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
For part a), the problem asks you to draw the direction field without solving the ODE. Writing it in this form:
$$y'=2e^{-t}-(1+\frac{1}{t})y$$
Try some $y$ values. For example, let $y=0,y'=2e^{-t}$. That will give you the direction field on $t$ axis. They are arrows pointing upward, but approaching 0 direction, toward some equilibrium value.
Let $y=1, y'=2e^{-t}-(1+\frac{1}{t})$. This gives you the direction field on the line $y=1$. They are arrows pointing downward, approaching $0$ direction, toward the same equilibrium value.
Especially you should look at the direction of the points when $t=1$.
So what is the critical value of $y$ at $t=1$, such that the field changed direction? I believe you can find that.
Best Answer
Note that
$$ y' = \frac{x}{4y} $$
is undefined at $y=0$. Thus, you cannot have a solution that contains $y=0$. Any solution must be either $y<0$, or $y>0$.
Since the given initial condition is $y(x_0)=1$, the solution must be $y > 0$, i.e.
$$ y(x) = \frac{\sqrt{x^2-x_0^2 + 1}}{2} $$
You can then find the domain of this function depending on $x_0$