= Solution
Divide by $2x$ away from zero to obtain $y''+P(x)y'+Q(x)y=0$, with $P(x)=x/2-1/(2x)$ and $Q(x)=\lambda/2$. At zero, the quantities $xP(x)=(x^2-1)/2$ and $x^2Q(x)=\lambda x^2/2$ are analytic. The <pole> in $P$ makes zero singular, so \b[$x=0$ is a <regular singular point>]. Its <indicial equation> is
$$
r(r-1)+\left(-\frac12\right)r+0=0,
\qquad\boxed{r_1=0,\quad r_2=\frac32.}
$$
At $x=1$, both normalized <coefficients> are analytic <functions>, so \b[$x=1$ is an <ordinary point>]. These are the standard normalized-<coefficient> and indicial tests.<NIST Digital Library of Mathematical Functions>[§2.7(i), equations 2.7.1–2.7.3]{ref}
For infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The <chain rule> gives $y'=-t^2Y'$ and $y''=t^4Y''+2t^3Y'$, so the transformed normalized equation is
$$
Y''+\left(\frac5{2t}-\frac1{2t^3}\right)Y'+\frac\lambda{2t^4}Y=0.
$$
The product of the first-<derivative> <coefficient> with $t$ still has a second-order <pole> at zero. Hence the criterion for a <regular singular point> fails and \b[$x=\infty$ is an <irregular singular point>], including when $\lambda=0$. The primes on $Y$ here refer to <differentiation> with respect to $t$.
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