How to solve this Sturm Liouville problem? $\dfrac{d^2\phi}{dx^2} + (\lambda - x^4)\phi = 0$
Would really appreciate a solution or a significant hint because I could find anything that's helpful in my textbook. Thanks!
 A: Hint:
Let $\phi=e^{ax^3}y$ ,
Then $\dfrac{d\phi}{dx}=e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}y$
$\dfrac{d^2\phi}{dx^2}=e^{ax^3}\dfrac{d^2y}{dx^2}+3ax^2e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y=e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y$
$\therefore e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y+(\lambda-x^4)e^{ax^3}y=0$
$\dfrac{d^2y}{dx^2}+6ax^2\dfrac{dy}{dx}+((9a^2-1)x^4+6ax+\lambda)y=0$
Choose $9a^2-1=0$ , i.e. $a=\dfrac{1}{3}$ , the ODE becomes
$\dfrac{d^2y}{dx^2}+2x^2\dfrac{dy}{dx}+(2x+\lambda)y=0$
Let $t=bx$ ,
Then $b^2\dfrac{d^2y}{dt^2}+\dfrac{2t^2}{b}\dfrac{dy}{dt}+\left(\dfrac{2t}{b}+\lambda\right)y=0$
$\dfrac{d^2y}{dt^2}+\dfrac{2t^2}{b^3}\dfrac{dy}{dt}+\left(\dfrac{2t}{b^3}+\dfrac{\lambda}{b^2}\right)y=0$
Choose $b^3=2$ , i.e. $b=\sqrt[3]2$ , the ODE becomes
$\dfrac{d^2y}{dt^2}+t^2\dfrac{dy}{dt}+\left(t+\dfrac{\lambda}{\sqrt[3]4}\right)y=0$
Which relates to Heun's Triconfluent Equation.
