Solving a linear ODE using Frobenius' method The given ODE is
$2y''$ - $y'/(x-1)$ + $y/(x-1)^2$ = 0.
Solving we get,
$\sum_{n=0}$$[2(n+r)(n+r-1)a_n - (n+r)a_n](x-1)^{(n+r-1)}$.
Now I have few basic doubts. Please bear with me.
Can't we directly equate the coefficient of $(x-1)^{(n+r-1)}$ to zero?
Because the solution given here shows that we need to first expand to get:
$[2r(r-1)-r+1]a_0(x-1)^{r-2} + [2r(r+1)-(r+1)+1]a_1(x-1)+...$
Then substituting the coefficient of $(x-1)^{r-2}$ to zero to get the indicial equation as:
$2r(r-1)-r+1=0$.
My second doubt is that, after solving, we get $r=1$ or $r=0.5$
Then what should we do after this?
The solution says substituting r, only coefficient of $a_0$ goes to zero, that means $a_1,a_2....= 0$
Therefore, the general solution should be $a_0(x-1)+a_2(x-1)^{0.5}$
Can someone elaborate on this please

Yes, that looks correct. The equation is of Euler-Cauchy type,
$$
2(x-1)^2y''-(x-1)y'+y=0
$$
this implies that basis solutions $(x-1)^r$ exist and $r$ solves the characteristic equation
$$
0=2r(r-1)-r+1=(2r-1)(r-1)
$$
that is identical to the indicial equation. So indeed this gives a complete basis $(x-1)^{1/2}, (x-1)$.

To the first point, in a more general series expansion the first terms would have coefficients with negative indices. One can avoid that by treating these terms separately or by expanding the coefficient sequence with $a_{k}=0$ for $k<0$.
