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$f(x)$ is continuous and differentiable function defined in $x \geq 0$. If $f(0)=1$ and $f'(x) >3 f(x)$, then prove that $$f(x) \geq e^{3x}, \forall x \geq 0$$

Could someone give me little hint as how to proceed in this question? I am not able to initiate.

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$g(x)=f(x)e^{-3x}, g'(x)=e^{-3x}(f'(x)-3f(x))>0$ so $g$ increases and $g(0)=1$.

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All you need is $$ (f(x)e^{-3x})'=e^{-3x}(f'(x)-3f(x)). $$

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From the given conditionwe obtain that $\frac{f'(x)}{f(x)} $ $>$ 3 . Now let $\frac {f'(x)}{f(x)}$ be $F(x)$. So we know that if $F(x) > G(x)$in some interval $(0,n)$, $\int_0^n $ $F(x) dx$ $>$ $\int_0^n $ $G(x) dx $ . So here we get that $ \int_0^n $ $\frac{f'(x)}{f(x)} dx$ > $\int_0^n $ $3$ $dx$ for some value of $n$ greater than $0$. Now making the substituion $f(x)$ $=$ $t$ and solve the integral to get $ln f(n) $ > $3n$, Now $n$ has a domain from (0,$\infty$).So for any value of $x$ within that domain it should satisfy $lnf(x)$>= $3x$ thus $f(x)$ > $e^{3 x}$ and since domain of $x$ = domain $n$ thus it can be told for all values of $ x$ > $0$ it will satisfy this relation.But from the given condition that $f(0)$ = $1$ we see that $f(0)$=$e^{3 \cdot 0}$ $=$ $1$ and hence it is proved that $f(x)$ $\ge$ $e^ {3x}$.

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