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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 $x^{2}q$ are analytic there. At infinity put $t=1/x$ and $Y(t)=y(1/t)$; the equation becomesThe fourth-order pole makes $t=0$ an irregular singular point, so infinity is irregular.

Exponents and recurrence. Substitution of a series for a Frobenius solution gives the indicial equationWith negative-index coefficients set to zero, the coefficient equation isFor $\sigma =\ell$, it gives $a_1=0$ andFor $\sigma =-\ell -1$, it givesFor $\ell \ge 1$, $a_1=0$. 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.

Normalized regular solution. Set $\sigma =\ell$ and $a_0=2^{\ell }\ell !/(2\ell +1)!$. The proposed even coefficientssatisfy $a_{2s}/a_{2s-2}=-1/[2s(2s+2\ell +1)]$, precisely the recurrence above. ThereforeThis 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 $(-1{)}^s/(2s+1)!$, givingDirectly, $x^2{j}_0^{\prime }=x\cos x-\sin x$, whose derivative is $-x\sin x$. Thus $(x^2{j}_0^{\prime }{)}^{\prime }+x^2{j}_0=0$, verifying the equation. The independent singular branch at $\ell =0$ can be chosen as $\cos x/x$.