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An indicial exponent is a root of the indicial equation at a regular singular point. It gives the leading local power in a Frobenius method; resonant roots can require logarithmic terms.

An indicial root is a root of the indicial equation obtained from a power-law ansatz. For a Cauchy-Euler differential equation, such roots give the exponents of its power solutions. Their sign determines which modes remain bounded near a singular endpoint.

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.

Work first on $x>0$, where the proposed logarithm is defined. Substitution of $y=x^m$ into the Cauchy-Euler differential equation gives the indicial equationIts distinct indicial roots are $m=\pm i$. Using $x^{\pm i}=e^{\pm i\log x}$ and Euler's formula, a real basis of solutions is $\cos (\log x),\sin (\log x)$. Their Wronskian is $1/x$, so they are linearly independent and span the two-dimensional solution space of a homogeneous linear differential equation. ThereforeFor the requested change of variables, set $Y(z)=y(e^z)$. The chain rule givesThe transformed second-order linear differential equation is $Y^{\prime \prime }+Y=0$. Its characteristic roots are again $\pm i$, giving $Y(z)=A\cos z+B\sin z$. Substituting $z=\log x$ recovers precisely the previous general solution.

If solutions on $x<0$ are wanted, the same calculation uses $z=\log |x|$ and gives independent constants on that interval. Zero is a singular point of a differential equation; the logarithmic oscillations do not define a nonzero continuous solution through it.

Multiply the growth equation by $2{\eta }^2(1+\eta )$:Use the Frobenius method, writing $D={\sum }_{n\ge 0}{b}_n{\eta }^{\alpha +n}$ with $b_0\ne 0$. The lowest power gives the indicial equationThus the leading powers are $\alpha =1$ and $\alpha =-3/2$. For $n\ge 1$, equating the coefficient of ${\eta }^{\alpha +n}$ givesFor the growing branch $\alpha =1$, $b_n=-2(n+1)b_{n-1}/(2n+5)$, so $b_1=-4b_0/7$ and $b_2=8b_0/21$. Taking $b_0=A_k$,Since $\eta \propto a$, the leading behavior is precisely the usual growing density contrast ${\delta }_C\propto a\propto t^{2/3}$ during matter domination. The negative first correction shows the beginning of suppressed growth as the string fraction increases. The other branch starts as $a^{-3/2}$, the familiar decaying matter mode; its exact form is proportional to $\sqrt{1+\eta }/{\eta }^{3/2}$.

With negligible cold-dark-matter velocity, the synchronous perfect-fluid density equation gives ${\delta }_C^{\prime }=-h^{\prime }/2$. Differentiate and use the trace Einstein equation, taking the cold matter sound speed to vanish and its density to dominate:In a spatially flat matter-dominated universe, $a\propto {\tau }^2$, $\mathcal{H}=2/\tau$, and the conformal Friedmann equation gives $4\pi G{a}^2{\rho }_C=3{\mathcal{H}}^2/2=6/{\tau }^2$. HencePut ${\delta }_C\propto {\tau }^s$. The indicial equation is $s(s-1)+2s-6=(s-2)(s+3)=0$. These two independent modes span the solution space, soEquivalently the growing mode is proportional to $a$ and the decaying mode to $a^{-3/2}$. The coefficients are set by initial data. Since synchronous gauge has residual coordinate modes, interpreting the density contrast physically requires fixing its residual freedom consistently, for example by the cold-matter rest frame and metric tensor integration convention; the mode equation alone is not a gauge-invariant observable.

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 origin is a regular singular point of this Kummer differential equation. Apply the Frobenius method, writing $y={\sum }_{n=0}^{\infty }{a}_n{x}^{n+r}$ with $a_0\ne 0$. The lowest power gives the indicial equationFor $n\ge 1$, comparison of the coefficient of $x^{n+r-1}$ givesSince $0<c<1$, both roots give valid independent branches. Normalize $a_0=1$:Here $(c{)}_n$ is the rising factorial, and the first series is the confluent hypergeometric function of the first kind $M(1,c,x)$. The two leading powers are distinct, proving linear independence. We describe real solutions on $x>0$, with their one-sided behavior at zero.

For the inhomogeneous equation a particular solution is $-(x+c)/2$, verified by direct substitution. Thus $y=A{y}_1+B{y}_2-(x+c)/2$. Since $y_2^{\prime }\sim (1-c)x^{-c}$, a finite derivative at zero forces $B=0$. Then $y(0)=0$ forces $A=c/2$, and the required solution isIn particular $y^{\prime }(0)=0$. The singular homogeneous branch cannot be added without violating the derivative condition.

Multiply the equation by $x$ to obtain $x{y}^{\prime \prime }-P(x)y^{\prime }-Q(x)y=0$. The leading terms after a Frobenius method trial $y=x^r$ arethe term involving $Q_0$ first appears at order $x^r$. The indicial equation is therefore

To obtain the first correction to a regular singular solution by reduction of order, start from the assumed solution $y_1=1+\beta x+O(x^2)$ and use the Abel identity:We work initially on $x>0$ near zero; since $P_0$ is a positive integer, the resulting expansion is an ordinary power series. Also $y_1^{-2}=1-2\beta x+O(x^2)$, so the reduction of order integral has integrandChoose its integration constant to be zero, eliminating an added multiple of $y_1$. Integrating and multiplying by $y_1$ givesWith $C=P_0+1$ the requested normalized answer isIts nonzero Wronskian makes it independent of $y_1$. The existence of the asserted $y_1$ also forces $P_0\beta +Q_0=0$; there is no need to introduce $Q_0$ into the normalized second coefficient.

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.

Classification and characteristic exponents. For a normalized second-order linear differential equation, an ordinary point at zero means $p$ and $q$ are real analytic functions there. A regular singular point is a singular point where $xp$ and $x^{2}q$ nevertheless extend as real analytic functions to zero. Write $p_0={\lim }_{x\to 0}xp(x)$ and $q_0={\lim }_{x\to 0}{x}^{2}q(x)$. The lowest-power coefficient in the Frobenius method gives the indicial equationIts roots are the characteristic exponents at a regular singular point. If their difference is not an integer, there are two linearly independent solutions $x^{{\sigma }_j}{A}_j(x)$, with $A_j$ real analytic and $A_j(0)\ne 0$. For distinct real roots ${\sigma }_1>{\sigma }_2$ with integer difference, a solution for the larger root exists, and a second has the form $x^{{\sigma }_2}B(x)+c{y}_1\log x$; the logarithmic coefficient can vanish. For a repeated root, the second solution has the form $y_1\log x+x^{{\sigma }_1}B(x)$ after normalization. This is a Logarithmic Frobenius solution. Complex conjugate roots $a\pm ib$ give real solutions whose leading behaviour is $x^a\cos (b\log x)$ and $x^a\sin (b\log x)$, with analytic-series corrections. All these descriptions are on $x>0$, or on a fixed complex branch.

For the present equation, division by $x$ gives $p=1+(1-m)/x$ and $q=(1-m)/x$. Therefore $xp=x+1-m$ and $x^{2}q=(1-m)x$ are real analytic. Zero is regular singular when $m\ne 1$; when $m=1$, it is ordinary after removing the common factor $x$. In that exceptional case the normalized equation is $y^{\prime \prime }+y^{\prime }=0$.

For the Frobenius method substitution, the lowest coefficient and the general linear recurrence relation areThese equations come respectively from the coefficients of $x^{\sigma -1}$ and $x^{n+\sigma -1}$. At $m=1$, the same formal exponents $0,1$ describe the two possible leading powers at an ordinary point, rather than singular behaviour.

Here the indicial equation has the repeated root $0$. The first Frobenius method solution is $y_1=e^{-x}$. Since $p=1+1/x$, the Abel identity gives $W=C{e}^{-x}/x$. Reduction of order with $C=1$ produceswhere the additive constant is absorbed into $y_1$. In particular the logarithmic coefficient is nonzero: an additional independent real analytic solution does not exist.

The general local form isThis is the repeated-root Logarithmic Frobenius solution. On the negative real side one may use $\log |x|$, or use a fixed logarithm branch for a complex solution.

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.

Here $q=\kappa /R$, so the equation from part (b) becomes the Euler-Cauchy equationThe power-law ansatz $B_z=R^s$ gives the indicial equationFor $0<\kappa <1/2$, the general field is thereforeThe polynomial discriminant changes sign atFor $\kappa >{\kappa }_c$, define $\mu =\sqrt{{\kappa }^2-1/4}$. A real form of the solution isThus the field has a log-periodic oscillation: its phase is periodic in $\ln R$, so it oscillates as the radius changes geometrically.

At $\kappa ={\kappa }_c=1/2$, the indicial equation has the multiple root $s=-1/2$. The general axial component isSince $B_{\varphi }=-2R{B}_z^{\prime }$, the azimuthal component isConsequentlyas $R\to \infty$; the same ratio is identically one when $D=0$. A magnetic field line has tangent parallel to $\mathbf{B}$, so its asymptotic angle $\vartheta$ with the $z$-axis satisfiesHence the field lines become helices making the constant anglewith the axis.