On the extension of the solution to a nonlinear ODE Consider the nonlinear ODE $$x' = (x^2 - e^{2t})f(t, x)$$ with $f$ continuous. Prove that for any $\tau > 0$, if $|x_0|$ is sufficiently small, the solution $x(t)$ to the ODE above can be extended to $\tau \leq t < \infty$. 
My general sketch:
1) If I can show that for any compact subset $\Omega \subset \mathbb{R}^n $, the solution $x(t)$ hits the boundary point of $\Omega$ then I can keep extending the solution to infinity; 
2) Equivalently, I was trying to show that the integral $$\int_{x_0}^{x_0 + \alpha} \frac{1}{F(t, x)}$$ diverges,  where $F(t, x) = (x^2 - e^{2t})f(t, x)$; 
3) The third way I was thinking is that I can use the fact $|F(t, x)| \leq K|x_0|$, the maximum interval of existence is $(-\infty, \infty)$. But I can't see to bound it. 
 A: The idea is to use the equation, and a variation on Gronwal's inequality to show the solutions $x(t)$ that start near $0$  remain bounded for al $t$.  
Not sure what to make of $\tau$. The way the problem is stated, it must have come  within a context in which referring to $\tau$ is significant.
Let $x_0$ be such that $|x_0|<1$, then, the solution $x(t)$ of the ODE that starts at $x_0$ satisfies $|x(t)|\ < 1 \le e^t$ for all small $t \ge 0$. I claim that $|x(t)| < e^t$ for all $t$. If that were not the case, there would be $T>0$ such that $|x(t)| < e^t$ for all $t \in [0,T)$ and $|x(T)| = e^T$. So far we have only used the continuity of $x$. Let's use the equation. For $t \in [0,T)$,
$$|x'(t)| = (e^{2t} - x^2)|f(t,x)| \le (e^{2T}-x^2)M_T$$ 
where $M_T = sup\{|f(t,u)| | 0 \le t \le T\text{ and } |u| \le e^t \}$, therefore
$$| \int_{0}^{t} \frac{x'(s)}{e^{2T}-x^2}ds| \le M_Tt.$$   
The integral can be simplified as follows:
$$\int_{0}^{t} \frac{x'(s)}{e^{2T}-x^2}ds = \int_{x_0}^{x(t)} \frac{1}{e^{2T}-x^2}dx = \frac{1}{e^T} ln(\frac{1+u}{1-u}){\LARGE|}_{e^{-T}x_0}^{e^{-T}x(t)} $$
so:  
$$|  ln(\frac{e^T + x(t)}{e^T - x(t)}) | \le 
|  ln(\frac{e^T + x_0}{e^T - x_0}) | + e^TM_Tt$$
By our choice of $T$, as $t \nearrow T$, $x(t) \rightarrow \pm e^T$, in either case, 
$|  ln(\frac{e^T + x(t)}{e^T - x(t)}) |  \rightarrow +\infty$, which contradicts the previous inequality. $\blacksquare$ 
