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Putting $Y=e^{z}V$ in the displayed Kummer differential equation gives $z{V}^{\prime \prime }+(z+b)V^{\prime }=0$. Thus $V^{\prime }=C{z}^{-b}{e}^{-z}$ on a chosen branch, yielding $Y=e^z[A+B\int z^{-b}{e}^{-z}dz]$. For $b=1/2$ and $z=x/2$, one solution is $e^{x/2}$ and a normalized second Frobenius solution is $\sqrt{x}{\sum }_{n\ge 0}{x}^n/(2n+1)!!$. The analytic factors in both solutions have infinite radius of convergence, since their coefficient ratios tend to zero. The second solution still has a square-root branch at zero; an entire series factor does not remove a nonintegral Frobenius exponent.

For a real pair of indicial roots $r_1\ge r_2$ at a regular singular point of a second-order linear differential equation, the second local solution is classified by the gap $\Delta$ and by the recurrence relation for the smaller indicial root. Case I has $\Delta \notin \mathbb{Z}$ and two pure Frobenius solutions. Case II has $\Delta =0$ and an independent Logarithmic Frobenius solution. With a positive integer gap, Case IIIa means the exceptional coefficient equation is inconsistent, so a nonzero multiple of the first Frobenius solution times a logarithm is necessary. Case IIIb means that equation is consistent and leaves a free coefficient, so there are two pure Frobenius solutions. This IIIa/IIIb convention is the one used in the Oxford course notes.^[ref]

The free coefficient in Case IIIb adds a multiple of the first Frobenius solution, as explained by the undetermined coefficient at Frobenius resonance. Stating whether the logarithm occurs is useful alongside the case name because other courses may choose different labels.

For $a>0$, $b\ne 0$, and a positive integer $\beta$, consider $x^2{y}^{\prime \prime }+x{y}^{\prime }+(-a^2+b{x}^{\beta })y=0$ near $x=0$. Its indicial roots are $a,-a$. The first Frobenius solution, normalized to leading coefficient one, is $y_{+}=x^a{\sum }_{k\ge 0}{c}_{k\beta }{x}^{k\beta }$, with $c_{k\beta }=-b{c}_{(k-1)\beta }/[k\beta (k\beta +2a)]$. If $N=2a$ is a positive integer, a second pure Frobenius solution with leading coefficient one would requireBefore $n=N$, induction makes $d_n=0$ for indices not divisible by $\beta$. If $\beta$ divides $N$, the coefficient $d_{N-\beta }$ is nonzero, and the exceptional equation $0=-b{d}_{N-\beta }$ is impossible. If $\beta$ does not divide $N$, the right-hand side is zero and $d_N$ is free. Thus, in the Frobenius case classification, a positive integer gap gives Case IIIa exactly when $\beta$ divides $2a$, and Case IIIb otherwise.

The coefficient of the necessary logarithm can also be calculated. Let $L=x^2{d}^2/d{x}^2+xd/dx-a^2+b{x}^{\beta }$. The product rule gives $L(y_{+}\log x)=2x{y}_{+}^{\prime }$. Therefore a second solution $x^{-a}\sum d_n{x}^n+K{y}_{+}\log x$, with $d_0=1$, hasFor $N=m\beta$ the preceding recurrence relation gives $d_{(m-1)\beta }=b^{m-1}/[{\beta }^{2m-2}((m-1)!{)}^2]$, soThis proves that the logarithm cannot disappear for a special nonzero value of $b$. For $a=0$, reduction of order instead produces the repeated-root Logarithmic Frobenius solution directly.

A Frobenius solution $y={\sum }_{k\ge 0}{a}_k{x}^{k+\sigma }$ has the repeated indicial root $\sigma =0$. Equating coefficients gives the displayed recurrence relation. The resulting analytic solution is an entire function: its consecutive term ratio has magnitude asymptotic to $|x|/k$. For $\nu =n\ge 0$ an integer, the recurrence terminates and $L_n(x)={\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})x^k/k!$. The second local solution is independent and has a logarithmic term at zero; repeated indicial roots do not supply two independent analytic series.

For the Gauss hypergeometric equation with parameters $\alpha ,\beta ,\gamma$, a Frobenius solution $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, has indicial roots $\sigma =0$ and $\sigma =1-\gamma$. Matching each successive power gives the displayed undivided recurrence. The usual ratio formula is valid only when its left coefficient is nonzero. At integer-separated roots, a zero coefficient requires a compatibility check as in an undetermined coefficient at Frobenius resonance; otherwise a logarithmic companion may be necessary. For example, $\gamma =0,\sigma =0$ gives $0=\alpha \beta a_0$ at $n=1$, obstructing that exponent-zero series when $\alpha \beta \ne 0$.

An indicial equation determines candidate leading powers in a Frobenius solution near a regular singular point. Fornear $x=0$, suppose $p(x)=xP(x)$ and $q(x)=x^{2}Q(x)$ extend analytically to zero. Substituting $y=x^r(a_0+a_{1}x+\cdots )$ with $a_0\ne 0$ and collecting the lowest power givesIts roots are the indicial exponents. Subsequent powers determine the coefficient recurrence; the leading polynomial alone does not decide whether the second solution contains a logarithm.

For the Cauchy-Euler equation $x^2{y}^{\prime \prime }+x{y}^{\prime }+{\mu }^{2}y=0$, $I(r)=r^2+{\mu }^2$. If $\mu >0$, its complex roots $\pm i\mu$ give real solutions $\cos (\mu \log x)$ and $\sin (\mu \log x)$ on $x>0$, exhibiting log-periodic oscillation.^[ref]

An indicial root is a root of the indicial equation for a regular singular point. For a leading term $x^r$ as $x\downarrow 0$, its magnitude is $x^{\mathrm{Re}r}$, so positive real part gives decay and negative real part gives divergence. At zero real part, logarithmic factors and oscillation must also be checked.

For $x^2{y}^{\prime \prime }+x{y}^{\prime }=0$, the repeated root is zero. Its independent solutions $1$ and $\log x$ have the same indicial root but different boundedness. For roots $r=\pm i\mu$, nonzero real solutions oscillate in $\log x$ with bounded magnitude and usually no limit at zero. Thus the sign of a real root is useful away from the borderline, but is not a complete boundedness rule for all Frobenius solutions.

Substitute a Frobenius series $y={\sum }_{n\ge 0}{a}_n{z}^{n+\rho }$ into the transformed equation. For $n\ge 1$, equating the coefficient of $z^{n+\rho -1}$ gives the recurrence relationAt the larger indicial root $\rho =2$, the coefficient of $a_n$ is $n(n+2)\ne 0$, so a Frobenius solution always exists. At the smaller indicial root $\rho =0$, the first two equations areSince $a_0\ne 0$, the exceptional equation is consistent exactly whenThese are both admissible values, including the zero value; the singular wording of the question does not remove either. This is the undetermined coefficient at Frobenius resonance: at either value $a_2$ is free, whereas otherwise the second independent solution requires a Logarithmic Frobenius solution.

Explicit pairs demonstrate sufficiency and linear independence. In the variable $z=x+1$ one may takeBoth displayed $u_2$ have leading term $z^2/2$, while $u_0(0)=1$. They therefore give independent Frobenius solutions with the two indicial exponents, and direct differentiation verifies the equation. The point is an apparent regular singular point in these two cases.

For the requested closed forms, particularly short choices areIn addition, $(x+2)e^{-x}$ solves the zero-parameter equation and $e^{-x}$ solves the equation with $c=-1$.

Move the expansion point to the origin with $z=x+1$. The transformed ordinary differential equation isAfter division by $z$, the functions $z(1-1/z)=z-1$ and $z^2(c/z)=cz$ are analytic functions at zero, so this is a regular singular point. For a Frobenius solution with leading term $a_0{z}^{\rho }$, $a_0\ne 0$, the lowest power is $z^{\rho -1}$ and its coefficient isTherefore the indicial equation and its indicial exponents areThe exponent $\rho$ denotes the quantity called $\alpha$ in the paper, and avoids confusion with other uses of that letter.

For the first Frobenius solution, take the larger indicial root $r=a$ and normalize the first coefficient to one:Substitution and comparison of the coefficient of $x^{n+a}$ giveEvery denominator is nonzero because $a\ge 0$.

For $\beta =1$, the recurrence relation is $c_n=-b{c}_{n-1}/[n(n+2a)]$, soFor $\beta =2$, $c_1=0$ and all odd coefficients vanish; the even coefficients satisfy $c_{2k}=-b{c}_{2k-2}/[2k(2k+2a)]$. HenceFor any positive integer $\beta$, induction in the recurrence relation shows that only indices divisible by $\beta$ survive:The empty product for $k=0$ equals one. Successive terms in the power series factor have ratio of magnitudesThus the ratio test gives convergence for every finite $x$; a nonintegral $a$ still requires a chosen side of zero or a complex branch for the factor $x^a$.

For a putative second Frobenius solution with the smaller indicial root $-a$, set $y_{-}=x^{-a}{\sum }_{n\ge 0}{d}_n{x}^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its recurrence relation isThe Frobenius case classification concerns the gap $2a$ between the indicial roots. In the course convention, Case IIIa is the obstructed recurrence requiring a Logarithmic Frobenius solution; Case IIIb permits two pure Frobenius series.

If $a>0$ and $2a\notin \mathbb{Z}$, this is Case I: no denominator vanishes. There are two linearly independent pure Frobenius solutions, withTheir different leading powers establish linear independence.

If $a=0$, this is Case II, with a repeated indicial root. There is only one pure Frobenius series with a nonzero leading coefficient. To see the second solution's form, reduction of order givesHere $y_{+}(t)=1+O(t^{\beta })$, so the integrand is $t^{-1}+O(t^{\beta -1})$. ConsequentlyThe logarithm has a nonzero coefficient and cannot be removed by adding a multiple of $y_{+}$.

If $N=2a$ is a positive integer, the only potentially obstructed coefficient is $n=N$. Below that index, $d_n=0$ unless $\beta$ divides $n$. At the exceptional index the equation readsIf $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_{\beta },\dots ,d_{(m-1)\beta }$ are nonzero because $b\ne 0$. The exceptional equation is inconsistent. This is Case IIIa, and the independent second solution has the formThe Frobenius resonance for an Euler equation with monomial potential explains why divisibility, rather than the integer gap alone, decides this outcome. For example, with $d_0=1$, applying the original differential operator to $y_{+}\log x$ gives $2x{y}_{+}^{\prime }$. Cancellation of the coefficient of $x^a$ requires $K=-b{d}_{N-\beta }/(2a)\ne 0$.

If $\beta$ does not divide $N$, then $d_{N-\beta }=0$; this includes $N<\beta$ through the negative-index convention. The exceptional equation is $0=0$, so $d_N$ is free. This is Case IIIb: set $d_N=0$ to select an independent pure Frobenius series starting with $x^{-a}$. Subsequent coefficients again follow the recurrence relation. Changing $d_N$ simply adds a multiple of $y_{+}$, the undetermined coefficient at Frobenius resonance.

The complete answer is thereforeIn particular, for $\beta =1$ every positive integer gap gives Case IIIa. For $\beta =2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\dots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne 0$ matters: at $b=0$ the two Cauchy-Euler equation powers already give independent solutions whenever $a>0$.

Normalize both leading coefficients to one. For the indicial root $a=1/2$, the recurrence relation from part (b) givesRecognizing the Taylor series of the sine gives the first Frobenius solutionFor the smaller indicial root $-1/2$, $n(n-1)d_n=-d_{n-2}$. The equation at $n=1$ is $0=0$; choose the free coefficient $d_1=0$ to avoid adding the first Frobenius solution. The remaining coefficients areThus the Taylor series of the cosine givesTheir leading terms are $x^{1/2}$ and $x^{-1/2}$, as required for Case IIIb. A direct check uses $y=x^{-1/2}u$: the differential equation becomes $x^{3/2}(u^{\prime \prime }+u)=0$, so these two functions solve it on $x>0$. Their Wronskian is $-1/x\ne 0$, proving linear independence. The general solution is $(C_1\sin x+C_2\cos x)/\sqrt{x}$.

Treat $F$ as a vector-valued solution; matrix solutions are handled columnwise. It is important to distinguish the coefficient convention from the indicial exponent. In a fixed Fuchsian frame, a finite singularity at $a$ meanswhere $B_a$ is holomorphic near $a$. The residue is $R_a$, equivalently the residue of the matrix-valued one-form $A(z)dz$. Such a Fuchsian linear differential system has a regular singular point: on sectors, its solutions have at most power and logarithmic growth. Intrinsically a regular singular system can be made Fuchsian by a meromorphic change of frame; a higher coefficient pole in some other frame does not itself prove irregularity. Residues here refer to the displayed fixed coefficient one-form.

At infinity use $w=1/z$ and $\tilde{F}(w)=F(1/w)$. The equation becomesA Fuchsian singularity at infinity means $\tilde{A}$ has at most a simple pole at $w=0$, and $R_{\infty }={\mathrm{Res}}_{w=0}\tilde{A}(w)$. The residue theorem applied entrywise on the Riemann sphere givesFor a direct calculation when all coefficient poles are simple, write $A(z)=P(z)+{\sum }_j{R}_j/(z-a_j)$ by rational partial fractions. Regularity at infinity in this frame requires $A(z)=O(1/z)$, so $P=0$. Transforming coordinates gives $R_{\infty }=-{\sum }_j{R}_j$. More generally the entrywise residue theorem remains valid even when a different meromorphic frame has higher-order poles.

The printed nonconstant-solution claim is false. A counterexample satisfying its spectral hypothesis isIts residue has eigenvalues $0,-1/3$, with zero of largest real part. The equations are $u^{\prime }=0$ and $v^{\prime }-v/(3z)=0$. If $v$ is analytic at zero, coefficient comparison gives $(n-1/3)v_n=0$ for every nonnegative integer $n$, so $v=0$. Hence every analytic solution is $(c,0{)}^T$, which is constant. This disproves nonconstancy, rather than silently replacing it with a weaker conclusion.

A useful correct conclusion is that a nonzero analytic solution exists under the stated hypotheses. Write $A=R/z+{\sum }_{j\ge 0}{B}_j{z}^j$, and let $m$ be the largest nonnegative integer with $-m\in \mathrm{spec}R$; such an $m$ exists because $0\in \mathrm{spec}R$. Choose a nonzero $h_0\in \ker (mI+R)$ and seekThe Frobenius method givesAll left-hand matrices are invertible by maximality of $m$. For large $n$ their inverse norm is $O(1/n)$ by a geometric-series inverse; enlarging the constant bounds every $n\ge 1$ by $C/n$. If $\Vert B_j\Vert \le M{r}^{-j}$ on a smaller disc, the scalar majorant with generating functionsatisfies $v^{\prime }=CM(1-t/r{)}^{-1}v$ and dominates the recurrence coefficientwise. It converges for $|t|<r$, proving convergence of $H$. Thus the analytic solution at a maximal integer Fuchsian exponent is nonzero, and is nonconstant if $m>0$. When $m=0$, nonconstancy requires additional information, as the counterexample shows.

There is also a sign problem in the printed exponent terminology. Substitution of $F=z^{\mu }H$ with $H(0)\ne 0$ into the actual equation givesThus the indicial exponents are eigenvalues of $-R$, not $R$. With the printed sign, an exponent of maximal real part admits a pure Frobenius solution, because the corresponding coefficient matrices are invertible at all subsequent positive integer steps. Lesser exponents can meet resonance.

The supplied example does not demonstrate such failure. Solving it directly givesEvery solution is analytic. The residue eigenvalue $-1$ corresponds to the true exponent $1$, and setting $c_1=0$ gives $F=z(-c_{2}z/2,c_2{)}^T$ with nonzero leading coefficient. Even with the literal printed factor $z^{-1}$, every displayed solution has the form $z^{-1}H$ with $H=zF$ analytic, though $H(0)=0$.

Nevertheless the last existence-of-a-counterexample request, read literally, can be fulfilled by the diagonal counterexample above. For its nonmaximal residue eigenvalue $\lambda =-1/3$, substitution of $F=z^{-1/3}H$ yieldsfor every integer $n\ge 0$. Neither diagonal entry, $n-1/3$ or $n-2/3$, vanishes, so $H=0$. Therefore there is no nonzero solution of that literal form, even without imposing $H(0)\ne 0$.

For the intended logarithmic resonance phenomenon with correctly signed exponents, take insteadIts equation gives $v=c_2/z$ and $u=c_1+c_2\mathrm{Log}z$ on a chosen branch. The exponent $-1$ would require $H(0)=(0,c_2{)}^T\ne 0$, but $z\mathrm{Log}z$ prevents $H=zF$ from being analytic when $c_2\ne 0$. The exponent zero has the constant solution $(1,0{)}^T$. This supplies an explicit resonance obstruction under the correct convention and completes the qualified resolution of the printed defects.

For two solutions $y_1,y_2$, define the WronskianDifferentiate and substitute the second-order linear differential equation:Integration proves the Abel identity, including the identically zero case:For $p=-2/x$, this is $W=C{x}^2$ on either interval excluding zero, since the integral uses $\log |x|$. The coefficient singularity at zero must not be ignored.

Multiply the particular equation by $x^2$ and substitute the Frobenius solution $y=x^{\lambda }{\sum }_{n\ge 0}{a}_n{x}^n$. With $a_{-1}=a_{-2}=0$, equating coefficients givesThe leading coefficient is nonzero, so the indicial equation is $\lambda (\lambda -3)=0$. Thus only $\lambda =0,3$ are possible, and the recurrences below construct both.

For $\lambda =0$, $a_1=0$, $a_2=-a_0/2$, and the $n=3$ recurrence reads $0=a_1=0$, leaving $a_3$ free. For all later indices,This is an undetermined coefficient at Frobenius resonance; choosing $a_3=0$ selects a convenient even basis solution but is not forced by the differential equation.

For $\lambda =3$, all odd-indexed coefficients vanish andThe factorial denominators show convergence for every finite $x$. Expanding the given hyperbolic functions directly yieldsConsequently the exponent-zero family is $a_0(\cosh x-x\sinh x)-3a_3(\sinh x-x\cosh x)$, and the normalized exponent-three solution is $-3a_0(\sinh x-x\cosh x)$. This verifies the coefficient formulas and supplies two independent basis functions.

Finally $y_1^{\prime }=-x\cosh x$ and $y_2^{\prime }=-x\sinh x$, soThis agrees with the Abel calculation. Its vanishing at zero does not contradict independence on regular intervals: the original equation is singular there.

For $z\ne 0$, division by $4z$ gives $p(z)=1/(2z)$ and $q(z)=1/(4z)$. Thus zero is singular, but $zp=1/2$ and $z^{2}q=z/4$ are analytic. Zero is a regular singular point.

Apply the Frobenius method with $y=z^r{\sum }_{n=0}^{\infty }{a}_n{z}^n$, $a_0\ne 0$. The lowest power $z^{r-1}$ has coefficient $[4r(r-1)+2r]a_0$, so the indicial equation is $2r(2r-1)=0$. Its characteristic exponents at a regular singular point are $0$ and $1/2$. At the next powers,With $a_0=1$, the two Frobenius solutions are thereforeBoth analytic power-series factors converge everywhere by the ratio test. Recognizing the trigonometric series givesOn a punctured neighbourhood choose a consistent square-root branch for the second solution; the first is entire in $z$. These solutions are independent, since their leading orders are $1$ and $z^{1/2}$. A direct square-root reduction of a regular-singular differential equation also verifies the closed forms: for $s=\sqrt{z}$ and $Y(s)=y(s^2)$, $4z{y}^{\prime \prime }=Y_{ss}-Y_s/s$ and $2y^{\prime }=Y_s/s$, so the equation reduces to $Y_{ss}+Y=0$.

The Möbius transformation $w=z/(z-1)$ sends $0\mapsto 0$, $\infty \mapsto 1$ and $1\mapsto \infty$. The hypergeometric function $F(a,c-b;c;w)$ has local exponent pairs $(0,1-c)$ at $w=0$, $(0,b-a)$ at $w=1$, and $(a,c-b)$ at $w=\infty$. Pulling back to $z$ and multiplying by $(z-1{)}^{-a}$ gives the pairsAt infinity the prefactor adds $a$ to both exponents, while at $z=1$ it subtracts $a$ from both pulled-back exponents. These are exactly the requested Riemann P-symbol exponents. Equivalently, the Pfaff transformation identifies this branch, up to a nonzero phase depending on the chosen branches, with $F(a,b;c;z)$.

When $c\notin \mathbb{Z}$, an independent Frobenius solution isIts exponent $1-c$ differs from the first branch's exponent zero, establishing linear independence. For resonant parameters, use a parameter limit giving the appropriate logarithmic branch. More directly, on a simply connected region avoiding the singularities and zeros of a nonzero first solution $y_1$, reduction of order givesThe Wronskian is a nonzero multiple of $z^{-c}(1-z{)}^{c-a-b-1}$, so this also produces an independent local solution in the resonant cases for which the first branch is defined.

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Write a second-order linear ordinary differential equation in monic form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$. At a finite point $x_0$, an ordinary point has analytic $P,Q$. A singular point is a regular singular point if $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically there. A singular point failing this condition is an irregular singular point. The point at infinity is classified by the reciprocal coordinate $t=1/x$.

For the Laguerre differential equation, $P=1/x-1$ and $Q=\nu /x$. Thus $xP=1-x$ and $x^{2}Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, the transformed equation isHere $t{P}_t=1+1/t$ is not analytic at zero, so infinity is an irregular singular point, for every real $\nu$.

For the Frobenius solution, the lowest power $x^{\sigma -1}$ has coefficient $a_0{\sigma }^2$. Hence the indicial equation is ${\sigma }^2=0$, and equating the remaining powers givesWith the repeated root $\sigma =0$, the Frobenius series for the Laguerre differential equation obeysIf the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the ratio test gives infinite radius of convergence. It is an entire function, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.

For $\nu =n\ge 0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})/k!$, giving the Laguerre polynomialsThe same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.

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 $\cos x/x$.