Given the differential equation
$2(1-x)y''-3y'+\frac{y}{x}=0$
and in standard form:
$y''-\frac{3}{2(1-x)}y'+\frac{1}{2x(1-x)}y=0$
I want to find the series solution for the larger root $σ = 1$ of the indicial equation which I have found and show that the solution is equal to
$y_1(x)=\frac{x}{\sqrt(1-x)}$
so differentiating $y=x^σ\sum_{n=0}^{\infty}a_nx^n$ twice and subbing into the DE then dividing by $x^{σ-2}$ then multiplying by $(1-x)$ and making $x^n$ terms vanish separately I get:
$-(n-1+σ)(n-2+σ)a_{n-1}+(n+σ)(n+σ-1)a_n-\frac{3}{2}(n-1+σ)a_{n-1}+\frac{1}{2}a_{n-1}=0$
however when I simplify that I get $a_n=\frac{3}{2}a_{n-1}$ which obviously will not give me the same $y_1$ as I need but I can't see where I've gone wrong...
If your expression is correct: $$-(n-1+σ)(n-2+σ)a_{n-1}+(n+σ)(n+σ-1)a_n-\frac{3}{2}(n-1+σ)a_{n-1}+\frac{1}{2}a_{n-1}=0$$ then putting in $\sigma=1$, I get $$a_n\left[n(n+1)\right] = a_{n-1}\left[n(n-1)+{3n\over2}-{1\over2}\right] $$ or $$a_n\left[n(n+1)\right] = a_{n-1}\left[n\left(n+{1\over2}\right)-{1\over2}\right]$$ or $$a_n = a_{n-1}\left[{n(n+{1\over2})-{1\over2}\over n(n+1)}\right]$$
This gives $a_1={1\over 2}a_0$, $a_2={3\over8}a_0$, $a_3={5\over16}a_0$ etc.