Consider the following ODE eigenproblem of $y(x)$ \begin{equation} y'' + [\varepsilon + b^2 x - (a + \frac{b^2}{2}x^2)^2 ] 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, although overall $y(x)$ decays well. So it's not clear to me whether this general solution helps the above question.