Why is $x_0$ a singular point of $A(x)y''(x)+B(x)y'(x)+C(x)y(x) = 0$ if $A(x_0)=0$? The question looks quite odd since this is the definition of a singular point for such differential equation. I was reading this (Frobenius Method) and I was wondering why if $A,B,C$ are analytic at $x_0$ then it is impossible to find two independent solutions for the DE. I know that if we write the DE as $y''(x) + p(x)y'(x)+q(x)y(x)=0$ then obviously $p(x) = B/A$ is not analytic at $x_0$. My question is why do we always have to write the DE in this last form?
If you can provide me any reference that treats this issue in a deeper level I would be very grateful. Most DE books are very engineering oriented and do not provide a clear answer for my question. Cheers.
 A: A linear DE of higher order can be canonically rewritten as a vector DE of the first order: introduce $z=(y,y')^T$ to get
$$
\left[\matrix{1 & 0\\0 & A}\right]z'=\left[\matrix{0 & 1\\-C & -B}\right]z.
$$
So you question is, basically, what happens with $Az'=Bz$ when $A$ is singular. An obvious comparison to $Ax=b$ tells us that if $A$ is invertible the solution exists for any $b$. If not then $b$ must be very special, i.e. belong to the image of $A$, otherwise there is no solution. Here is the same, if $A$ is invertible at a point $x_0$, one can write $z'=A^{-1}Bz$ and solve the DE around $x_0$ for any initial condition. If $A$ is not invertible then $\exists v\colon v^TA=0$, so $v^TBz=0$ (unless $z'=\infty$). It means that $z$ is restricted to the hyperplane, otherwise no solution. This additional restriction sets the constraint on how the initial condition may be defined there. They cannot be arbitrary any longer, but must be on the hyperplane. The fact that we are limited in how to choose the initial value makes the singular point important. The standard way to avoid troubles is to assume that the DE is solvable wrt the highest derivative (non-singularity).
P.S. The similar notion in PDE is a characteristic surface.
