Show that the Initial Value Problem does not cross its equilibrium points. Let $x'=(x-1)(x-4)(x+5)$ and $x(0)=3$ be an Initial Value Problem. Prove  the existence and uniqueness of the solution and that the solution will be trapped in between $x=1$ and $x=4$ lines and that it can be extended in the entire line.
Any help is appreciated.
Thank you,
Klara
 A: Note: I misinterpreted the question as looking for a solution for all initial conditions, whereas only $x(0) = 3$ is of interest, so this answer is a bit of overkill.
Let $f(x) = (x-1)(x-4)(x+5)$. A plot of $f$ is given below.

$f$ has zeros at $\{-5,1,4\}$, is strictly positive on $(-5,1)\cup(4,\infty)$ and strictly negative on $(-\infty, -5)\cup (1,4)$. $f$ is smooth, so is locally Lipschitz on bounded intervals. Hence solutions exist locally and are unique and smooth. We have $x(t) = x(0)$ is the unique solution for initial conditions $x(0) \in \{-5,1,4\}$.
The state space is $\mathbb{R}$, hence if $x(0)$ lies between two zeros of $f$, it must remain between the two zeros. To see this, suppose $z_1 \leq x(0) \leq  z_2$, where $z_1,z_2$ are zeros of $f$ and suppose for the sake of contradiction that $x(t_1)$ lies outside the interval $[z_1,z_2]$. Continuity implies that for some $t_0 \in [0,t_1)$, $x(t_0) \in \{z_1,z_2\}$. However uniqueness of solution would imply that $x(t) \in \{z_1,z_2\}$ for all $t \geq t_0$, which is a contradiction. Hence $x(t)$ remains in the interval $[z_1,z_2]$. Since a uniform Lipschitz constant applies to $f$ for $x \in [z_1,z_2]$, we see that there exists a unique solution defined for all $t \geq 0$ (and $t<0$, but that is not a concern here).
This establishes the existence of a solution starting from $x(0) \in [-5,4]$.
However, it turns out that if $x(0)$ lies outside this interval, the solution 'blows up' in finite time. To see this, we again proceed by contradiction. Suppose a solution $x(t)$ exists for all $t \geq 0$, and $x(0) \notin [-5,4]$.
Note that $x(t)$ is increasing if $x(0) > 4$, and decreasing if $x(0) < -5$. So, if $x(0) < -5$, then $x(t) \leq x(0)$ for all $t \geq 0$ and if $x(0) > 4$, then $x(t) \geq x(0)$ for all $t\geq 0$.
Using partial fractions, we can rewrite the differential equation as $(\frac{1}{x-5} - \frac{3}{x-1}+\frac{2}{x-4}) \dot{x} = 54$. This can be written as (cf. separation of variables)
$\frac{d}{dt} (\ln \phi(x)) = 54$
where
$\phi(x) = \frac{(x+5)(x-4)^2}{(x-1)^3}$.
Note that $\phi(-5) = \phi(4) = 0$, and $\lim_{|x|\to \infty} \phi(x) = 1$. A tedious analysis shows that $\phi'(x) = \frac{54(x-4)}{(x-1)^4}$. It follows that $\phi(x) \in (0,1)$ on $(-\infty, -5) \cup (4,\infty)$. It also follows that  $t \to \phi(x(t))$ is increasing.
We can integrate the differential equation to get $\phi(x(t)) = \phi(x(0)) e^{54 t}$, which contradicts the boundedness of $\phi$ on $(-\infty, -5) \cup (4,\infty)$.
