Regular and singular points in a second order differential equation. For a linear homogeneous second order differential equation
$$y'' + p(x)y'+q(x)y=0$$
A point $x_0$ is regular point of the equation if the functions $p,q$ are analytic at $x_0$, and a singular point if they are not. If the functions $p,q$ satisfy:
$$\lim_{x\to x_0} (x-x_0)p(x) <\infty$$
$$\lim_{x\to x_0} (x-x_0)^2 q(x)<\infty$$
then the singularity is said to be regular, otherwise it's irregular.
Two questions:
Firstly, what motives this particular bound on the order of growth (linear, quadratic) of the functions?
Secondly, I am wondering how (or indeed, if) those definitions can be adapted to the general case, where we drop linearity and homogeneity, so the equation is of the form:
$$y''=f(x,y,y')$$
 A: For the equation you have given,
$$
\begin{align}
    \frac{d}{dx}\left[\begin{matrix}y \\ (x-x_{0})y'\end{matrix}\right] & =
                  \left[\begin{matrix}y'\\(x-x_{0})y''+y'\end{matrix}\right] \\
     & = \left[\begin{matrix}y' \\  (1-(x-x_{0})p)y'-(x-x_{0})qy\end{matrix}\right]
 \\
     & =
                \frac{1}{x-x_{0}}\left[\begin{matrix}0 & 1 \\ -(x-x_{0})^{2}q & 1-(x-x_{0})p\end{matrix}\right]\left[\begin{matrix}y \\ (x-x_{0})y'\end{matrix}\right].
\end{align}
$$
So the vector equation is
$$
          (x-x_{0})Y'(x) = A(x)Y(x).
$$
The matrix $A(x)$ has a removable singularity at $x=x_{0}$ with
$$
        A(x_{0}) = \left[\begin{matrix}0 & 1 \\ -q_{0} & 1-p_{0}\end{matrix}\right].
$$
This procedure generalizes to higher-order equations of the form
$$
         y^{(n)}+p_{1}(x)y^{(n-1)}+\cdots+p_{n-1}(x)y^{(1)}+p_{n}(x)y^{(0)}=0,
$$
where these limits exist:
$$
         \lim_{x\rightarrow x_{0}}(x-x_{0})^{k}p_{k}(x),\;\;\; k=1,2,3,\cdots,n.
$$
This is a fairly specific form of equation leading to $(x-x_{0})Y'(x)=A(x)Y(x)$, where the components of $Y(x)$ are $y,(x-x_{0})y',\cdots,(x-x_{0})^{n-1}y^{(n-1)}$. This form of equation doesn't extend well to general functions $f$ as you have asked.
