The recurrence relation I ended up getting doesn't match with the final answer. Did a couple of revisions of the thing but nothing changes.
I get the recurrence relation as $${a_{r+2}=\frac{-[(r^2-2r+p)a_r+(r+1)(2r-1)a_{r+1}]}{(r+2)(r+1)}}$$
My attempt at a solution
The original equation can be written as $${y''-\frac{xy'}{x^{2}}+\frac{py}{x^{2}}=0}$$
Put $(x-1)=t$
Notice that $y''$ and $y'$ don't make a difference
Therefore we have $${y''-\frac{(t+1)y'}{(t+1)^{2}}+\frac{py}{(t+1)^{2}}=0} \qquad (1)$$
where $P(t)=-\frac1{(t+1)}$ and $Q(t)=\frac{p}{(t+1)^{2}}$, which have $t=0$ as an ordinary point
Now we can write $$y =\sum{a_rt^r},$$ $$y' =\sum{ra_rt^{r-1}},$$ $$y'' =\sum{(r-1)ra_rt^{r-2}}$$
Using these in the (1) we get $${\sum{(t+1)^{2}(r-1)ra_rt^{r-2}}-\sum{(t+1)ra_rt^{r-1}}+p\sum{a_rt^r}=0}$$ $$\implies{\sum{(t^2+2t+1)(r-1)ra_rt^{r-2}}-\sum{(t+1)ra_rt^{r-1}}+p\sum{a_rt^r}=0}$$ $${\implies\sum{(t^2+2t+1)(r-1)ra_rt^{r-2}}-\sum{(t+1)ra_rt^{r-1}}+p\sum{a_rt^r}=0}$$ $${\implies\sum{(r-1)ra_rt^{r}}+2\sum{(r-1)ra_rt^{r-1}}+\sum{(r-1)ra_rt^{r-2}}-\sum{ra_rt^{r}}-\sum{ra_rt^{r-1}}+p\sum{a_rt^r}=0}$$
Equating coefficients of $t^r$ on both sides, we have $${(r-1)ra_r+2{(r+1-1)(r+1)a_{r+1}}+(r+2-1)(r+2)a_{r+2}-ra_r-(r+1)a_{r+1}+pa_r=0}$$ $$\implies{((r-1)r-r+p)a_r+{(2r(r+1)-(r+1))a_{r+1}}+(r+1)(r+2)a_{r+2}=0}$$ $$\implies{(r^2-2r+p)a_r+{(r+1)(2r-1)a_{r+1}}+(r+1)(r+2)a_{r+2}=0}$$ $$\implies{(r+1)(r+2)a_{r+2}=-[(r^2-2r+p)a_r+{(r+1)(2r-1)a_{r+1}}]}$$
which gives the recurrence relation
$${a_{r+2}=\frac{-[(r^2-2r+p)a_r+(r+1)(2r-1)a_{r+1}]}{(r+1)(r+2)}}$$
However when I use the above mentioned relation for the solution, the coefficients don't come out to be correct as per the answer key.
The solution supposedly is http://s11.postimg.org/4zm9oouf6/IMG_20150708_190227.jpg
I can't post images yet, so the link.
