What does it mean when it is said that a solution to a differential equation only exists on an interval? I solved a DE and got $$y=\frac{1}{1-x}$$ the book states that this solution only exists on $\mathbb{R}-\{1\}$ I guess because otherwise it is undefined but what does it mean by solution because say if we pick a value where it is defined we will just end up with a number how exactly is this an answer?
Also could someone explain what it means by unique solution? Does this just mean that there is only one solution to the DE?
Are there any simple cases where there are many solutions to a DE, I can't think of any of the top of my head.
Thanks.
 A: The initial value problem
$$y' = \sqrt{y}, \,\,\,  y(0) = 0,$$
has multiple solutions.  
For example $y(x) = 0$ and $y(x) = x^2/4$.
In general, the initial value problem 
$$y' = f(x,y), \,\,\,  y(x_0) = y_0,$$
is guaranteed to have a unique solution on some open interval containg $x_0$ when the function $f$ satisfies certain conditions.  Look up the Picard-Lindelof Theorem. The notion of uniqueness is relevant only in the context of an accompanying initial condition. The differential equation may have many solutions, in general. However, we say the solution is unique for an initial value problem when given any two functions $y_1$ and $y_2$ that satisfy both the differential equation and initial condidition, then $y_1(x) = y_2(x)$ for all $x$ where solutions exist. 
The interval on which the solution exists need not be extendable to $\mathbb{R}$ as in your example where $f(x,y) = y^2$.
A: Okay, the function y= 1/(1- x) is defined for all x except x= 1.  But I really don't understand what you mean by "if we pick a value where it is defined we will just end up with a number".  That, of course, is true of ANY function!  But a differentiable function can be thought of as more than just a collection of number- there is also the way the different numbers are "connected".  Here, your function is $y= (1- x)^{-1}$.  It's derivative is $-(1- x)^{-2}= -\frac{1}{(1- x)^2}= -y$ so satisfies the differential equation $y'= -y^2$.  Was that the equation you solved?
 You also ask about "unique" solutions.  Surely you know that solving a differential equation is equivalent to integrating and integrating leads to an arbitrary "constant of integration"!   For example, the very simple differential equation y'= x has the "general solution" y= (1/2)x^2+ C.  What is true is that if f(x,y) is continuous in x and "Lipschitz" in y ("differentiable" with respect to y is sufficient but not necessary) in some neighborhood of $(x_0, y_0)$ then there exist a neighborhood of $(x_0, y_0)$ in which there is a unique (exactly one) function, y, that satisfies the differential equation [b]and[/b] $y(x_0)= y_0$.
An example of a differential equations problem where the solution is NOT unique is $y'= y^{1/2}$ with intial condition y(0)= 0  (Notice that tex'= (1/2)y^{-1/2} which is not defined at y= 0).  We can write the equation as y^{-1/2}dy= dx and, integrating, $2y^{1/2}= x+ C$ which is the same as $y= (1/4)(x+ C)^2$.  Setting Both x and y equal to 0 in that gives C= 0 so a solution $y= (1/4)x^2$ for all x.  But y(x)= 0 for all x also satisfies $y'= y^{1/2}$.  If y= 0 for all x, then y'= 0 and $y^{1/2}= 0$ so the equation becomes 0= 0.  In fact, we can get an infinite number of distinct solutions to the equation that also satisfy the initial condition by taking $y= (1/4(x+ a)^2$ for some a< 0, y= 0 for a< x< b and $y= (1/4)(x+ b)^2$ for x> b.
