= Solution
\b[Singular points.] Normalizing the equation gives $p=2/x$ and $q=1-\ell(\ell+1)/x^2$. The only finite singular point is $x=0$, and it is regular because $xp$ and $x^2q$ are analytic there. At infinity put $t=1/x$ and $Y(t)=y(1/t)$; the equation becomes
$$
Y''+\left[\frac1{t^4}-\frac{\ell(\ell+1)}{t^2}\right]Y=0.
$$
The fourth-order pole makes $t=0$ an <irregular singular point>, so \b[infinity is irregular].
\b[Exponents and recurrence.] Substitution of a series for a <Frobenius solution> gives the <indicial equation>
$$
\sigma(\sigma+1)-\ell(\ell+1)=0,\qquad\boxed{\sigma=\ell\quad\text{or}\quad\sigma=-\ell-1.}
$$
With negative-index coefficients set to zero, the coefficient equation is
$$
\boxed{[(j+\sigma)(j+\sigma+1)-\ell(\ell+1)]a_j=-a_{j-2}.}
$$
For $\sigma=\ell$, it gives $a_1=0$ and
$$
a_j=-\frac{a_{j-2}}{j(j+2\ell+1)},\qquad j\geq2.
$$
For $\sigma=-\ell-1$, it gives
$$
a_j=-\frac{a_{j-2}}{j(j-2\ell-1)},\qquad j\geq1,\quad j\neq2\ell+1.
$$
For $\ell\geq1$, $a_1=0$. At the exceptional odd index $j=2\ell+1$, the recurrence becomes $a_{2\ell-1}=0$, which is already satisfied. Thus $a_{2\ell+1}$ is free; choosing it to be zero gives an independent pure-even series times $x^{-\ell-1}$. A nonzero choice adds a multiple of the regular solution. For $\ell=0$, this freedom occurs at $j=1$ itself. This <Frobenius resonance in the spherical Bessel equation> is compatible, so no logarithmic term is needed despite the integer separation of exponents.
\b[Normalized regular solution.] Set $\sigma=\ell$ and $a_0=2^\ell\ell!/(2\ell+1)!$. The proposed even coefficients
$$
a_{2s}=\frac{2^\ell(-1)^s(s+\ell)!}{s!(2s+2\ell+1)!}
$$
satisfy $a_{2s}/a_{2s-2}=-1/[2s(2s+2\ell+1)]$, precisely the recurrence above. Therefore
$$
\boxed{j_\ell(x)=2^\ell x^\ell\sum_{s=0}^\infty\frac{(-1)^s(s+\ell)!}{s!(2s+2\ell+1)!}x^{2s}.}
$$
This is the regular <Spherical Bessel function>. The coefficient ratio tends to zero, so its series converges for every finite $x$; the other even-series factor also has infinite radius, although the prefactor has a pole at zero.
For $\ell=0$, the coefficients reduce to $(-1)^s/(2s+1)!$, giving
$$
\boxed{j_0(x)=\frac{\sin x}{x},\qquad j_0(0)=1.}
$$
Directly, $x^2j_0'=x\cos x-\sin x$, whose <derivative> is $-x\sin x$. Thus $(x^2j_0')'+x^2j_0=0$, verifying the equation. The independent singular branch at $\ell=0$ can be chosen as $\cos x/x$.
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