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Can someone suggest a way to solve the following differential equation \begin{equation} (1+\frac{z^2}{ 2q})^3F''(z)+(\frac{3-4p}{2q}) (1+\frac{z^2}{2q})^2 z F'(z) +\epsilon F(z)=0 \end{equation} where $-\infty<z<+\infty$. The solution $F$ when multiplied by $(1+\frac{z^2}{2q})^{-p}$ is expected to converge to zero when $z$ goes to infinity. In the present form this equation does not admit a polynomial solutions. Any suggestion is appreciated.

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  • $\begingroup$ What are $p$, $q$, and $\epsilon$? $\endgroup$ – Nick Feb 5 at 22:45
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The only case for which I have been able to find a solution corresponds to $\epsilon=0$.

Reducing the order, we have $$F'(z)=c_1 \left(2 q+z^2\right)^{\frac{4 p-3}{2} }$$ from which $$F(z)= c_1\,(2 q)^{2 p-\frac{3}{2}}\, z\, _2F_1\left(\frac{1}{2},\frac{3}{2}-2 p;\frac{3}{2};-\frac{z^2}{2 q}\right)+c_2$$

Could the problem be relevant of perturbation theories ?

In any manner, without boundary conditions, I do not see what else I could do.

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  • $\begingroup$ Actually this equation arise from a problem in quantum mechanics. The case $(p=q)\rightarrow \infty$ yields the Hermite equation for which polynomial solutions exist for $\epsilon=2n$. For arbitrary $p$ and $q$ still one expects a discreet set of values for $\epsilon$ (much like Hermite equation) but I don't see how that naturally comes about $\endgroup$ – o h Feb 8 at 20:44

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