Frobenius method differ by integer When the roots of the indicial equation differ by an integer the equation is of the form:
$$y_2 (z)=cy_1 (z) \ln(z)+z^{\sigma_2 } \sum_{n=0 }^\infty(b_n z^n )$$
Here is what is bothering me. The last term on the RHS (namely $z^{\sigma_2 } ∑_{n=0 }^\infty(b_n z^n )$) is the Frobenius series that we would usually (if the roots did not differ by an integer) substitute into the original ODE. But the reason that we did not do this in the first place is because it leads to something going wrong (such as dividing by 0 for one of the terms). Thus when we sub $y_2$ into the ODE since derivatives are like linear operators we will get exactly the same problem as we had before hand i.e. for one of the terms we will need to divide by 0, say. 
So how do we get around this problem?
To demensionstrate my point consider example 7 in this document http://www.most.gov.mm/techuni/media/EM_03011_p2chap34.pdf

The ODE is $(x^2-x)y''-xy'+y=0$
  The roots to the indicial equation are $r_1=1$ and $r_2=0$. And it can be easly shown that for a given root $r$ the reccurence relationship is:
  $$a_{s+1}=\frac{a_n(n+r-1)^2}{(n+1+r)(n+r)}$$
  when $r=0$ when have
  $$a_{s+1}=\frac{a_n(n-1)^2}{(n+1)(n)}$$
  Thus when $n=0$ we end up dividing by 0 and we can not use this method.
  When $r=1$ this can be solved to get $y_1=a_0 x$.
  Thus using $$y_2 (x)=cx \ln(x)+x^{r_2 } \sum_{n=0 }^\infty(b_n x^n )$$
  and differentiating it twice and subbing it back into the ODE we get:
  $$-xc+\sum (n(n-1)b_n+(n+1)nb_{n+1}-b_nn+b_n)x^n=0$$
  so when equating coefficients of $x^n$ in an effort to find $b_n$ we will get exactly the same recurrence relationship as before and thus not be able to find $b_1$ since it will involve dividing by 0

 A: I'll present a solution of the ODE $$(x^2-x)y''-xy'+y=0$$ via the method of Frobenius in the manner that seems clearest to me. First, we take a solution with $y\sim x^r$ and observe that 
$$(x^2-x)y''-xy'+y = -r(r-1) x^{r-1}+O(x^r)=0\quad\text{as }x\to 0$$ implies the indicial equation $r(r-1)=0$ with roots $r=0,1$. For the latter, we substitute a power series of the form $y_1=x+\sum_{n=1}^\infty a_n x^{n+1}$ which yields $y_1=x$ as a solution. (I take $y_1 \sim x$ to fix the overall constant.)
As noted in the question, such a method fails for $r=0$. Instead we construct the second solution $y_2=cx\ln x+b(x)$ where $b(x)$ has leading order $x^0$. To eliminate an overall multiplicative factor, I'll assume $b(0)=1$. But I will also assume $b'(0)=0$. To justify this, observe that if we picked a non-zero $b'(0)$ then we could form the solution $y_2-b'(0)y_1$ which would have no strictly linear term by construction. Hence $b'(0)$ is superfluous and I shall assume $b(x)=1+\sum_{n=2}^\infty b_n x^n$.
All that remains is the tedious step of subbing in this solution; to make things easier, I'll organize the ODE as the equality $x^2 y''-xy'+y=x y''$. Then
\begin{align}
x^2 y_2''-xy_2'+y_2
&=1+\sum_{n=2}^\infty (n-1)^2 b_n x^n,\\
=xy''&=c+2b_2 x+\sum_{n=2}^\infty (n+1)b_{n+1} x^{n}  
\end{align}
Comparing these, we require $c=1$, $b_2=0$, and $(n-1)^2b_n=(n+1)b_{n+1}$. Since the last recurrence never terminates we conclude all $b_{n\geq 2}=0$ i.e. $b(x)=1$. Thus $y_2=x\log x+1$.
With this, we conclude that the general solution is of the form $\boxed{y(x)=A x+B (x\log x+1)}$.
