Frobenius Method Indicial Equation I need to verify that the indicial equation only has one root.
$xy''+(1-x)y'+\frac{1}{2}y=0$
Attempt:
$y=\sum\limits_{m=0}^\infty {a_mx}^{m+r}$
$y'=\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1}$
$y'=\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2}$
Substitute: 
$x\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2}+(1+x)\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1}+\frac{1}{2}\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$
Reduce:
$\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-1}+\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1} -\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r}+\frac{1}{2}\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$

$\sum\limits_{m=0}^\infty[(m+r-1)(m+r)+(m+r)]a_{m}x^{m+r-1}-\frac{1}{2}\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$
Shift Index:
$\sum\limits_{m=0}^\infty[(m+r-1)(m+r)+(m+r)]a_{m}x^{m+r-1}-\frac{1}{2}\sum\limits_{m=1}^\infty {a_{m-1}x}^{m+r-1}=0$

$[r^2]a_0x^{r-1}+\sum\limits_{m=1}^\infty([(m+r-1)(m+r)+(m+r)]a_m-\frac{1}{2}a_{m-1})x^{m+r-1}=0$
Therefore r is the double root of $0$, but I am not sure if this is correct.How would I go about verifying there is only one root?

The indicial must be at $x=0$ in this case. Dividing by the coefficient of the highest order derivative gives
$$
          y'' + \left(\frac{1}{x}+1\right)y'+\frac{1}{2x}y = 0.
$$
The indicial equation is found by keeping the $\frac{1}{x}y'$ terms and $\frac{1}{x^{2}}y$ terms:
$$
               y''+\frac{1}{x}y' = 0.
$$
Set $y=x^{p}$. The conditions to have such solutions is that $p$ satisfies the indicial equation
$$
              p(p-1)+p = 0, \\
              p^{2}=0.
$$
