 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.