# Second Order Differential Equation with Polynomial Coefficient

Can someone suggest a way to solve the following differential equation $$$$(1+\frac{z^2}{ 2q})^3F''(z)+(\frac{3-4p}{2q}) (1+\frac{z^2}{2q})^2 z F'(z) +\epsilon F(z)=0$$$$ where $$-\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.

• What are $p$, $q$, and $\epsilon$? – Nick Feb 5 at 22:45

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 $$F(z)= c_1\,(2 q)^{2 p-\frac{3}{2}}\, z\, _2F_1\left(\frac{1}{2},\frac{3}{2}-2 p;\frac{3}{2};-\frac{z^2}{2 q}\right)+c_2$$
• 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 – o h Feb 8 at 20:44