Why we can not define the solution of the differential equation beyond a particular $t$? I was studying Ordinary Differential Equations, and my book was trying to explain the maximal interval of solution of a differential equation. If we consider this differential equation
$$
\frac{dx}{dt} = x^2 \\
\text{with initial condition}~x(0)= a,~~~a \gt 0
$$
Then, the solution to this equation is
$$
x(t) = \frac{1}{ a^{-1} - t}$$
Now, if $t$ starts to increase from $0$ the denominator will decrease and consequently $x(t)$ will increase, when $t =a^{-1}$ the denominator is $0$ and $x(t) = + \infty$. As we let $t$ to decrease from $0$, our denominator will increase and finally when $t = -\infty$, $x(t) = 0$. So, the solution of the differential equation is defined on the interval $(-\infty, a^{-1})$. But the problem comes when the book makes this statement

but there is no way to define the solution that extends further into the future beyond $t=a^{-1}$.

Why the function is not defined for $t \gt a^{-1}$? There is a discontinuity at $t=a^{-1}$ but beyond that the function is nice, why the solution is not defined after $t = a^{-1}$?
If compare $x(t)$ with some other simple functions like, for example, $f(x) = \frac{1}{2-x}$, $f(x)$ is well defined after $x = 2$, and here we have its graph:
.
Why the book says that beyond $t=a^{-1}$ the solution is not defined?
 A: Essentially, it depends on what we mean by a "solution", and what is meant by "extending" a solution. What you define as a solution of an initial value problem is somewhat a matter of taste, but here's a common approach.
For a first order initial value problem
$$
\frac{dx}{dt}=f(x,t)\ \ \ \ \ \ \ x(t_0)=x_0
$$
we may define a solution as a function $x:I\to\mathbb{R}$ where $I=(t_\min,t_\max)$ is an open interval in $\mathbb{R}$ containing $t_0$, which satisfies the initial condition and satisfies the differential equation wherever it is defined.
Why do we demand that solutions be defined on an open interval, and not some other domain? There are many reasons; one is that we would like solutions to be unique. Your IVP gives an example of what can go wrong if we don't do this: we could define $x(t)=(a^{-1}-t)^{-1}$ for $t\neq 0$, which satisfies the IVP. We could just as easily, however, define
$$
x(t)=\begin{cases}
\frac{1}{a^{-1}-t} & t<a^{-1} \\
\frac{1}{b^{-1}-t} & t>a^{-1}
\end{cases}
$$
for any $b\ge a$. This is just as much a "solution" as the first choice, and so there are infinitely many distinct "solutions" in this sense. If we consider even more disconnected domains all kinds of different functions will be possible "solutions". This happens in general if we allow disconnected domains.
Of course, even by requiring solutions be defined on an open interval, there is still some redundancy, since we may always restrict to a smaller interval still containing the initial condition and obtain a new solution. If we want to define a solution which can be unique, we can look for the "largest possible" interval. To this end, given a solution $x:I\to\mathbb{R}$, we say another solution $y:I'\to\mathbb{R}$ is an extension of $x$ if $I\subset I'$ (that is, $I'$ contains $I$ and is strictly larger that $I$) and $x$ and $y$ agree on $I$. A maximally extended solution is one which has no extensions. Maximally extended solutions don't always exist (and they aren't always unique), but they do exist (and are unique) in many cases, such as when $f$ is sufficiently smooth.
In the sense defined above, the solution given by the text for your IVP $x:(-\infty,a^{-1})\to\mathbb{R}$ is maximally extended. We cannot choose a larger interval $(-\infty,c)$ with $c>a^{-1}$, since the function defined on that interval would fail to be differentiable at $a^{-1}$, and thus not satisfy the differential equation on its domain.
