Apply the power series method of solution to the following initial value problem in the neighborhood of the given point $x_0$. $$xy''+x^2y'-2y=0, \qquad x_0=1, \qquad y(1)=0, \qquad y'(1)=\frac12$$
My attempt:
$y(x)=\sum\limits_{m=0}^\infty a_mx^m$
Substitute:
$x[\sum\limits_{m=2}^\infty m(m-1)a_mx^{m-2}]+x^2[\sum\limits_{m=1}^\infty ma_mx^{m-1}]-\sum\limits_{m=0}^\infty 2a_mx^m=0$
$[\sum\limits_{m=0}^\infty a_{m+2}(m+2)(m+1)x^{m-1}]+\sum\limits_{m=1}^\infty ma_{m}x^{m+1}-\sum\limits_{m=0}^\infty 2a_mx^m=0$
$(2a_2-2a_0)+\sum\limits_{m=1}^\infty [(m+2)(m+1)a_{m+2}+(m+2)a_m]x^{m+1}=0$
This gives:
$2a_2-2a_0=0$
$(m+2)(m+1)a_{m+2}+(m+2)a_m=0$
Therefore:
$a_2=a_0$
$a_{m+2}=-\frac{1}{m+1}a_m$
How do I verify that $x_0$ is ordinary point and what do I do with $y(1)=0$ and $y'(1)=\frac{1}{2}$
First of all please read this and make an observation that the given point $x_0 = 1$ is an ordinary point whereas $x = 0$ is a regular singular point. If you have studied the power series method then it must have been mentioned that for regular singular points we need to apply Frobenious Method. Fortunately here the given point $x_0 = 1$ is ordinary and the method you have used is find except a simple change which is the fact that the series must be written around the point $x_0 = 1$ i.e. $$y(x)=\sum\limits_{m=0}^\infty a_m (x-x_0)^m = \sum\limits_{m=0}^\infty a_m (x-1)^m$$
The procedure you have shown in the question looks correct and you can follow the same for the above series. The unknown values of $a_0$ and $a_1$ at the end of the procedure can be obtained using the initial conditions easily due the the series expansion around $1$.