= Solution
Use a <Frobenius solution> $y=x^\sigma\sum_{n\geq0}a_nx^n$, with $a_0\ne0$. The lowest power $x^{\sigma-1}$ has coefficient $2\sigma(2\sigma-1)a_0$, giving the <indicial equation>
$$
\boxed{2\sigma(2\sigma-1)=0,\qquad\sigma=0,\frac12.}
$$
For $n\geq1$, matching the coefficient of $x^{n+\sigma-1}$ gives
$$
2(n+\sigma)(2n+2\sigma-1)a_n-(2n+2\sigma-1)a_{n-1}=0.
$$
For both candidate exponents the cancelled factor is nonzero in this range, so the recurrence is
$$
\boxed{a_n=\frac{a_{n-1}}{2(n+\sigma)}.}
$$
With $a_0=1$, the exponent-zero solution is
$$
\boxed{y_1(x)=\sum_{n=0}^\infty\frac{x^n}{2^nn!}=e^{x/2}.}
$$
For $\sigma=1/2$, the coefficients are $a_n=1/(2n+1)!!=2^nn!/(2n+1)!$, giving
$$
\boxed{y_2(x)=\sqrt{x}\sum_{n=0}^\infty\frac{x^n}{(2n+1)!!}=\sqrt{x}\left(1+\frac x3+\frac{x^2}{15}+\frac{x^3}{105}+\cdots\right).}
$$
The <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 $zY''+(1/2-z)Y'-(1/2)Y=0$, 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.
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