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For $\ x(0)\equiv x_0>0\ $ and a system governed by $\dot x(t)=-k\ x(t),$ I find that $x(t)>0\ \ \ \forall\ t.$ (Because the solution is $x(t)=e^{-kt}x_0$.)

For which $f$ and $\dot x(t)=f(x(t)),$ does this property hold?

I don't know how difficult this is, but the question might be generalized to functions $f(x(t),t)$ and/or higher order ODEs.

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    @Rahul Narain: Thanks for the remark. Turns out that the solution of $\dot x(t)=(-\frac \alpha T)\ x(t)^{1-\frac{1}{\alpha}}$ is $x(t)=(1-\frac t T)^\alpha$ for all finite $\alpha$ and these are zero for $t=T$.2012-01-16

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If uniqueness holds, then $f(0)=0$ is enough, since $x(t)\equiv0$ is a solution, and two solutions cannot cross.

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    One possibility (again assuming existence and uniqueness) is "There exists $a$ such that $f(a)=0$". If $a\ge0$ and x_0>a, then the solution of $x'=f(x)$, $x(0)=x_0$ is such that x(t)>a\ge0 for all $t$. Similarly, if a<0 and x_0, then the solution of $x'=f(x)$, $x(0)=x_0$ is such that x(t) for all $t$.2012-01-17