Question

If $P\left(1\right)=0$ and $\frac{dP\left(x\right)}{dx}>P\left(x\right)$ for all $x\ge 1$, then prove $P\left(x\right)>0$ for all $x>1$

Answer 1

Given that $\frac{dP\left(x\right)}{dx}>P\left(x\right),\forall x\ge 1$ and $P\left(I\right)=0$
$\Rightarrow \frac{dP\left(x\right)}{dx}-P\left(x\right)>0$
Multiplying by $e^{-x}$, we get
$e^{-x}\frac{dP\left(x\right)}{dx}-e^{-x}P\left(x\right)>0$
$\Rightarrow \frac{d}{dx}\left[e^{-x}P\right(x\left)\right]>0$
$\Rightarrow e^{-x}P\left(x\right)$ is an increasing function
$\therefore \forall x>1,e^{-x}P\left(x\right)>e^{-1}P\left(1\right)=0$ [Using $P\left(1\right)=0$]
$\Rightarrow e^{-x}P\left(x\right)>0,\forall x>1$
$\Rightarrow P\left(x\right)>0,\forall x>1$. $[\because e^{-x}>0]$