= Solution
\b[Classification and existence.] For $y''+p(x)y'+q(x)y=0$, an <ordinary point> $x_0$ is one where $p,q$ are analytic. There are two independent analytic <power series> solutions, with arbitrary initial values $y(x_0),y'(x_0)$. A <regular singular point> is a singular point where $(x-x_0)p(x)$ and $(x-x_0)^2q(x)$ 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,
$$
xp(x)=\frac{1-x}2,\qquad x^2q(x)=-\frac x4.
$$
Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a <regular singular point>.
\b[Indicial equation and recurrence.] Multiply by $4x$ and insert $y=x^\sigma\sum_{n\geq0}a_nx^n$, $a_0\ne0$:
$$
4xy''+2(1-x)y'-y=0.
$$
The lowest power $x^{\sigma-1}$ has coefficient $[4\sigma(\sigma-1)+2\sigma]a_0$, giving
$$
\boxed{2\sigma(2\sigma-1)=0,\qquad\sigma=0,\ \frac12.}
$$
The coefficient of $x^{n+\sigma}$ is
$$
2(n+\sigma+1)(2n+2\sigma+1)a_{n+1}-(2n+2\sigma+1)a_n.
$$
For either exponent the common factor is nonzero for $n\geq0$, so the <recurrence relation> reduces to
$$
\boxed{a_{n+1}=\frac{a_n}{2(n+\sigma+1)},\qquad n\geq0.}
$$
Normalizing each leading coefficient to one, the two <Frobenius solutions> are
$$
\boxed{y_1(x)=\sum_{n=0}^\infty\frac{x^n}{2^nn!}=e^{x/2},\qquad y_2(x)=\sqrt{x}\left(1+\frac x3+\frac{x^2}{15}+\frac{x^3}{105}+\cdots\right).}
$$
More explicitly, $y_2=\sqrt{x}\sum_{n\geq0}2^nn!x^n/(2n+1)!$. The distinct leading powers establish <linear independence>, so $y=C_1y_1+C_2y_2$ on the positive real side of the singular point. The displayed <exponential function> is the requested closed-form solution; substitution gives $y_1'=y_1/2,y_1''=y_1/4$, which makes the differential expression zero.
\b[Convergence.] For either exponent the ratio of successive terms in the coefficient <power series> is
$$
\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|=\frac{|x|}{2|n+\sigma+1|}\longrightarrow0.
$$
Thus 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.
Back to article page
