Mathematics - Differential Equation Question with Solution | TestHub

Given that,

$f^{\text{'}}\left(x\right)>2 f\left(x\right) \forall x\in R$

$\Rightarrow f^{\text{'}}\left(x\right)-2f\left(x\right)>0 \forall x\in R$

$\therefore e^{-2x} \left(f^{\text{'}}\left(x\right)-2f\left(x\right)\right)>0 \forall x\in R$

$\Rightarrow \frac{d}{dx}\left(e^{-2x} f\left(x\right)\right)>0 \forall x\in R$

Let  $g\left(x\right)=e^{-2x} f\left(x\right)$

Now,  $g^{\text{'}}\left(x\right)>0 \forall x\in R$

$\Rightarrow g\left(x\right)$ is strictly increasing $\forall x\in R$

Also  $g\left(0\right)=1$

$\Rightarrow g\left(x\right)>g\left(0\right)=1$

$\therefore e^{-2x}. f\left(x\right)>1 \forall x\in \left(0, \infty \right) \Rightarrow f\left(x\right)>e^{2x} \forall x\in \left(0, \infty \right)$

As,  $f^{\text{'}}\left(x\right)>2 f\left(x\right)>2e^{2x}>2\forall x\in \left(0, \infty \right)$

$\Rightarrow f\left(x\right)$  is strictly increasing on $x\in \left(0, \infty \right)$