Solve $x(1-x)y''+2(1-2x)y'-2y=0$ by the Frobenius Method 
Find the second solution. First solution is $\dfrac 1 {1-x}$.  Solve by Frobenius Method: $$x(1-x)y''+2(1-2x)y'-2y=0\,.$$ 

The first solution I am able to get is $\dfrac{1}{1-x}$. Other solution is $\dfrac{1}{x}$, but I am getting $-\dfrac{1}{x(1-x)}$. Where am I going wrong?
 A: As for your attempt, you did nothing wrong.  Note that
$$\frac{1}{x}=(-1)\left(\frac{1}{1-x}\right)+(-1)\left(-\frac{1}{x(1-x)}\right)\,.$$
That is, you can take the two linearly independent solutions of your differential equation to be 
$$y(x)=\frac1{1-x}\text{ and }y(x)=\frac{1}{x}\,,$$
but you can take them to be
$$y(x)=\frac1{1-x}\text{ and }y(x)=-\frac{1}{x(1-x)}$$
as well.

Here is a full solution as requested, although this seems unnecessary, because you have actually solved the problem.  Suppose that $y:U\to\mathbb{R}$ is a twice differentiable function on an open set $U\subseteq \mathbb{R}$ which satisfies
$$x(1-x)\,y''(x)+2(1-2x)\,y'(x)-2\,y(x)=0$$
for all $x\in U$.  We make an Ansatz that $$y(x)=\sum_{n=0}^\infty a_nx^{n+r}$$
where $r$, $a_0$, $a_1$, $a_2$, $\ldots$ are real numbers.
We have
$$y'(x)=\sum_{n=0}^\infty\,(n+r)\,a_n\,x^{n+r-1}\,,$$
and
$$y''(x)=\sum_{n=0}^\infty\,(n+r)(n+r-1)\,a_n\,x^{n+r-2}\,.$$
Thus,
$$x(1-x)\,y''(x)=\small r(r-1)\,a_0\,x^{r-1}+\sum_{n=0}^\infty\,\big((n+r+1)(n+r)\,a_{n+1}-(n+r)(n+r-1)\,a_n\big)\,x^{n+r}\,,$$
and
$$2(1-2x)\,y'(x)=2r\,a_0\,x^{r-1}+\sum_{n=0}^\infty\,\big(2(n+r+1)\,a_{n+1}-4(n+r)\,a_n\big)\,x^{n+r-1}\,.$$
Ergo,
$$\begin{align}0&=x(1-x)\,y''(x)+2(1-2x)\,y'(x)-2\,y(x)
\\&=r(r+1)\,a_0\,x^{r-1}+\sum_{n=0}^\infty\,(n+r+1)(n+r+2)\,\big(a_{n+1}-a_n\big)\,x^{n+r}\,.
\end{align}$$
It follows immediately that
$$r(r+1)=0\text{ and }(n+r+1)(n+r+2)\,\big(a_{n+1}-a_n\big)=0$$
for $n=0,1,2,\ldots$.  
If $r=0$, then $(n+1)(n+2)\,(a_{n+1}-a_n\big)=0$ for all $n=0,1,2,\ldots$.  Thus, $a_n=a_0$ for every $n=0,1,2,\ldots$, and we may write
$$y(x)=a_0\,y_1(x)\,,\text{ where }y_1(x):=\sum_{n=0}^\infty\,x^{n+0}=\frac{a_0}{1-x}\,.$$
The power series solution is valid only for $U=(-1,+1)$, but $y_1(x)=\dfrac{1}{1-x}$ works even if $U=\mathbb{R}\setminus\{1\}$.
If $r=-1$, then $n(n+1)\,\big(a_{n+1}-a_n\big)=0$ for all $n=0,1,2,\ldots$.  Thus, $a_n=a_1$ for every $n=1,2,3,\ldots$.  We may assume that $a_1=0$ (otherwise, we may subtract an appropriate multiple of $y_1(x)$ out of $y(x)$).  This means
$$y(x)=a_0\,y_2(x)\,,\text{ where }y_2(x):=\frac{1}{x}\,.$$
Note that we may take $U$ to be $\mathbb{R}\setminus\{0\}$ in this case.  
The general solutions are 
$$y(x)=\frac{A}{1-x}+\frac{B}{x}$$
which are defined on $U=\mathbb{R}\setminus\{0,1\}$, although $A$ and $B$ must be treated as local constants (i.e., they are constant on three different intervals: $(-\infty,0)$, $(0,1)$, and $(1,\infty)$).
