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The homogeneous equation is a Cauchy-Euler equation. With the power-law ansatz $y=x^r$ on $x>0$, its indicial equation isThus $r=(1-\alpha )/2$ is a repeated indicial root. The two independent solutions are $x^r$ and $x^r\log x$, givingTheir Wronskian, with $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, is $x^{2r-1}=x^{-\alpha }\ne 0$, verifying linear independence.

A finite point $x_0$ is an ordinary point when the normal-form coefficients $p,q$ are analytic at $x_0$. Then arbitrary values $y(x_0),y^{\prime }(x_0)$ determine a convergent Taylor series, and two independent choices give two linearly independent local analytic solutions.

A singular point is a regular singular point when $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$. The Frobenius method then provides at least one solutionwith a locally convergent analytic factor and an exponent fixed by the indicial equation. If its two indicial roots differ by a noninteger, there are two independent Frobenius solutions. For integer-separated roots a second solution may include a logarithm; a repeated root gives a Logarithmic Frobenius solution accompanying the single pure series. Such solutions need not be finite or ordinary Taylor series at the singular point.

Seek a Frobenius solution $y=z^r{\sum }_{n\ge 0}{a}_n{z}^n$ with $a_0\ne 0$. Since $(z+z^3{)}^2=z^2+2z^4+z^6$, its lowest coefficient gives the indicial equationThus the repeated indicial root is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ givesFor $n\ge 1$ this becomes the linear recurrence relationEvery coefficient is fixed by $a_0$, since the denominator never vanishes for $n\ge 1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary Taylor series with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.

A second independent local solution is a logarithmic solution from a repeated Frobenius exponent. Indeed, Abel's identity gives $W=C{z}^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, reduction of order givesAfter choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times \text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure Frobenius solution.

Set $t=x-m$ and use the Frobenius solution $y=t^{\nu }(a_0+a_{1}t+\cdots )$ with $a_0\ne 0$. From part (a), the leading coefficients are $P\sim 1/t$ and $Q\sim m^2{\omega }^2/t^2$. The coefficient of the lowest power $t^{\nu -2}$ is thereforeThe indicial equation reduces to ${\nu }^2+m^2{\omega }^2=0$, givingSince $m\omega >0$, the two indicial roots differ by a nonzero imaginary number, not an integer. Thus the two local Frobenius solutions exist independently on a chosen branch of $\log (x-m)$.

At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.

Set $t=y-1$. Then $y^2-1=t(t+2)$, and a Frobenius solution has the formIn the differential equation, the lowest power is $t^{\sigma -1}$, whose coefficient isThe indicial equation and its indicial roots are consequentlyTheir difference is not an integer, so the Frobenius method produces two independent series without a logarithmic term:Choose a branch of $(y-1{)}^{1/2}$, or take $y>1$ for real solutions. A general local solution is $C_1{x}_1+C_2{x}_2$ by linear independence and the superposition principle. The analytic factors converge at least for $|y-1|<2$, up to the next singular point at $y=-1$.

In the shifted variable $t=y-1$, substitute the Frobenius solution $x={\sum }_{j\ge 0}{a}_j{t}^{j+\sigma }$ into $t(t+2)x^{\prime \prime }+x^{\prime }+x=0$. Equating the coefficient of $t^{j+\sigma }$ givesThus the linear recurrence relation, valid for $j=0,1,\dots$, isFor the indicial root $\sigma =0$, this becomesFor the indicial root $\sigma =1/2$, rename the coefficients $b_j$; thenWith $a_0=b_0=1$, the first terms of the two Frobenius solutions are

For $u_{xx}+y^2{u}_{yy}=0$ with zero vertical-side values on $0<x<\pi$, the Fourier coefficient $b_n(y)$ obeys the Cauchy-Euler differential equation $y^2{b}_n^{\prime \prime }=n^2{b}_n$. Boundedness at $y=0$ removes the negative indicial root, leaving $b_n(y)=A_n{y}^{(1+\sqrt{1+4n^2})/2}$. Bounded top coefficients give absolute convergence of the Fourier sine series below the top boundary.

When two indicial roots differ by a positive integer, the recurrence for the smaller-root Frobenius solution can have a zero coefficient at the index of the larger root. If the corresponding right-hand side also vanishes, that coefficient is free and adds the larger-root solution; if it does not vanish, a pure Frobenius series is obstructed and a logarithmic term may be required. For $x^2{y}^{\prime \prime }-2x{y}^{\prime }-x^{2}y=0$, the recurrence $n(n-3)a_n=a_{n-2}$ has $a_1=0$ and leaves $a_3$ free. Choosing $a_3=0$ selects the even solution, not the only solution with exponent zero.