If $f$ have solution, then $g$ too Lets $f$ and $g$ two positive functions on $\mathbb{R}$, with $g\leq f$ in all point. Assume that the initial value problem $$x'=f(x),\mbox{ } x(0)=0$$ Admits solution over $\mathbb{R}$. Show that the initial value problem $x'=g(x)$, $x(0)=0$, Also supports solution on $\mathbb{R}$
Both equations may have a solution but I do not understand how they can relate to inequality and how one assures the other, it does not tell me anything about $g$ or $f$, I can not apply any theorem I know, help
 A: If, as I suppose, the assumptions are $f,g\in C(\mathbb{R})$, $0 < g(x) \leq f(x)$ for every $x\in\mathbb{R}$, then the stated result is true.
Namely, let us define the functions
$$
F(x) := \int_0^x \frac{1}{f(s)}\, ds,
\qquad
G(x) := \int_0^x \frac{1}{g(s)}\, ds,
\qquad x\in\mathbb{R}.
$$
These two functions are of class $C^1$, strictly increasing, and
$$
(1) \quad F(0) = G(0) = 0,\qquad
G(x) \geq F(x) > 0\ \forall x>0,
\qquad
G(x) \leq F(x) < 0\ \forall x<0.
$$
By assumption the Cauchy problem $x'=f(x)$, $x(0) = 0$ admits a global solution $u$ defined in all $\mathbb{R}.$
This can happen if and only if the image of $F$ is $\mathbb{R}$, i.e. if and only if
$$
(2) \qquad
\lim_{x\to +\infty} F(x) = +\infty,
\qquad
\lim_{x\to -\infty} F(x) = -\infty.
$$ 
In this case the solution $u$ to the Cauchy problem is unique, globally defined, and $u(t) = F^{-1}(t)$, $t\in\mathbb{R}$.
On the other hand, from the inequalities in (1) and from (2),
by comparison we get
$$
\lim_{x\to +\infty} G(x) = +\infty,
\qquad
\lim_{x\to -\infty} G(x) = -\infty,
$$
hence also the Cauchy problem $x'=g(x)$, $x(0) = 0$ admits a unique global solution given by $v(t) = G^{-1}(t)$, $t\in\mathbb{R}.$
