Catagorising a Differential Equation I have
$$
\frac{d^{2}}{d\epsilon^{2}}g^{\star}+\frac{\left(R^{2}+3\epsilon^{2}\right)}{\epsilon\left(R^{2}-\epsilon^{2}\right)}\frac{d}{d\epsilon}g^{\star}+\frac{\left(5R^{2}+3\epsilon^{2}\right)}{\left(R^{2}-\epsilon^{2}\right)^{2}}g^{\star}=0
$$
and I wonder if anyone could tell me some facts about it.
I know its linear, second order etc, that its Fuchsian (with its three regular singular points) and that it can be transformed (therefore) into the hypergeometric differential equation. Perhaps there's a DE expert who can tell me a bit more about its characterisation. I have (I think) both linearly independent solutions.
 A: Assume $R\neq0$ for the key case:
Let $r=\epsilon^2$ ,
Then $\dfrac{dg^\star}{d\epsilon}=\dfrac{dg^\star}{dr}\dfrac{dr}{d\epsilon}=2\epsilon\dfrac{dg^\star}{dr}$
$\dfrac{d^2g^\star}{d\epsilon^2}=\dfrac{d}{d\epsilon}\left(2\epsilon\dfrac{dg^\star}{dr}\right)=2\epsilon\dfrac{d}{d\epsilon}\left(\dfrac{dg^\star}{dr}\right)+2\dfrac{dg^\star}{dr}=2\epsilon\dfrac{d}{dr}\left(\dfrac{dg^\star}{dr}\right)\dfrac{dr}{d\epsilon}+2\dfrac{dg^\star}{dr}=2\epsilon\dfrac{d^2g^\star}{dr^2}2\epsilon+2\dfrac{dg^\star}{dr}=4\epsilon^2\dfrac{d^2g^\star}{dr^2}+2\dfrac{dg^\star}{dr}=4r\dfrac{d^2g^\star}{dr^2}+2\dfrac{dg^\star}{dr}$
$\therefore4r\dfrac{d^2g^\star}{dr^2}+2\dfrac{dg^\star}{dr}+\dfrac{2(R^2+3r)}{R^2-r}\dfrac{dg^\star}{dr}+\dfrac{5R^2+3r}{(R^2-r)^2}g^\star=0$
$\dfrac{d^2g^\star}{dr^2}+\left(\dfrac{1}{2r}-\dfrac{3r+R^2}{2r(r-R^2)}\right)\dfrac{dg^\star}{dr}+\dfrac{3r+5R^2}{4r(r-R^2)^2}g^\star=0$
$\dfrac{d^2g^\star}{dr^2}+\left(\dfrac{1}{r}-\dfrac{2}{r-R^2}\right)\dfrac{dg^\star}{dr}+\left(\dfrac{5}{4R^2r}-\dfrac{5}{4R^2(r-R^2)}+\dfrac{2}{(r-R^2)^2}\right)g^\star=0$
Let $g^\star=(r-R^2)u$ ,
Then $\dfrac{dg^\star}{dr}=(r-R^2)\dfrac{du}{dr}+u$
$\dfrac{d^2g^\star}{dr^2}=(r-R^2)\dfrac{d^2u}{dr^2}+\dfrac{du}{dr}+\dfrac{du}{dr}=(r-R^2)\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$
$\therefore(r-R^2)\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}+\left(\dfrac{1}{r}-\dfrac{2}{r-R^2}\right)\left((r-R^2)\dfrac{du}{dr}+u\right)+\left(\dfrac{5}{4R^2r}-\dfrac{5}{4R^2(r-R^2)}+\dfrac{2}{(r-R^2)^2}\right)(r-R^2)u=0$
$(r-R^2)\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}+\left(\dfrac{r-R^2}{r}-2\right)\dfrac{du}{dr}+\left(\dfrac{1}{r}-\dfrac{2}{r-R^2}\right)u+\left(\dfrac{5(r-R^2)}{4R^2r}-\dfrac{5}{4R^2}+\dfrac{2}{r-R^2}\right)u=0$
$(r-R^2)\dfrac{d^2u}{dr^2}+\dfrac{r-R^2}{r}\dfrac{du}{dr}-\dfrac{u}{4r}=0$
$r(r-R^2)\dfrac{d^2u}{dr^2}+(r-R^2)\dfrac{du}{dr}-\dfrac{u}{4}=0$
Which reduces to Gaussian hypergeometric equation.
