solution of first order differential equation and maximal interval Find the solution of $x' = x^2t$ with initial value $x(0) = x_{0}$. Determine the
maximal interval where it exists, depending on $x_{0}$
Please help me find the maximal interval!
 A: By separating and solving, the solution may be expressed as
$$\frac{1}{x_0} - \frac{1}{x} = \frac{1}{2} t^2$$
The LHS must be $>0$ as the RHS is.  Therefore the maximal range of $x(t)$ is $\{x:x>x_0\}$.  The maximal interval for $t$ that meets this condition on $x$ is then
$$ \left\{t:0<t<\sqrt{\frac{2}{x_0}} \right \}$$
A: An initial value problem $x' = f(x)$ with $x(0) = x_0$, has a unique solution defined on some interval $(-a, a)$.
This IVP has a unique solution $x(t)$ defined on a maximal interval of existence $(\alpha,~\beta)$.
Furthermore, if $\beta \lt \infty$ and if the limit 
$$x_1 = \lim_{t\rightarrow \beta^{-}} x(t)$$
exists then $x_1 \in \dot E$, the boundary of $E$. The boundary of the open set $E$, $\dot E = \overline E$ ~ $E$ where $\overline E$ denotes the closure of $E$.
On the other hand, if the above limit exists and $x_1 \in E$, then $\beta = \infty$, $f(x_1) = 0$ and $x_1$ is an equilibrium point of the IVP.
The domain of a particular solution to a differential equation is the largest open interval containing the initial value on which the solution satisfies the differential equation.
Theorem (Maximal Interval of Existence). An IVP has a maximal interval of existence, and it is of the form $(t^{-}, t^{+})$, with $t^{-} \in [-\infty, \infty)$ and $t^{+} \in (-\infty, \infty]$. There is a unique solution $x(t)$ on $(t^{-}, t^{+})$, and $(t, x(t))$ leaves every compact subset $\mathcal K$ of $\mathcal D$ as $t \downarrow t^{-}$ and as $t \uparrow t^{+}$.
Proof See ODE Notes.
More Examples of Domains See the very readable section More Examples of Domains.
Example
$$x' = x^2, ~x(0) = 1$$
has the solution 
$$x(t) = \frac{1}{1-t}$$
defined on its maximal interval of existence $(\alpha, ~ \beta) = (-\infty, 1)$.
Furthermore, $x_1 = \displaystyle \lim_{t\rightarrow 1^{-}} x(t) = \infty$. You can do the other side.
Original Problem
For your problem, we have:
$$\tag 1 x' = x^{2}t, ~ x(0) = x_0$$
Solving $(1)$, yields: $x(t) = \large -\frac{2}{c + t^{2}}$
Using the IC, $x(0) = x_0$, yields,
$$\tag 2 \large x(t) = -\frac{2}{t^{2} - \frac{2}{x_0}}$$
Depending on $c = \large \frac{2}{x_0}$, where $c \in \mathbb{R}$, there are several cases:


*

*if $c \lt 0$, then $x(t) = \large x(t) = -\frac{2}{t^{2} - \frac{2}{x_0}}$ is a global solution on $\mathbb{R}$,

*if $c \gt 0$, the solutions are defined on $(-\infty, -\sqrt{c})$, $(-\sqrt{c}, \sqrt{c})$, $(\sqrt{c}, \infty)$. The solutions are maximal solutions on $\mathbb{R}$, but are not global solutions.

*if $c = 0$, then the maximal non global solutions on $\mathbb{R}$ are defined on $(-\infty, 0)$ and $(0, \infty)$.
Note for completeness, that there is another solution $x(t) = 0$, which is a global solution on $\mathbb{R}$.
I would strongly suggest:
$(1)$ That you review the solution and the different results for varying "c" and plot those to make sure you understand them.
$(2)$ That you use the initial definition I gave above, which works for $t^{-}$ and $t^{+}$ and make sure you can do it the way I showed and using that argument as per the theorem.
Regards
