global solutions for ODEs I'm searching for related results for the following problem:

Consider the ODE:
  $$
x'=-3x+ x^2\log x,\quad x(0)=3/2. 
$$
  Does a solution $x(t)$ exist on $t\in[0,\infty)$?

A quick research on Google returns mainly the local theory such as Picard's existence theorem. 
Can anybody come up with some references about global existence of solutions to ODEs?
In the problem quoted above, the function 
$$
f(x)=-3x+ x^2\log x
$$
is locally Lipschitz. I'm not sure if this can guarantee the existence of solution on $[0,\infty)$. 
 A: The main tool is still Picard; we just have to keep controlling the solutions all the way to $t=\infty$, not letting it escape via a vertical asymptote. For example, if $f:\mathbb R\to\mathbb R$ is globally Lipschitz, then the solution of $x'=f(x)$ is global, because the intervals we get from Picard's theorem have infinite total length. (Sketch: starting with $(t_0,x_0)$, we have safety intervals of size roughly $\sim |f(x_0)|^{-1}$, within which the solution will exist and will not increase by more than $1$. Repeat and use the fact that the harmonic series diverges.)
The parenthetical is so sketchy because it does not apply to your situation: your $f$ is not globally Lipschitz. The function $f(x)=-3x+x^2\log x$ looks like this: 

If you started with $x_0$ such that $f(x_0)>0$, the solution would keep on increasing, and since $f$ grows faster than $x^p$ for some $1<p<2$, it would blow up in finite time. 
But, you start with $3/2$, where $f$ is negative. So, the solution moves to the left on the $x$-axis. Now we should be worried about hitting $x=0$. But this will not happen. Indeed, extending $f$ by $0$ to the negative semiaxis, we have a Lipschitz function on $(-\infty,3/2]$. There is an equilibrium solution $x(t)\equiv 0$. By the uniqueness theorem, a nonzero solution cannot reach  this equilibrium in finite time; if it did, uniqueness would fail going backward in time. 
Thus, the solution never reaches $0$, staying positive and decreasing all the time.
