Varying an action I am stuck varying an action.
This is the action $$S=\int\mathrm d^{4}x \frac{a^{2}(t)}{2}(\dot{h}^{2}-(\nabla h)^2)$$
And this is the solution, $\ddot{h} + 2 \frac{\dot{a}}{a}\dot{h} - \nabla^{2}h $.
This is what I get $$\partial_{0}(a^{2}\partial_{0}h)-\partial_{0}(a^{2}\nabla h)-\nabla(a^{2}\partial_{0}h)+\nabla^{2}(ha^{2})$$
I don't really see my mistake, perhaps i am missing something. (dot represents $\partial_{0}$)
It is this problem (see Lectures on the Theory of Cosmological
Perturbations, by Brandenburger)

$\quad$To quadratic order in the fluctuating fields, the action separates into separate terms involving $h_+$ and $h_x$. Each term is of the form $$S^{(2)}=\int\mathrm d^4x\dfrac{a^2}{2}\left[h'^2-(\nabla h)^2\right]\,,\tag{91}$$lading to the equation of motion 
  $$h_k''+2\dfrac{a'}{a}h_k'+k^2h_k=0\,.\tag{92}$$
  The variable in terms of which the action $\text{(91)}$ has canonical kinetic term is $$\mu_k\equiv ah_k\,,\tag{93}$$
  and its equation of motion is $$\mu_k''+\left(k^2+\dfrac{a''}{a}\right)\mu_k=0\,.\tag{94}$$

Going from step 91 to 92
 A: Varying the action we find the Euler-Lagrange equation of motion
$$\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h)} = \frac{\partial \mathcal{L}}{\partial h}$$
or
$$\frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial \dot h}
- \frac{\partial}{\partial x^i} \frac{\partial \mathcal{L}}{\partial (\partial_i h)}
= \frac{\partial \mathcal{L}}{\partial h},$$
where 
$$\mathcal{L} = \frac{1}{2} a(t)^2 \left(\dot h^2 - (\nabla h)^2\right)$$
is the Lagrangian (also called the Lagrangian density). 
Boundary terms can typically arise but are here ignored on physical grounds.
We find 
$$\begin{eqnarray*}
\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\partial \dot h} 
    &=& \frac{\partial}{\partial t}(a^2 \dot h) \\
    &=& a^2\ddot h + 2 a \dot a \dot h \\
\frac{\partial}{\partial x^i}\frac{\partial \mathcal{L}}{\partial (\partial_i h)}
    &=& \frac{\partial}{\partial x^i}(a^2 \partial_i h) \\
    &=& a^2 \nabla^2 h \\
\frac{\partial \mathcal{L}}{\partial h} 
    &=& 0.
\end{eqnarray*}$$
Therefore, 
$$a^2 \ddot h + 2 a\dot a \dot h - a^2 \nabla^2 h = 0.$$
Dividing by $a^2$ we find the claimed result, 
$$\ddot h + 2 \frac{\dot a}{a} \dot h - \nabla^2 h = 0.$$
