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?
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$ 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: $$y_1(x)=\sqrt x\sum_{n=0}^\infty \dfrac {a_0(-1)^n x^{2n}}{(2n+1)!}=\dfrac {a_0 } {\sqrt x}\sum_{n=0}^\infty \dfrac {(-1)^n x^{2n+1}}{(2n+1)!}$$ $$ \boxed {y_1(x)=\dfrac {a_0} {\sqrt x}\sin x }$$