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I've found that the algebra of this differential equation

$$\frac{d^2y}{dz^2}-(3z^2+\gamma)\frac{dy}{dz}+(cz+\alpha)y=0$$

is in $sl(2)$ because it is possible to use the generators of the $sl(2)$ group $$J^+ =z\frac{d}{dz}-2jz, \quad J^0=z\frac{d}{dz}-j, \quad J^-=\frac{d}{dz}$$

to recover the aforementioned differential equation, if write it as the following combination I get the operator (note:$j$ represents the spin of a particle, and $\gamma, c$ and $\alpha$ $\in \mathbb{R}$):

$$a_{--}J^-J^- + b_{-}J^-+b_{+}J^+ = a_{--}\frac{d^2}{dz^2}+(b_{+}z^2+b_-)\frac{d}{dz}-b_+jz$$

where $a_{--},b_+, b_- \in \mathbb{R}$. I was wondering if I made a transformation of the original differential equation say, $y(z) = h(z)$ would be true to say that this differential equation also follows this algebra as well??

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Cross-posted from physics.stackexchange.com/q/97861/2451 –  Qmechanic Feb 5 at 14:31
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