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Work first on $x>0$. For $b=0$, the power-law ansatz $y=x^r$ gives the indicial equation $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the Cauchy-Euler equation has general solutionFor $a=0$ the repeated indicial root supplies the logarithm rather than two copies of the constant solution. Alternatively, the change of variables $t=\log x$ turns the differential equation into $Y^{\prime \prime }-a^{2}Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log |x|$, with independent constants on the two sides of zero.

For general $b$, normalize the second-order linear differential equation asThe coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^{2}Q(x)=-a^2+b{x}^{\beta }$ are analytic functions there because $\beta$ is a positive integer. The regular singular point criterion for a second-order equation therefore proves that zero is a regular singular point for every allowed $a,b,\beta$. The indicial equation remains $r^2-a^2=0$.

Put $N=n(n+1)$. After division by $1-x^2$, the Legendre differential equation has coefficientsBoth are analytic functions at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm 1$, $p$ has a genuine simple pole; hence these are singular points even when $n=0$ makes $q$ vanish. However,and both products extend analytically across $x_0$. By the regular singular point criterion for a second-order equation,None of the three specified points is an irregular singular point.

Normalize the second-order equation as $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$, with $P,Q$ holomorphic at all finite nonzero points. The regular singular point criterion for a second-order equation at zero says that $zP(z)$ and $z^{2}Q(z)$ extend holomorphically across zero.

To impose the condition at infinity, set $w=1/z$ and $Y(w)=y(1/w)$. Since $y^{\prime }=-w^2{Y}^{\prime }$ and $y^{\prime \prime }=w^4{Y}^{\prime \prime }+2w^3{Y}^{\prime }$, the transformed equation isRegular singularity at $w=0$ makes $2-P(1/w)/w$ and $Q(1/w)/w^2$ bounded and holomorphic there. Equivalently $zP(z)$ and $z^{2}Q(z)$ are bounded at infinity. They are entire functions by the zero-endpoint condition, so Liouville's theorem makes both constants. The Euler-Cauchy classification from two regular singular points is thereforeConversely these equations have ordinary coefficients everywhere on ${\mathbb{C}}^{\ast }$ and meet both regular-singularity bounds. If “regular singular” is required to exclude an ordinary point, omit $(A,B)=(0,0)$, which makes zero ordinary, and $(2,0)$, which makes infinity ordinary. Under the common “at worst regular singular” convention these two cases are included. Multiplication by a nonvanishing coefficient does not change the normalized equation or this classification.

With $t=1-x^2$ the chain rule gives $y^{\prime }=-2x{u}^{\prime }$ and $y^{\prime \prime }=-2u^{\prime }+4x^2{u}^{\prime \prime }$. Substitution, initially for $x>0$, givesThe transformed coefficients are $P(t)=1/t$ and $Q(t)=1/[4t(1-t)]$. At $t=0$, both $tP$ and $t^{2}Q$ are analytic; at $t=1$, both $(t-1)P$ and $(t-1{)}^{2}Q$ are analytic. Each point is singular because a normalized coefficient has a pole, but satisfies the regular singular point criterion for a second-order equation. Thus both points are regular singular points.

For a Frobenius method substitution $u=t^{\rho }{\sum }_{n\ge 0}{a}_n{t}^n$, the lowest power gives the indicial equationThe repeated indicial root is zero. Setting $\rho =0$ and equating the coefficient of $t^n$ givesConsequently the analytic power-series solution of a differential equation isTaking $a_0=2\pi$ reproduces the ellipse perimeter expansion.

For a second independent solution, normalize $u_1(0)=1$. Reduction of order and $P=1/t$ giveSince $u_1(s{)}^{-2}=1+O(s)$, this has the Logarithmic Frobenius solution formThe logarithm is $\log |t|$ on the real negative side, or a chosen complex logarithm on a slit complex neighborhood. Its nonzero logarithmic coefficient establishes independence. The physical perimeter is finite at $t=0$, so it selects the analytic branch rather than this logarithmic one.

After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ are analytic functions, so $0$ is a regular singular point. Its indicial equation is $r(r-1)+2r=r(r+1)=0$. There are no other finite singular points of a differential equation. At infinity the regular singular point criterion for a second-order equation would require $P=O(z^{-1})$, but $P\to -1$; hence infinity is an irregular singular point.

Substituting $y=f/z$ reduces the equation to $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two characteristic exponents at a regular singular point, $0,-1$. The integer difference does not require a logarithm. At infinity $e^z/z$ has an essential singularity, consistent with the irregular singular point; the existence of the simpler solution $1/z$ does not make infinity regular.

For$x_0$ is an ordinary point when $P$ and $Q$ are analytic there, and a singular point otherwise. A singular point is regular singular when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary-point and regular-singular criteria.

For Kummer's equation, substituteEquating the coefficient of $x^m$ givesTaking $c_0=1$,where $(a{)}_m=a(a+1)\cdots (a+m-1)$ is the rising factorial. Thus

Putting $y=x^{1-b}u$ and simplifying givesThereforeFor nonintegral $b$, the powers $x^0$ and $x^{1-b}$ at zero are distinct, so these solutions are linearly independent.

When $b\to 1$, both solutions tend to $M(x,a,1)$. Differentiate their difference with respect to $b$. Writing $M_a,M_b$ for derivatives with respect to the second and third arguments,Hence at $b=1$ one may takeas two linearly independent solutions. Their independence follows from the logarithmic term in the second solution.

The point $z_0$ is a regular singular point when it is singular butextend holomorphically to $z_0$.

For a normalized equation$x_0$ is an ordinary point when $P,Q$ are analytic there. It is a regular singular point when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary point criterion for a second-order equation and regular singular point criterion for a second-order equation.

Hereso $x=0$ is regular singular. Seek a power-series solution of a differential equationEquating the coefficient of $x^n$ givesand henceThereforeThe series terminates exactly when $\lambda =N$ is a nonnegative integer: the factor with $j=N$ then makes $a_{N+1}=0$. Thus the polynomial solutions occur for

After division by $x^2$, the equation becomesThe regular singular point criterion for a second-order equation would require $x(1/x)$ and $x^2(-1/x^4)$ to be analytic at zero. The second expression is $-x^{-2}$, so the criterion fails:

The Papperitz symboldescribes the two-dimensional solution space of a second-order homogeneous Fuchsian differential equation on the Riemann sphere with exactly three regular singular points $z_1,z_2,z_3$. Near $z_i$, its two independent Frobenius behaviours are locallysubject to logarithmic modifications in resonant cases. The six characteristic exponents obey the Fuchs relation

For the Gauss hypergeometric equation, the exponents at infinity are $a$ and $b$. Put $w=z^{-1}$. Transforming the Papperitz symbol from $z$ to $w$ shows thatandare local solutions near $z=\infty$, with respective leading behaviours $z^{-a}$ and $z^{-b}$. When $a-b\notin \mathbb{Z}$, these behaviours are distinct and the two solutions are linearly independent. They therefore form a basis of the solution space on any simply connected common domain with compatible branch choices.

The solution $F(a,b;c;z)$ normalized at zero can be analytically continued into that domain. Since it solves the same second-order equation, it must be a constant linear combination of the basis at infinity:The constants depend on $a,b,c$ and the branch convention, but not on $z$. This is the hypergeometric connection formula at infinity.

After division by $z$, the coefficient functions areBoth $zP(z)=1-z$ and $z^{2}Q(z)=\lambda z$ are analytic at zero, so the regular singular point criterion for a second-order equation shows that $z=0$ is regular singular. Substitution of $y=z^r$ into the leading terms gives the indicial equationThe exponent is repeated. Directly substituting $y={\sum }_{n\ge 0}{a}_n{z}^n$ givesThus $a_0$ determines the entire power series, so there is only one such solution $y_1$ up to scale. The logarithmic solution from a repeated Frobenius exponent predictsso its leading nonanalytic term is proportional to $\log z$.

For the contour ansatz, differentiation under the integral and one integration by parts giveThe integral vanishes whenwhose solution is the Laguerre contour-integral amplitudeThe contour and branches must make $f$ single-valued along the traversed path and must kill the endpoint term

Now suppose $\lambda <0$ is nonintegral. Near $t=0$, $f(t)=O(t^{-\lambda -1})$ is integrable and the endpoint factor is $O(t^{-\lambda })\to 0$. We may therefore choosea finite loop based at the branch point $0$ and avoiding $1$ except by encirclement. A second choice isalong one bank of the negative real axis, with a consistent branch. At $-\infty$, $e^{zt}$ kills the algebraic endpoint factor because $\mathrm{Re}z>0$.

The ${\gamma }_1$ integral is analytic in $z$ because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of $y_1$. This is the Finite Laguerre contour solution construction.

Fora finite point $z_0$ is an ordinary point exactly when $p$ and $q$ are holomorphic there. It is a regular singular point exactly whenare holomorphic at $z_0$.

Put $\zeta =1/z$ and $W(\zeta )=w(1/\zeta )$. Direct differentiation givesso the transformed equation isConsequently $z=\infty$ is ordinary precisely whenare holomorphic at $\zeta =0$. It is a regular singular point at infinity precisely when $zp(z)$ and $z^{2}q(z)$ are holomorphic functions of $1/z$ near infinity.

If zero and infinity are regular singular and every nonzero finite point is ordinary, the Laurent series of $p$ and $q$ can contain only the terms compatible with both endpoint bounds. Hencefor constants $a,b$. The equation is a Cauchy-Euler differential equation. Its indicial equation isFor distinct roots ${\lambda }_1,{\lambda }_2$, the general solution on a domain with a chosen logarithm branch isFor a repeated root $\lambda$, the general solution of an Euler-Cauchy equation is

Finally require infinity to be ordinary. In the transformed equation its coefficients becomeBoth are holomorphic at zero exactly when $a=2$ and $b=0$. Thus the further restriction is

Divide by $2x^2$ to put the ordinary differential equation into normalized form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$:The coefficients are singular at zero, butare both analytic there. This is precisely the regular singular point criterion for a second-order equation, so $x=0$ is a regular singular point.

After division by the leading coefficient, write $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, whereThe only finite singular points of a differential equation are $\boxed{x=0\text{ and }x=-1}$. At zero, $xP=-1/(1+x)$ and $x^{2}Q=(x+3)/[4(1+x)]$ are analytic. At $-1$, $(x+1)P=-1/x$ and $(x+1{)}^{2}Q=(x+1)(x+3)/(4x^2)$ are analytic. The regular singular point criterion for a second-order equation therefore makes both points regular.

If infinity is included, it is a regular singular point at infinity as well: $P(x)=O(x^{-2})$ and $Q(x)=1/(4x^2)+O(x^{-3})$. Equivalently, with $t=1/x$ the transformed coefficients are $\tilde{P}=2/t+1/(1+t)$ and $\tilde{Q}=(1+3t)/[4t^2(1+t)]$, which satisfy the same regular-singular criterion at $t=0$.