$\frac{dx}{dt} =x^2$ with $x(0)=x_0$.
$x(t) = \frac{1} {x_0^{−1} −t}$ satisfies the differential equation above (provided that $x_0\neq 0$). Clearly when $t = 0$, we have $x(0) = x_0$.
Our solution has some interesting properties. If $x_0 > 0$ then the denominator is initially positive (at $t = 0$), but decreases as $t$ increases until it reaches zero at time $t = x^{−1}_0$ . This means that the solution, $x(t)$, has become infinite by the time $t = x^{−1}_ 0$; we say that it ‘blows up’ in a finite time.
Things are much nicer, though, if we want to see where our solution came from in the past. We can decrease $t$ (from zero) as much as we like, since as $t$ decreases the denominator becomes larger, and so the solution itself tends to zero as $t →−∞$. So when $x_0 >0$, we can define the solution of the differential equation on the interval $(−∞,x_0^{-1})$, but there is no way to define the solution on an interval that extends further into the future beyond the time $t = x^{−1}_ 0$ . We refer to $(−∞,x_0^{−1})$ maximal interval of existence for the differential equation.
I think this statement is completely wrong "but there is no way to define the solution on an interval that extends further into the future beyond the time $t = x^{−1}_ 0$". I think the solution is also defined even after $t = x^{−1}_ 0$ and here only $t = x^{−1}_ 0$ is discontinuity and it undefined at this point we can still get solution beyond $t = x^{−1}_ 0$.I have graphed the solution using positive $x_0$ and found that after $x^{-1}_0$ there are negative solution. So how am I wrong guys? I don't think book is right.We can go beyond $x_0^{-1}$ without touching $x_0^{-1}$.
Look up your definition of solutions of a differential equation.
It will be something like: a differentiable function $x(t):I\to \mathbb{R}$ where $I$ is an interval, satisying the differential equation at each point $t\in I$.
In your case, if you extend the solution past $t=x_0^{-1}$ then either the function will not be differentiable everywhere (it's not in $t=x_0^{-1}$ as it blows up) or the domain of definition is not an interval. Thus it doesn't satisfy the definition.
The reason why we don't like solutions as being defined on disconnected domains is that you lose uniqueness unless you specify new conditions on each disconnected component.
For instance in your case, suppose you want to extend your solution $x(t)=\frac{1}{x_0^{-1}-t}$, defined for $t<x_0^{-1}$, past $t=x_0^{-1}$.
Then you get an infinite family of possible extensions, in the form $x_c(t)=\frac{1}{c-t}$ (for $t>x_0^{-1}$) where $c\in \mathbb{R}$ is completely arbitrary. They don't care at all about what happens before $t=x_0^{-1}$. Unless you specify $x(t)$ for some $t>x_0^{-1}$, you can't narrow it down to a single solution, which is what we expect with IVPs.