Hi I'm trying to understand the most general conditions under which I can conclude finite time blow up of an ODE of the form $\dot{x} = f(x)$ with initial condition $x_0 > 0$ and $f(x) \geq 0$ for all $x \geq 0$.

If I re-write this in a separable way so that $dt = \frac{dx}{f(x)}$ then I want to determine if there is some finite time $T$ for which $x(T) = \infty$. Ok so it is clear to me that if $\int_{x_0}^{\infty} \frac{dx}{f(x)} = \infty$ then there is **no finite time blowup** for the initial condition $x_0$ (since we would need some finite $T$ for which $x(T)=\infty$ and this precludes that). I'm confused however if this is a necessary condition. In other words, if I know that $\int_{x_0}^{\infty} \frac{dx}{f(x)} < \infty$ does this necessarily tell me that I **DO** get finite time blowup? This is my question. It's not entirely clear to me why this should *prove* finite time blowup.

Two examples to keep in mind in all of this are $f(x) = x^2$ (finite time blowup) and $f(x) = x^{1/2}$ (no finite time blowup).

Osgood’s conditionand known since 1898. See Walter'sOrdinary Differential Equations(1998 ed., Springer GTM 182), p. 147. – Fizz Jan 29 '15 at 14:16