Differential equations, theoretical question $$x(t)={1 \over (c-t)}\tag{$**$}$$is the general solution of $$ x'=x^2$$ 
Why is the area existence and uniqueness of the solution $R_{tx}^2??$
Its the general solution in : $G_1=\{(t,x): t\in R , x<0\}$ and $G_2=\{(t,x): t\in R , x>0\}$
I'm aware that $(**)$ isn't defined at $x=0$ and is for every $t$, what I don;t understand is the $R_{tx}^2$ area of existence and uniqueness , what why this is so? 
 A: The existence and uniqueness theorem of initial value theorem guarantees that the initial value problem 
$$x'=f(x,t), x(t_0)=x_0$$ 
has a unique solution $x(t)$ in an interval of $(t_0-\delta,t_0+\delta)$ if $f$ and $\frac{\partial f}{\partial x}$ is continuous in a rectangular region of $\mathbb{R}^2_{tx}$.
This tells you that your problem has unique solution at least in some open interval around the initial point.
Now you know that your ode ends up with this solution:
$$x=\frac{1}{\frac{1}{x_0}-t}$$
It is valid whenever $x\ne 0$.
Also notice that $x=0$ is a fixed point of the ODE. So $x\equiv 0$ is also a solution. 
If you draw the direction field:

You see that $x$ would go to $0$ if the initial value is negative; $x=0$ if the initial value is $0$. $x$ goes to infinity if the initial value is positive.
You can also draw the solution curve using your solution $x=\frac{1}{\frac{1}{x_0}-t}$. The following are the pictures with positive and negative initial conditions, respectively:

Hence wherever your initial value is, there exists a unique solution. However, the interval of validity is $t\in(-\infty,\frac{1}{x_0})$ and $t\in(\frac{1}{x_0},\infty)$, for $x_0>0, x_0<0$, respectively.
