Can this differential equation be transformed into an hypergeometric equation? $$(1+x^2)y'' -4xy' + 6y = 0 $$
Can this be transformed into an hypergeometric equation of the form $x(1-x)y'' + (c - (a + b + 1)x)y' -aby = 0$?
I know that we can do the transform is the term before y'' a polynomial of degree 2, the term before y' of degree 1 and before y is a constant.  Another condition is that the polynomial term of degree two has two distincts roots.  Nothing is said about complex numbers, so I'm not sure.  $ 1 + x^2 $ has two complex roots.  Can I actually transform the equation into an hypergeometric equation?
Thank you.
 A: First, introduce $\xi = -i x$, and write $y(x) = \eta(\xi)$, then the quadratic in front of the second derivative has two real roots in $\xi$; the ODE changes to
\begin{equation}
 (1-\xi^2)\eta'' +4 \xi \eta' -6 \eta = 0.
\end{equation}
Then, you just have to rescale $\xi$ such that the parabola in front of the second derivative has roots at $0$ and $1$ instead of at $\pm 1$; you can do this by introducing $z = \frac{1+\xi}{2}$, and writing $\eta(\xi) = F(z)$. Then, the ODE transforms to
\begin{equation}
 z(1-z) F'' + (4 z-2) F' -6 F = 0. \tag{1}
\end{equation}
Note that $(1)$ can be solved in terms of elementary functions, yielding
\begin{equation}
 F(z) = c_1 (1-z)^3 + c_2 \left(\frac{1}{3}-z(1-z)\right).
\end{equation}
A: Try let $y=px^2+qx+r$ ,
Then $y'=2px+q$
$y''=2p$
$\therefore(1+x^2)2p-4x(2px+q)+6(px^2+qx+r)\equiv0$
$2qx+2p+6r\equiv0$
$\therefore\begin{cases}2q=0\\2p+6r=0\end{cases}$
$\begin{cases}q=0\\p=-3r\end{cases}$
$\therefore$ The ODE has one group of linear independent solution $y=C_1(3x^2-1)$
Try let $y=sx^3+tx^2+ux+v$ ,
Then $y'=3sx^2+2tx+u$
$y''=6sx+2t$
$\therefore(1+x^2)(6sx+2t)-4x(3sx^2+2tx+u)+6(sx^3+tx^2+ux+v)\equiv0$
$6sx+2t+6sx^3+2tx^2-12sx^3-8tx^2-4ux+6sx^3+6tx^2+6ux+6v\equiv0$
$(6s+2u)x+2t+6v\equiv0$
$\therefore\begin{cases}6s+2u=0\\2t+6v=0\end{cases}$
$\begin{cases}u=-3s\\t=-3v\end{cases}$
$\therefore$ The ODE has another group of linear independent solution $y=C_2(x^3-3x)$
$\therefore$ The general solution is $y=C_1(3x^2-1)+C_2(x^3-3x)$
