$f:\mathbb R \to \mathbb R$ 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 $f'(x)&gt;f(x),\forall x \in \mathbb R$ ; then is it true that $f(x)&gt;0,\forall x&gt;0$ ?
Because $f'(x)&gt;f(x)$ we have that $ e^{-x} f(x)$ is strictly increasing.That is the conclusion.
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 $x&gt; 0, y'(x)&gt;0$.

Therefore $y$ is positive for $x \ge 0$. 

Now lets suppose it exists $t&gt;0$ such that $y(t)=0$. Then for  $\epsilon &gt;0 $ small enough, because $y'(t) &gt;0$, for $x \in ]t-\epsilon,t[$ , $y(x) &lt; y(t) $, which is in contradiction with the previous point.