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Apply the Frobenius method. For $y=x^{\sigma }{\sum }_{n=0}^{\infty }{a}_n{x}^n$, the coefficient of the lowest power $x^{\sigma }$ comes only from $2x^2{y}^{\prime \prime }-x{y}^{\prime }+y$; the remaining term $xy$ starts one power higher. As $a_0\ne 0$, this lowest coefficient gives the indicial equationTherefore the two possible characteristic exponents at a regular singular point areTheir difference is not an integer, so the two corresponding Frobenius solutions do not require a logarithmic correction.

For $n\ge 1$, the coefficient of $x^{n+\sigma }$ isFactoring the bracket gives the recurrence relationFor $\sigma =1/2$ this is $a_n=-a_{n-1}/[n(2n-1)]$. Iteration and $(2n)!=2^{n}n!(2n-1)!!$ yieldFor $\sigma =1$, the recurrence is $a_n=-a_{n-1}/[n(2n+1)]$, givingHere the double factorial products are one at $n=0$, consistent with the chosen normalization. Recognizing the cosine and sine power series, for positive $x$ one may also write the two Frobenius solutions asThe constants multiplying the two independent solutions may be chosen separately.

For either exponent, the recurrence from part (c) impliesfor every finite complex $x$. The ratio test therefore givesThe factor multiplying $x^{\sigma }$ is an entire function in each case. For $\sigma =1/2$, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_n{x}^n$, not to a Taylor expansion of the branched full solution.

Use a Frobenius solution $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$. The lowest power $x^{\sigma -1}$ has coefficient $2\sigma (2\sigma -1)a_0$, giving the indicial equationFor $n\ge 1$, matching the coefficient of $x^{n+\sigma -1}$ givesFor both candidate exponents the cancelled factor is nonzero in this range, so the recurrence isWith $a_0=1$, the exponent-zero solution isFor $\sigma =1/2$, the coefficients are $a_n=1/(2n+1)!!=2^{n}n!/(2n+1)!$, givingThe ratio test gives infinite radius of convergence for both series factors, since the successive ratios in magnitude are $|x|/(2n+2)$ and $|x|/(2n+3)$. The second full solution still needs a square-root branch in the complex plane; on $x>0$ it is a real solution. The distinct leading behaviors $1$ and $\sqrt{x}$ prove linear independence.

For an independent closed-form check on the structure, put $z=x/2$. The equation becomes $z{Y}^{\prime \prime }+(1/2-z)Y^{\prime }-(1/2)Y=0$, a Kummer differential equation with equal parameters. Its exponential reduction of a confluent hypergeometric equation with equal parameters explains the elementary first solution and the branched companion.

For the Frobenius solution, differentiate term by term:The unshifted terms in the equation have coefficient $[4(n+\sigma )(n+\sigma -1)-4(n+\sigma )+3]a_n$. The extra factors of $x$ give the shifted contribution $[4(n+\sigma )(n+\sigma -1)+1]a_n{x}^{n+\sigma +1}$. Equating the coefficient of $x^{k+\sigma }$ for $k\ge 1$ gives the linear recurrence relationFactoring both coefficients is especially useful:The separate lowest-power equation is $(4{\sigma }^2-8\sigma +3)a_0=0$. Do not cancel $2k+2\sigma -3$ when it vanishes: that is the possible undetermined coefficient at Frobenius resonance.

For the larger indicial root $\sigma =3/2$, the recurrence in part (b) becomesNo factor being cancelled is zero in this range, soThus the resulting Frobenius solution isThere is exactly one free parameter, $a_0$, setting the overall scale. The analytic series factor has radius of convergence one. For a real-valued expression take $x>0$; a complex version requires a local square-root branch.

The Taylor series of the logarithm givesTherefore the closed form of the larger-root Frobenius solution isThe series derivation holds initially for $0<x<1$; the closed form extends as a real solution to all $x>0$. To check it without series, put $y=\sqrt{x}v(x)$ in the original equation. All undifferentiated $v$ terms cancel and the remaining expression is $4x^{5/2}[(1+x)v^{\prime \prime }+v^{\prime }]$. Hence $[(1+x)v^{\prime }{]}^{\prime }=0$, and $v=\log (1+x)$ indeed solves it. Its logarithm is analytic near zero, so this is not a forced $\log x$ companion term at the Frobenius point.

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.

Put $t=1-x$ and $Y(t)=y(x)$. The differential equation becomesSubstitute the Frobenius solution $Y=t^{\sigma }{\sum }_{n\ge 0}{a}_n{t}^n$. The lowest power has coefficient ${\sigma }^2{a}_0$, soEquating successive coefficients gives, with $a_{-1}=0$,Thus $a_0\ne 0$ is the only free coefficient: $a_1=-\lambda a_0$, $a_2=({\lambda }^2-c^2)a_0/4$, and so on. Since every denominator is nonzero, there is exactly one independent pure Frobenius solution of this form. The repeated root supplies no second free coefficient; the other solution is logarithmic, as parts (iii) and (iv) demonstrate.

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.

Normalize the free leading coefficient to $a_0=1$. The linear recurrence relation from part (b) givesConsequently the first four nonzero coefficients areand the single pure Frobenius solution isFor clarity about the second independent solution, a choice with no added multiple of $y_1$ isThese companion coefficients follow by reduction of order with the normalized Wronskian $W=z^{-3}(1+z^2{)}^{3/2}\exp [3z^2/(2(1+z^2))]$. The logarithm requires a local branch; it does not supply a second ordinary power series 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 $\log (x-m)$.

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 $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$, assume $z_0$ is a singular point, rather than an ordinary point. It is a regular singular point exactly whenextend holomorphically across $z_0$. If $P,Q$ themselves extend holomorphically, the point is ordinary and should not be classified as singular merely because the displayed products are also analytic.

Writing $p_0=p(z_0)$ and $q_0=q(z_0)$, substitution of $y=(z-z_0{)}^r$ into the lowest-order terms gives the indicial equation $r(r-1)+p_{0}r+q_0=0$. A Frobenius solution then accounts for the analytic higher-order coefficients.

For $y^{\prime \prime }+a{y}^{\prime }/z+by/z^2=0$ at zero, the equation is regular singular unless both coefficients vanish, with indicial polynomial $r(r-1)+ar+b$. A coefficient such as $P(z)=z^{-2}$ instead violates the criterion and gives an irregular singular point.

With $s=\sqrt{z}$ and $Y(s)=y(s^2)$, the displayed linear differential equation becomes $Y_{ss}+Y=0$ away from zero. Its Frobenius solutions have exponents $0$ and $1/2$: $\cos \sqrt{z}={\sum }_{n\ge 0}(-1{)}^n{z}^n/(2n)!$ and $\sin \sqrt{z}=z^{1/2}{\sum }_{n\ge 0}(-1{)}^n{z}^n/(2n+1)!$. The first is an entire function of $z$; the second needs a local square-root branch. This illustrates how a regular singular point can arise from a degenerate change of independent variable.

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.