 Solution

After division by $x^2$, the equation is $y^{\prime \prime }+x^{-1}{y}^{\prime }-(1+{\alpha }^2{x}^{-2})y=0$. Every $x\ne 0$ is an ordinary point. The origin is singular, but it is a regular singular point because $xP(x)=1$ and $x^{2}Q(x)=-(x^2+{\alpha }^2)$ are analytic there. An ordinary point has analytic normalized coefficients $P,Q$; a singular point failing the displayed regularity test is irregular.

The Frobenius method ansatz $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$ gives the indicial equation $r^2-{\alpha }^2=0$, $a_1=0$, andFor nonintegral $\alpha$, two independent solutions are

For integral $\alpha$, put $m=|\alpha |$. In the $r=-m$ recurrence the denominator vanishes at $n=2m$, so that series fails or coincides in the exceptional $m=0$ case. Up to scale the single Frobenius series isFor $\alpha =1$, $y_1=x(1+x^2/8+\cdots )$, soIts integral contains both $s^{-2}$ and $\log s$; multiplying by $y_1$ leaves a pole and a logarithmic term. Thus the reduction-of-order solution is not a power series at zero.