$f:\mathbb R \to \mathbb R$ be differentiable such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is $f(x)>0,\forall x>0$? Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is it true that $f(x)>0,\forall x>0$ ?

Let $y(x)=e^{-x}f(x)$. Then $ f$ (strictly) positive $ \iff  y$ (striclty) positive. 
$\forall x $,  $   y'(x)=e^{-x}(f'(x)-f(x)) \ge 0$ and if $x> 0, y'(x)>0$.
Therefore $y$ is positive for $x \ge 0$. 
Now lets suppose it exists $t>0$ such that $y(t)=0$. Then for  $\epsilon >0 $ small enough, because $y'(t) >0$, for $x \in ]t-\epsilon,t[$ , $y(x) < y(t) $, which is in contradiction with the previous point.
