> y L[t], {y, 0, 1}, 
  PlotRange -> {-5, 10}], {t, 0, 50}]

The solution grows larger in time and this makes me anxious. Is this how it's supposed to be or is there something wrong with my code?

I have also solved this problem without the use of pdetoode (check out my previous question ) and I got different result :

I greatly appreciate anyone who would point out my error in both methods.