Estimate asymptotics of an ODE Consider the following ODE eigenproblem of $y(x)$
\begin{equation}
    y''  + \left[\varepsilon + b^2 x - \left(a + \frac{b^2}{2}x^2 \right)^2 \right] y=0
\end{equation}
with eigenvalue $\varepsilon$, real constants $a,b$. The boundary condition is $y(\pm\infty)=0$. Numerically, this turns out to have well-behaved eigensolutions.
My question is how to see the typical length scale of the eigensolution $y(x)$, i.e., how it asymptotically decays. For instance, if $y(x)\sim e^{-x^2/c^2}$, $c$ is the length scale I mean.

This ODE can also be shown to have the following general solution
\begin{equation}
    y(x)= \sum_{s=\pm} C_s\, e^{-arx_s - \frac{x_s^3}{2}} \mathscr{H}_\mathrm{T}(\alpha,\beta_s,\gamma,x_s)
\end{equation}
with integration constants $C_\pm$, $r=(\frac{3}{b^2})^{\frac{1}{3}}$,  $\alpha=r^2\varepsilon,\beta_\pm=\pm3,\gamma=2ra,x_\pm=\pm x/r$ and $\mathscr{H}_\mathrm{T}$ the triconfluent Heun's function. However, its asymptotics is not solely determined by the exponential factor, because $\mathscr{H}_\mathrm{T}$ is not truncated to be a finite polynomial for these $\beta$'s and actually diverges beyond the exponential suppression, although overall $y(x)$ decays well. So it's not clear to me whether this general form helps the above question.
 A: I will assume that $b \neq 0$. Write the equation in the form
$$
\frac{{d^2 y(x)}}{{dx^2 }} = \left( {\frac{{b^4 }}{4}x^4  + ab^2 x^2  - b^2 x + a^2  - \varepsilon } \right)y(x).
$$
Take
$$
f(x) = \frac{{b^4 }}{4}x^4  + ab^2 x^2  - b^2 x,\quad g(x) = a^2  - \varepsilon 
$$
and choose an interval $(x_*,+\infty)$ on which $f$ is positive. Applying the results in http://dlmf.nist.gov/2.7.iii with $a_1=x_*$ and $a_2=+\infty$, it is seen that there is a unique solution such that
$$
y(x) = \left( \frac{{b^4 }}{4}x^4  + ab^2 x^2  - b^2 x \right)^{ - 1/4} \exp \left( { - \int_{x_*}^x {\sqrt {\frac{b^4}{4}t^4  + ab^2t^2  - b^2 t} dt} } \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{x}} \right)} \right)
$$
as $x\to +\infty$. After some binomial expansions and elimination of unnecessary constant factors, we can see that there is a unique solution such that
$$
y(x) = \exp \left( { - \frac{b^2}{6}x^3  - ax} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{x}} \right)} \right)
$$
as $x\to +\infty$. You can do a similar analysis for an interval of the form $(-\infty,x_{**})$ on which $f$ is positive.
