This is a statement of a problem with $4$ sections. I've solved the other ones, but this I couldn't solve it:
Let $\dot{x}=\arctan(x(t)\cdot t)$, $x(t_0)=x_0$ be an IVP. Prove that if $x_0<0$, then $x(t)<0$ for all $\mathbb{R}$.
Can we say something about the solution without actually having it? How can I say how's the solution if I can't find it?
This is a standard argument in comparing solution to ODE, using the uniqueness theorem: since $f(t, x) = \arctan(xt)$ has $\frac{\partial f}{\partial x} = \frac{t}{1+t^2x^2}$ continuous everywhere, the solution to the initial value problem (IVP)
$$\begin{cases} x' = \arctan(xt), \\ x(t_p) = x_p\end{cases}$$
has unique solution.
Now assume that $x(t)$ satisfies the differential equation with $x(t_0) = x_0<0$. If $x(t) \ge 0$ for some $t>0$ (the case for $t<0$ is similar, then there is $t_p$ so that
$$ x(t_p) = 0, \ \ x_p(s) <0 \ \ \ \forall 0<s<t_p.$$
But note that $\tilde x(t) \equiv 0$ also satisfies the differential equation and $\tilde x(t_p) = 0$. But $x \neq \tilde x$ and thus it contradicts with the uniqueness theorem. This implies $x(t) <0$ for all $t\in \mathbb R$.