 Solution

The Wronskian is $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. If constants $A,B$ obey $A{y}_1+B{y}_2=0$, differentiating gives $A{y}_1^{\prime }+B{y}_2^{\prime }=0$. At any point where $W\ne 0$, 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 $y=x^r{\sum }_{n=0}^{\infty }{a}_n{x}^n$, $a_0\ne 0$. The lowest power gives the indicial equationThe coefficient of $x^{n+r}$ givesFor either root the common factor is nonzero for $n\ge 0$, soWith leading coefficient one, the two independent solutions areTheir analytic series factors both have 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 $C_1{y}_1+C_2{y}_2$. This is the $b=1/2$, $z=x/2$ case of the exponential reduction of a confluent hypergeometric equation with equal parameters.