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Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.

By the regular singular point at infinity criterion, infinity is regular singular precisely whenare holomorphic at $t=0$, but the ordinary-point conditions fail. Equivalently,with convergent expansions in $1/z$, excluding the ordinary case.

Fora finite point $z_0$ is an ordinary point exactly when $p$ and $q$ are holomorphic there. It is a regular singular point exactly whenare holomorphic at $z_0$.

Put $\zeta =1/z$ and $W(\zeta )=w(1/\zeta )$. Direct differentiation givesso the transformed equation isConsequently $z=\infty$ is ordinary precisely whenare holomorphic at $\zeta =0$. It is a regular singular point at infinity precisely when $zp(z)$ and $z^{2}q(z)$ are holomorphic functions of $1/z$ near infinity.

If zero and infinity are regular singular and every nonzero finite point is ordinary, the Laurent series of $p$ and $q$ can contain only the terms compatible with both endpoint bounds. Hencefor constants $a,b$. The equation is a Cauchy-Euler differential equation. Its indicial equation isFor distinct roots ${\lambda }_1,{\lambda }_2$, the general solution on a domain with a chosen logarithm branch isFor a repeated root $\lambda$, the general solution of an Euler-Cauchy equation is

Finally require infinity to be ordinary. In the transformed equation its coefficients becomeBoth are holomorphic at zero exactly when $a=2$ and $b=0$. Thus the further restriction is