Let $ E, J \subset \mathbb R$ be open intervals and let functions $h:J \to \mathbb R$ and $g: E \to \mathbb R$ be continuous. let $\xi \in E$ and assume that $g(\xi)=0$. Define $f:J \times E \to \mathbb R$ by setting $f(t,z)=h(t)g(z)$ and consider the differential equation $\dot x(t) = f(t,x(t))=h(t)g(x(t)))$. Let $x: I \to E$ be a solution.
I have some assumptions:
(1) $g(z)\neq 0$ for all $z \in [\xi, \xi +\delta]$ where $\delta > 0$ and that the improper interal $\int_ \xi^{\xi + \delta} \frac{dz}{g(z)} $ is divergent
(2) There exists $\tau \in I$ such that $x(\tau)>\xi$.
How can I show that $x(t) >\xi$ for all $t \in I$??
Any help would be appreciated.
Here are some thoughts (I don't really know about these kinds of differential equations, however.)
Suppose that there is some $t$ such that $x(t) \leq \xi$; then, since $x$ is continuous on $J$, there is a $c \in J$ such that $x(c) = \xi$. More specifically, if we assume (without loss of generality) that $\delta \leq x(\tau)$, the interval $(\xi, \xi+\delta)$ has a nonempty open preimage under $x$; let $(c,d)$ be one of the disjoint open intervals making up this preimage such that $x(c) = \xi$ and $x(d) = \xi + \delta$ (or vice versa, in which case modify the following accordingly).
Then for $t$ in the interval $(c,d)$, $g(x(t)) \neq 0$, so $h(t) = \dfrac{\dot{x}(t)}{g(x(t))}$.
So $$ \int_c^d h(t) dt = \int_c^d \frac {\dot{x}(t)}{g(x(t))} dt = \int_{\xi}^{\xi+\delta} \frac{1}{g(u)} du $$ where we have used substitution with $u = x(t)$. This is a divergent integral, which is a contradiction, since $h$ is continuous on $[c,d]$ and thus integrable.