Solve $y''(x)=[a(x^2-1)^2+b]y(x)$ While I was trying to solve a problem, I've found an equation like $y''(x)=[a(x^2-1)^2+b]y(x)$. I've tried everthing I know (like Riccati's algorithm and homogenous treatment, obtaining $u'(x)=a(x^2-1)^2+b-u(x)^2$). 
However, none of the methods works (I've also tried Wolfram Mathematica with both expressions). I guess the solution is related to some polynomials (like Hermite polynomials are related to simple quantum oscillator), but I don't know the name of this polynomials (maybe they have never been studied before).
Thanks
PS: I know that $y(x)$ must be symmetric ($y(x)=y(-x)$) and finite $y(\infty)\to 0$. I also know that the "ground state" of this function (is a quantum mechanics problem) is something like two gaussian distribution added together (with $\mu=\pm 1$). 
 A: Hint:
$y''(x)=(a(x^2-1)^2+b)y(x)$
$y''(x)-(ax^4-2ax^2+a+b)y(x)=0$
Let $y(x)=e^{nx^3}u(x)$ ,
Then $y'(x)=e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x)$
$y''(x)=e^{nx^3}u''(x)+3nx^2e^{nx^3}u'(x)+3nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)=e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)$
$\therefore e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)-(ax^4-2ax^2+a+b)e^{nx^3}u(x)=0$
$u''(x)+6nx^2u'(x)+((9n^2-a)x^4+2ax^2+6nx-a-b)u(x)=0$
Choose $9n^2-a=0$ , i.e. $n=\dfrac{\sqrt a}{3}$ , the ODE becomes
$u''(x)+2\sqrt ax^2u'(x)+(2ax^2+2\sqrt ax-a-b)u(x)=0$
Let $u(x)=e^{kx}v(x)$ ,
Then $u'(x)=e^{kx}v'(x)+ke^{kx}v(x)$
$u''(x)=e^{kx}v''(x)+ke^{kx}v'(x)+ke^{kx}v'(x)+k^2e^{kx}v(x)=e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)$
$\therefore e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)+2\sqrt ax^2(e^{kx}v'(x)+ke^{kx}v(x))+(2ax^2+2\sqrt ax-a-b)e^{kx}v(x)=0$
$v''(x)+2(\sqrt ax^2+k)v'(x)+((2\sqrt ak+2a)x^2+2\sqrt ax+k^2-a-b)v(x)=0$
Choose $k=-\sqrt a$ , the ODE becomes
$v''(x)+2\sqrt a(x^2-1)v'(x)+(2\sqrt ax-b)v(x)=0$
Which relates to Heun's Triconfluent Equation.
A: ran some numerical integration on this ... looks like the "period" of oscillations is decreasing exponentially while the envelope amplitude is exponentially increasing.

