Why are the solutions of differential equations given with restricted domains? I solved $y^2=y'$ with the initial condition that $y(0)=1$, and I got $\displaystyle y(x)=\frac{1}{1-x}$. 
My book says it's only the portion of this function from $-\infty$ to $1$ that counts. 
I don't understand why?
Now, if the initial condition were $y(2)=-1$, would that make it the portion from $1$ to $+\infty$? 
What if both initial conditions were given?
 A: Let us look at a more familiar problem.  We want to find a function $y$ such that
$$\frac{dy}{dt}=-\frac{1}{(t-3)^2} \qquad \text{and} \qquad y(4)=17$$
Integrate.  If we do it in the familiar first year calculus way, we get $1/(t-3)+C$.  To evaluate $C$, put $t=4$.  Quickly we get $y(t)=-1/(t-3)+18$.
For what $t$ is this correct?  I claim that it is only for $t >3$.  To see why, assume that $y$ represents the displacement of a particle at time $t$.  Then $\frac{dy}{dt}$ represents its velocity.  
The velocity (and displacement) are not defined at time $t=3$.  From the information that the velocity at any time $t\ne 3$ is $-1/(t-3)^2$, and the information that the displacement at $t=4$ is $17$, we can find the displacement at any time $t>3$.  
But we cannot go backwards past the singularity at time $t=3$ to reach any conclusion about displacement at any time $t<3$.
Similarly, if we had been told the displacement at say $t=1$, we could find the displacement at any time $t<3$, but we could get no information about $t>3$.
If we want a "general" formula for the $\int\frac{-dt}{(t-3)^2}$ that holds for as wide a domain as possible, we would have to say that $y=-1/(t-3)+C_1$ for $t>3$ and $y=-1/(t-3)+C_2$ for $t<3$, where $C_1$ and $C_2$ are constants, possibly different constants.
But more usually we restrict the domain, and like in our example with the initial condition $y(4)=17$, we simply say that $y(t)=-1/(t-3)+18$ for $t>3$.
Roughly speaking the same sort of thing happens in your differential equation example.  You presumably solved the equation by rewriting as 
$$\frac{dy}{y^2}=dx$$
then integrating, and applying the initial condition.
Look at what you got for $y$.  It (and its derivative) blows up at $x=1$.  So like in the integration problem I started with, initial information about $x=0$ can't give any knowledge about the function past its singularity at $x=1$.
Suppose now we take your suggested $y(2)=-1$.  When solving the DE, we get
$-1/y=x+C$, so $C=-1$, I hope. That gives $y=-1/(x-1)$, valid for $x>1$.  
But change things to $y(2)=1$. We get $C=-3$. So the position of the singularity can be affected by the initial condition. (This cannot happen in simple integration examples.)
