If $\vert F(t,x)\vert \leq \alpha (t)\vert x\vert + \beta(t)$ the maximal solutions are global for an ODE Let $F: \mathbb{R} \times\mathbb{R}^2 \rightarrow \mathbb{R}$ localy lipschitz in its second variable. Let the Cauchy problem be:
$$x'=F(t,x),\\  x(0) = x_0 $$
Let $\alpha : \mathbb{R} \rightarrow \mathbb{R}$ and $\beta : \mathbb{R} \rightarrow \mathbb{R}$. Show that if $\forall x \in \mathbb{R}^2, \forall t \in \mathbb{R}$ $$|F(t,x)| \leq \alpha(t)|x| + \beta (t) $$
then any maximal solution is global.
I am unable to do this. I suppose I have to either show that $F$ is uniformly bounded or that it is uniformly Lipschitz, but I have no clue how to do either. Any help will be appreciated.
 A: See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|\le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution,  and why the bound prevents finite boundaries of the maximal domain.
A: Suppose $x(t)$ is not global, namely $\exists t_0>$ such that
$$ \lim_{t\to\to t_0^-}|x(t)|=\infty. \tag{1}$$
Integrating from $0$ to $t<t_0$ gives
$$ x(t)=x(0)+\int_0^tF(s,x)ds $$
which implies
$$ |x(t)|\le |x(0)|+\int_0^t|F(s,x)|ds. $$
Using $F(t,x)\le\alpha(t)|x|+\beta(t)$, one has
$$ |x(t)|\le |x(0)|+\int_0^t(\alpha(s)|x(s)|+\beta(s))ds.\tag{2}$$
Define
$$ g(t)=\int_0^t(\alpha(s)|x(s)|+\beta(s))ds \tag{3} $$
and then
$$ g'(t)=\alpha(t)|x(t)|+\beta(t). \tag{4} $$
By (1), it is not hard to see
$$ \lim_{t\to\to t_0^-}g(t)=\infty. \tag{5}$$
Using (3) and (4) in (2), one has
$$ g'(t)-\beta(t)\le|x_0|\alpha(t)+\alpha(t)g(t) $$
or
$$ g'(t)-\alpha(t)g(t)\le\beta(t)+|x_0|\alpha(t). $$
It is easy to see
$$ \left(e^{-\int_0^t\alpha(s)ds}g(t)\right)'\le (\beta(t)+|x_0|\alpha(t))e^{-\int_0^t\alpha(s)ds}. $$
Integrating from $0$ to $t$, one has
$$ g(t)\le g(0)e^{-\int_0^t\alpha(s)ds}+ e^{-\int_0^t\alpha(s)ds}\int_0^t(\beta(r)+|x_0|\alpha(r))e^{-\int_0^r\alpha(s)ds}dr. $$
Using this, one has
$$ g(t)\le C, t\in[0,t_0]$$
for some constant $C$, which is against (5).
