How to solve or transform this 2nd-order linear ODE? Is it possible to solve the eigenvalue problem of $$y''(x) - 2\gamma\, y'(x) + [\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]\, y(x)=0$$ where $\lambda$ is the eigenvalue and $\alpha,\gamma$ are parameters. The boundary condition is $y(\pm\infty)=0$.
Or instead of the eigenvalue problem, can I just solve it with freely running $\lambda$? Then probably I can tackle the eigenproblem by imposing the boundary condition. Hopefully, it is related to or can be transformed into some classical form with special function?
 A: $y''(x)-2\gamma y'(x)+\left(\lambda^2+\gamma^2-\left(\dfrac{x^2}{2}+\alpha\right)^2+x\right)y(x)=0$
$y''(x)-2\gamma y'(x)-\left(\dfrac{x^4}{4}+\alpha x^2-x+\alpha^2-\lambda^2-\gamma^2\right)y(x)=0$
Let $y(x)=e^{nx^3}u(x)$ ,
Then $y'(x)=e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x)$
$y''(x)=e^{nx^3}u''(x)+3nx^2e^{nx^3}u'(x)+3nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)=e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)$
$\therefore e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)-2\gamma(e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x))-\left(\dfrac{x^4}{4}+\alpha x^2-x+\alpha^2-\lambda^2-\gamma^2\right)e^{nx^3}u(x)=0$
$u''(x)+(6nx^2-2\gamma)u'(x)+\left(\dfrac{(36n^2-1)x^4}{4}-(6\gamma n+\alpha)x^2+(6n+1)x-\alpha^2+\lambda^2+\gamma^2\right)u(x)=0$
Choose $36n^2-1=0$ , i.e. $n=-\dfrac{1}{6}$ , the ODE becomes
$u''(x)-(x^2+2\gamma)u'(x)-((\alpha-\gamma)x^2+\alpha^2-\lambda^2-\gamma^2)u(x)=0$
Let $u(x)=e^{kx}v(x)$ ,
Then $u'(x)=e^{kx}v'(x)+ke^{kx}v(x)$
$u''(x)=e^{kx}v''(x)+ke^{kx}v'(x)+ke^{kx}v'(x)+k^2e^{kx}v(x)=e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)$
$\therefore e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)-(x^2+2\gamma)(e^{kx}v'(x)+ke^{kx}v(x))-((\alpha-\gamma)x^2+\alpha^2-\lambda^2-\gamma^2)e^{kx}v(x)=0$
$v''(x)-(x^2+2\gamma-2k)v'(x)-((\alpha-\gamma+k)x^2+\alpha^2-\lambda^2-\gamma^2-k^2+2\gamma k)v(x)=0$
Choose $k=\gamma-\alpha$ , the ODE becomes
$v''(x)-(x^2+2\alpha)v'(x)+\lambda^2v(x)=0$
Which relates to Heun's Triconfluent Equation.
Alternatively, choose $n=\dfrac{1}{6}$ , the ODE becomes
$u''(x)+(x^2-2\gamma)u'(x)-((\alpha+\gamma)x^2-2x+\alpha^2-\lambda^2-\gamma^2)u(x)=0$
Choose another $k=\alpha+\gamma$ , the ODE simplify to
$v''(x)+(x^2+2\alpha)v'(x)+(2x+\lambda^2)v(x)=0$
Which relates to Heun's Triconfluent Equation.
A: Starting with
$$
   y''(x) - 2\gamma\, y'(x) + [\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]\, y(x)=0,
$$
let $y = e^{\gamma t} f$.  Then the above is reduced to potential form
$$
           (e^{\gamma t}f''+2\gamma e^{\gamma t}f'+\gamma^2e^{\gamma t}f)
      -2\gamma(e^{\gamma t}f'+\gamma e^{\gamma t}f)
      +[\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]e^{\gamma t}f=0 \\
      f''+(\lambda^2-(\frac{x^2}{2}+\alpha)^2+x)f=0.
$$
This may be written as an eigenfunction problem in standard potential form:
$$
                 -f''+\left((\frac{x^2}{2}+\alpha)^2-x\right)f=\lambda^2 f.
$$
