Find two linearly independent series solutions to $$2xy''+y'+xy=0$$ ($x_0$ is a regular singular point),
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So, $$y=\sum^\infty_{n=0}a_nx^{n+r}$$$$y'=\sum^\infty_{n=0}(n+r)a_nx^{n+r-1}$$ and $$y''=\sum^\infty_{n=0}(n+r)(n+r-1)a_nx^{n+r-2}$$
So sum works out to; $$\sum^\infty_{n=0}2(n+r)(n+r-1)a_nx^{n+r-1}+\sum^\infty_{n=0}(n+r)a_nx^{n+r-1}+\sum^\infty_{n=0}a_nx^{n+r+1}$$ $$=$$ $$\sum^\infty_{n=0}2(n+r)(n+r-1)a_nx^{n+r-1}+\sum^\infty_{n=0}(n+r)a_nx^{n+r-1}+\sum^\infty_{n=2}a_{n-2}x^{n+r-1}$$ Equating coeffiecents, when $n=0$, $$2r(r-1)+r=0$$ $$r=0, r=\frac{1}{2}$$ When $n=1$ $a_1=0$ for $r=\frac{1}{2}$ and $a_1$ is $0$ for $r=0$ Since $ a_1$ is equal to zero, then there are no odd term in the series. I then got $ a_n = \frac {a_02^{\frac {n}{2}}(-1)^n}{(2n+1)! (2n)!}$ But I can't think of a what I'd do next...
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i've answered the qs