How to plot a phase space I was trying to study the function $x(t)=\int_0^t e^{-s^2}ds$, which I did successfully using high school maths. After that, I decided that I wanted to try to study it using some multivariable calculus, but couldn't really do it, so I'm asking for the proper way to face this: the problem is equivalent to the ODE $x'(t)=e^{-t^2}$ given some initial value. Let $y(t)=t$ and therefore $y'(t)=1$. Now I want to study the phase space of the system $\begin{cases} x'=e^{-y^2}\\ y'=1 \end{cases}$. How can I do it? There are no equilibrium points and the Jacobian's only eigenvalue is $0$. Also, the linearized system is $\begin{cases}x'=(-2y_0e^{-y_0^2})\cdot y \\ y'=0\end{cases}\;$ which makes no sense since in $(0, 0)$ it gives $x'=0, y'=0$ whereas in the original one it is $x'=1, y'= 1$. What's the proper way to study this system?

Yes, there are no equilibrium points, so it has no meaning to compute the Jacobian at the fixed points. Also no meaning of linearization around a fixed point.
We have $\dot{x} = e^{-y^2}$ and $\dot{y} = 1$, we see that the $\dot{y}=1$ is independent of $x$ and $y$, so we have $y = t + c$, $c$ an integration constant. So substituting the expression of $y$ in $\dot{x} = e^{-y^2}$, we have $x = D \int_{0}^{t} e^{-s^2} ds$, $D$ is integration constant. This definite integral reminds me of the error function, whose plot if you have in mind then it would be handy.
In order to plot the phase space that is plot $x$ vs $y$ we need to have a plot of $D$ erf(t) vs $t$, same as the graph of erf(x).
So we have the vector field, phase plane as below for different initial conditions.


