Singular Points in first order differential In a first order differential where $y'=f(x,y)/g(x,y)$ why is any point that have $g(x,y)=0$ is not considered a singular point? I mean isn't singular point the points at which $y'$ is not defined?
 A: This is the question about how we understand what differential equation is. Note that any equation
$$
\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}
$$
can be written as
$$
\frac{dx}{dy}=\frac{g(x,y)}{f(x,y)}
$$
to emphasize that variables $x$ and $y$ have equal rights. Even better, one can write
$$
f(x,y)dx-g(x,y)dy=0,
$$
showing the relation of the solutions with the vector field $(f,-g)$. It is this last form is understood when one speaks about singular points (where the vector field is not defined), and for this both $f$ and $g$ must be zero.
A: Think in terms of the plane (x,y) embedded in the space (x,y,z,...) and the architecture and shape of the solution manifold as seen from that plane.  In two dimensions both f(x,y) and g(x,y) have to be undefinied because a variable that is in the slope's numerator in one part of the space may be in the denominator in some other slope calculation.  An undefined point is where every infinitesimal increase in horizontal displacement results in an infinite increase in vertical displacement, positive or negative.  From the standpoint of a plane composed of two variables, approaching an undefined point is like approaching a pole.  Imagine a plane created by $S_0$ and $S_1$.  Let $S_0=\frac{y}{x^\beta}$, $S_1=\frac{\dot{y}}{x^{\beta -1}}$, $S_2=\frac{\ddot{y}}{x^{\beta-2}}$, and so on.  These are the stabilizers of the Lie group $X=\lambda x$ and $Y=\lambda ^\beta y$, where $\lambda _0 =1$ represents the unitary conversion. These stabilizers form an embedded sub-manifold.  Singularities, saddle points and the separatrices that connect them must remain invariant under any invariant group transformation which preserves the structure of a smooth manifold (a Lie group).  
The slope field for the variables $S_0$ and $S_1$ is given by
$$\frac{dS_1}{dS_0}=\frac{x\frac{dS_1}{dx}}{x\frac{dS_0}{dx}}=\frac{S_2-(\beta -1)S_1}{S_1-\beta S_0}
$$Likewise, in the Lie plane created by $S_1$ and $S_2$ the slope is given by $$
\frac{dS_2}{dS_1}=\frac{x\frac{dS_2}{dx}}{x\frac{dS_1}{dx}}=\frac{S_3-(\beta -2)S_2}{S_2-(\beta-1)S_0}$$This pattern is repeated throughout this space, and this is the whole point: every $\frac{dS_n}{dx}=0$ at the undefined points!
Since the undefined points are invariant under group transformation (i.e., the poles are stable) it must be true that $S_1=\beta S_0$, $S_2=(\beta-1)S_1=\beta(\beta -1)S_0$, $S_3=(\beta-2)S_2=\beta(\beta -1)(\beta -2)S_0$, and so on.  This is the Lie algebra between the stabilizers, and it is part of the kernel of the map.  It can be used to solve differential equations.
An example:  $\ddot{y}+\frac{\dot{y}}{x}+y^2=0$ becomes $\frac{d}{dx'}\frac{dy'}{dx'}+\frac{dy'}{dx'}+(y')^2=0$ under the map  of the Lie group.  $$\lambda^{\beta -2}\ddot{y}+\lambda^{\beta -2}\frac{\dot{y}}{x}+\lambda ^{2\beta}=0$$For invariance, $\beta=-2$ and $S_0=x^2 y$, $S_1=x^3 \dot{y}$ and $S_2=x^4 \ddot{y}$, and so on.  Multiplying the DEQ by $x^4$ yields $S_2+S_1+S_0^2=0$  Applying the Lie algebra, $(-2)(-3)S_0+(-2)S_0+S_0^2=0$  Since $S_0=0$ implies $y=0$, we ignore that solution and choose $S_0=-4$.  This implies a special solution for the DEQ of $y=\frac{-4}{x^2}$, which is easily verified.  
