Find $y_1(x)$ and $y_2(x)$ from the recurrence relation of $r_2$ only, by using the Frobenius method The given original equation is:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
The series I have is:
$$x^r\left[\left(r^2-r+r+x^2-\frac{1}{4}\right)C_0+\left(r^2+r+1+r+x^2-\frac{1}{4} \right)C_1x\right]+\sum^\infty_{n=2}\left[(n+r)(n+r-1)+(n+r)+\frac{3}{4}\right]C_nx^n$$
The $r$ values I have are $$r_1=\frac{1}{2}, \space r_2=-\frac{1}{2}$$ Meaning that this problem is Case $2$, when $r_1-r_2$= a positive integer. The general formula for $y_1(x)$ and $y_2(x)$ in Case $2$ are:
$$y_1(x)=\sum^\infty_{n=0}C_nx^{n+r_1}$$
$$y_2(x)=Cy_1(x)\ln(x)+\sum^\infty_{n=0}b_nx^{n+r_2}$$
So when $r=-\frac{1}{2}$ the series should be 
$$x^{-\frac{1}{2}}\left[(r^2+x^2-\frac{1}{4})C_0+(r^2+2r+x^2+\frac{3}{4})C_1x \right]+x\sum^\infty_{k=2}\left[((k+r)(k+r-1)+(k+r)+\frac{3}{4})C_kx^{n}\right]$$
The amount of terms and rigorous syntax required for these kinds of problems leads me to believe that I couldn't have gone this far without making a mistake, nor can I proceed as my professor is unreachable for help at the moment. My question is am I right so far and where do I go from here?  
 A: $$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
1) This is Bessel's equation of order $\dfrac  12$
2) Indicial equation:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
Substitute $y=x^a$
$$a(a-1)x^a+ax^a+x^{a+2}-\frac{1}{4})x^a=0$$
Take the coefficient of the lowest power of $x$:
$$P(a)=a^2-\dfrac 14$$
The indicial equation:
$$(a-\frac 12)(a+\frac 12)=0 \implies S_a=\{\frac 12,-\frac 12\}$$
So you have done this correctly.
3) Case 2: as you noted for the roots of the indicial polynomial. The formula for the second solution you have to use is correct too.
4) Recurrence formula:
$$ x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
$$(n+r)(n+r-1)a_n+(n+r)a_n-\dfrac 14a_n+a_{n-2}=0$$
$$(n+r)^2a_n-\dfrac 14a_n =-a_{n-2}$$
$$a_n =-\dfrac {a_{n-2}}{(n+r)^2-\frac 14}$$
For $r=\frac 12$ the recurrence formula becomes:
$$a_n =-\dfrac {a_{n-2}}{n(n+1)}$$
Calculate some coefficients and deduce the pattern:
$$a_{2n}= \dfrac {(-1)^na_0}{(2n+1)!}$$
Therefore one of the solution to the DE is:
$$y_1(x)=\sqrt x\sum_{n=0}^\infty \dfrac {a_0(-1)^n x^{2n}}{(2n+1)!}=\dfrac  {a_0          } {\sqrt x}\sum_{n=0}^\infty \dfrac {(-1)^n x^{2n+1}}{(2n+1)!}$$
$$ \boxed {y_1(x)=\dfrac {a_0} {\sqrt x}\sin x }$$ 
