= Solution
The <Wronskian> is $W=y_1y_2'-y_1'y_2$. If constants $A,B$ obey $Ay_1+By_2=0$, differentiating gives $Ay_1'+By_2'=0$. At any point where $W\ne0$, these form an invertible two-by-two system, forcing $A=B=0$. Thus the two solutions are linearly independent.
For the given equation, the normalized coefficients are $p=(1-x)/(2x)$ and $q=-1/(4x)$, so zero is a <regular singular point>. Substitute a <Frobenius series> $y=x^r\sum_{n=0}^\infty a_nx^n$, $a_0\ne0$. The lowest power gives the <indicial equation>
$$
\boxed{2r(2r-1)=0,\qquad r=0,\ \tfrac12.}
$$
The coefficient of $x^{n+r}$ gives
$$
2(n+r+1)(2n+2r+1)a_{n+1}=(2n+2r+1)a_n.
$$
For either root the common factor is nonzero for $n\geq0$, so
$$
\boxed{a_{n+1}=\frac{a_n}{2(n+r+1)}.}
$$
With leading coefficient one, the two independent solutions are
$$
\boxed{y_1=\sum_{n=0}^\infty\frac{x^n}{2^nn!}=e^{x/2},\qquad y_2=x^{1/2}\sum_{n=0}^\infty\frac{x^n}{(2n+1)!!}=\sqrt{x}\left(1+\frac x3+\frac{x^2}{15}+\frac{x^3}{105}+\cdots\right).}
$$
Their analytic series factors both have \b[infinite radius of convergence], because the ratio of consecutive terms tends to zero for every fixed $x$. The factor $x^{1/2}$ still gives the second solution a branch point at zero; it is not an entire function or an ordinary Taylor series there. On $x>0$, the general real solution is $C_1y_1+C_2y_2$. This is the $b=1/2$, $z=x/2$ case of the <exponential reduction of a confluent hypergeometric equation with equal parameters>.
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