+a_2 h_1^2+a_3
\Bigl[ \dot h_1-h_0'+\frac{C'}{C}h_0 \Bigr]^2\,. \label{eq:Lagr}$$ We now introduce an auxiliary field q and define the following Lagrangian $$\begin{aligned}
\label{eq:Lagq}
L&=\Bigl[ a_1-\frac{(C' a_3)'}{C} \Bigr] h_0^2
+a_2 h_1^2+a_3
\Bigl[ 2q \left( {\dot h_1}- h_0'+\frac{C'}{C} h_0 \right) \nonumber\\
& -q^2\Bigr]\,.\end{aligned}$$ It is easy to see that both the Lagrangians give the same equations of motion, in fact, performing a variation with respect to $q$ gives $$\begin{aligned}
q={\dot h_1}- h_0'+\frac{C'}{C} h_0\,,\end{aligned}$$ which when replaced in (\[eq:Lagq\]) gives (\[eq:Lagr\]). Therefore, varying the Lagrangian (\[eq:Lagq\]) with respect to $h_0$ and $h_1$, leads to $$h_0=-\frac{\left( C a_3 q \right)'}{C a_1-\left( C' a_3 \right)'}\,,~~~h_1=\frac{a_3}{a_2} {\dot q}\,,\label{hhq}$$ respectively. Plugging this result back to (\[eq:Lagq\]), we obtain a quadratic Lagrangian solely in terms of the master variable $q$, which constitutes the only degree of freedom$-$the sole propagating odd mode. Any other function, such as $h_0$ and $h_1$, can be determined once $q$ is known using the relations (\[hhq\]). Solving few integration by parts, we arrive to a canonical form of the action $$\begin{aligned}
\label{eq:final2}
S^{(2)}=\int {\rm d} t\,{\rm d} r\,\Bigl[\alpha \dot q^2+\beta q'^2+\gamma q^2 \Bigr]\,,\end{aligned}$$ where $$\begin{aligned}
\alpha&=\frac{l(l+1)}{4(l-1)(l+2)}\sqrt{\frac{B}{A}}\frac{C}{A}\frac{{\cal H}^2}{{\cal G}}\,,\\
\beta&=-\frac{l(l+1)}{4(l-1)(l+2)}\sqrt{\frac{B}{A}}BC\frac{{\cal H}^2}{{\cal F}}\,,\end{aligned}$$ while $\gamma$ has a longer expression and will not be necessary at this level. In 