The Liouville transformation works as follows. Take the differential operator
\begin{equation}
L = \frac{d^2}{d x^2} + a_1(x) \frac{d}{d x} + a_2(x),
\end{equation}
where $a_1 \in C_1$ and $a_2 \in C_0$. Then, defining
\begin{equation}
A(x) = \text{exp}\left(-\frac{1}{2} \int^x a_1(\xi)d\xi\right)
\end{equation}
and
\begin{equation}
b(x) = -\frac{1}{2} \frac{d a_1}{d x} - \frac{1}{4} a_1^2 + a_2,
\end{equation}
we can transform the equation $L\,y(x) = f(x)$ into
\begin{equation}
A(x) \left[\frac{d^2}{d x^2} + b(x)\right] z(x) = f(x),
\end{equation}
where $y(x) = A(x) z(x)$.

Your equation is of the form $L y = f$ when you divide by the function multiplying $\alpha''$. You can then calculate
\begin{equation}
A(u) = (2\pi)^{-\frac{1}{4}}\left(\sigma\phi(u) + \phi(u/\sigma)\right)^{-\frac{1}{2}}.
\end{equation}
Calculating the function $b(u)$ is more involved, but should pose no problem (it's a little long to post here). Dividing by $A(x)$ yields
\begin{equation}
\left[\frac{d^2}{d x^2} + b(x)\right] z(x) = \frac{f(x)}{A(x)} \quad(*).
\end{equation}

So far, so good: $b(x)$ and $f(x)/A(x)$ are both smooth functions. However, the problem is that $b(x)$ is unbounded: it behaves as $-x^2$ for $|x|\gg1$. Note that $f(x)/A(x)$ decays superexponentially as $x \to \pm \infty$, so that's no problem.

Roughly speaking, that means that in the `far field' $|x|\gg1$, the solution $z(x)$ will behave like the solutions to the homogeneous equation
\begin{equation}
\frac{d^2 z}{d x^2} - x^2 z = 0,
\end{equation}
which in turn behave as a linear combination of $e^{\pm x^2}$. If their behaviour would *not* be superexponential, then we would have an exponential dichotomy. This would allow us to choose the two linearly independent solutions such that one decays as $x \to -\infty$ and the other one decays as $x \to +\infty$. Using the method of variation of parameters (or using Green's function), one could then formally express the solution to the inhomogeneous problem $(*)$ in terms of those functions, and investigate whether it would be possible for this inhomogeneous solution to be bounded.

However, for superexponential growth/decay, I'm not sure whether this works. You could always try to mimic the approach sketched above, but I'm not sure which problems you could encounter. I think that the expertise of @ChristianRemling will be very useful here, and I hope he is able to be of further assistance.