Singularity of Differential Equations Determine the singularities of the differential equation
$(x-1)^3x^2y''-2 (x-2)xy'-3y=0$
My Attempt: 
 $y''-\frac {2 (x-2)}{(x-1)^3x}y'-\frac {3}{(x-1)^3x^2}y=0$

at $x_0 =0$, we have
$(x-x_0)p=(x-x_0)\frac{-2 (x-2)}{(x-1)^3x}$,
$(x-x_0)^2q=(x-x_0)^2\frac {-3}{(x-1)^3x^2}$
$xp=\frac {-2 (x-2)}{(x-1)^3}$,
$x^2q=\frac {-3}{(x-1)^3}$
What does this tell me? I followed what I learned in class, but I am having trouble understanding if that is a singular point.
 A: The singularities are just the values of $x$ that will make the denominator zero, because you can't divide by zero (and doing so creates a singularity). Since you are dividing through by the coefficient of $y''$, whatever value for $x$ makes the coefficient $(x-1)^3x^2=0$ will cause a singularity. If either $(x-1)^3=0$ or $x^2=0$, there will be a singularity so solving for $x$ in both of these equations gives you $x=1$ and $x=0$.
You'll then have to use both of these singularity points as your $x_0$. This will give a total of four limits that you need to solve for in order to make sure the differential equation has a power series solution. 
For the first singularity ($x_0=0$) the two limits you need to check are:
$$1.) \lim_{x\to x_0}\left[(x-x_0)\frac{B(x)}{A(x)}\right] = \lim_{x\to0}\left[(x-0)\frac{-2(x-2)x}{(x-1)^3x^2}\right]=\lim_{x\to0}\left[\frac{-2(x-2)}{(x-1)^3}\right]=-4$$
$$2.) \lim_{x\to x_0}\left[(x-x_0)^2\frac{C(x)}{A(x)}\right] = \lim_{x\to0}\left[(x-0)^2\frac{-3}{(x-1)^3x^2}\right]=\lim_{x\to0}\left[\frac{-3}{(x-1)^3}\right]=3$$
The solutions are regular because they are finite, as opposed to irregular if the limits were $\pm \infty.$
