The given ODE is:
$x(x+1)y''+xy'-y=0$ , with $y_1(x)=x$ and the point of expansion $x=0$
I solved the ODE using reduction of order, which gave me:
$$y(x)= ax + b(x(ln(x+1)-ln(x)) - 1)$$
Knowing what to look out for, I started by putting the ODE into standard form:
$$y''+(1/(x+1))y'-(1/x(x+1) )y=0$$
from which I found that,
$p(x)=(1/(x+1))$ and $q(x)=-(1/x(x+1))$
I then found the terms $p_o$ and $q_o$ of the indicial equation:
$p_o= lim_{x→0}〖xp(x)〗=0$ and $q_o= lim_{x→0}〖x^2 q(x)〗=0$
So my indicial equation was: $r(r-1) = 0$ , which gave me the roots $r=0,1$
From there I checked the condition $r_1-r_2 = 1$ (a positive integer), so my second linearly independent solution had the form:
$$y_2 (x)=cy_1 (x)ln(x)+∑_{n=0}^∞b_n x^{n+1}$$
I plugged this into my original ODE, and after a little series manipulation I got:
$$2cx+c+6b_2 x^3-b_o x+∑_{n=2}^∞x^n [b_{n-1} (n^2-1)+b_n (n^2+n)]=0$$
I get a recurrence relation of:
$$b_n=-\frac{(n-1)b_{n-1}}{n}$$
And 3 other relations: $c=0$ , $2c-b_o=0$ , $6b_2=0$
After this part I am unsure how to proceed to the correct answer I found using reduction of order: $b(x(ln(x+1)-ln(x)) - 1)$
You know one solution $y_1(x)=x$ and the other solution can be easily obtained by assuming $y_2=v(x).y_1(x)$ as its second solution.
Plugging it in the given ODE will reduce it to a second order ODE in dependent variable $v$, which on simplification yields the well known formula
$v(x)=\int $$\frac{e^{-\int P(x).dx}}{y_1^2}dx$. In your case $P(x)=1/(x+1)$ and $y_1(x)=x$.
So, $v(x)=In(1+\frac{1}{x})-\frac{1}{x}$ gives $y_2=xIn(1+\frac{1}{x})-1$ as the second solution.
Added:-If $r_1\ne r_2$ and $r_1-r_2$ is an integer with $r_1>r_2$, then the solution is of the form:
$y=A.y_1+B.y_2$, where $y_1=(y)_{r=r_1}$ and $y_2=\frac{\partial{(r-r_2).y(r)} }{\partial r}|_{r=r_2}$