<s>[INST] Find the indicial equation for $r_1$ and $r_2$ The given expression is:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
First we need to convert the given into its indicial equation:
$$x^2\sum^\infty_{n=0}(n+r)(n+r-1)C_nx^{n+r-2}+x\sum^\infty_{n=0}(n+r)C_nx^{n+r-1}+x^2\sum^\infty_{n=0}C_nx^{n+r}-\frac{1}{4}\sum^\infty_{n=0}C_nx^{n+r}=0$$
Now factoring through the $x$ coefficients 
$$\sum^\infty_{n=0}(n+r)(n+r-1)C_nx^{n+r}+\sum^\infty_{n=0}(n+r)C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}-\frac{1}{4}\sum^\infty_{n=0}C_nx^{n+r}=0$$
Now bringing first, second and fourth series into a single series
$$\sum^\infty_{n=0}\left[(n+r)(n+r-1)+(n+r)-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}=0$$
Now to evaluate the expression with $n=0$ and $n=1$; And just as a personal preference, I like to factor out the $x^r$ during this step:
$$x^r\left[C_0\left(r^2-r+r+x^2-\frac{1}{4} \right) x^0\right]+x^r\left[(1+r)(r)C_1x+(1+r)C_1x+C_1x^3-\frac{1}{4}C_1x \right]+\sum^\infty_{n=0}...=0$$
Now from here I calculated my $r$ values and this is where I got confused. For the first bracket group of the expression above I got $r=\frac{1}{2}$ but when I tried to calculate $r$ for the second bracket group I got $r=-\frac{1}{2}$ and $r=-\frac{3}{2}$. This is where I know I messed up somewhere as I am only supposed to get two values of $r$. I re-did the problem from the start and got the same answers, where did I go wrong?
 [/INST] $$\sum^\infty_{n=0}\left[(n+r)(n+r-1)+(n+r)-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}=0$$
$$\sum^\infty_{n=0}\left[(n+r)^2-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=2}C_{n-2}x^{n+r}=0$$
Changing the indices:
$$ \color {blue} {C_0x^r(r^2-\frac 14)}+C_1x^{r+1}\left(r^2+2r+\frac 34 \right)+\sum^\infty_{n=2}(\left[(n+r)^2-\frac{1}{4}\right]C_n+C_{n-2})x^{n+r}=0$$
Your indicial polynomial is  the coefficient of the lower power of $x$:
$$P(r)=r^2-\frac 14$$
The indicial equation is:
$$\left(r^2-\frac 14 \right)=0$$
$$\implies S_r=\left  \{ -\frac 12, \frac 12 \right \}$$
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