If $f\left(0\right)=1$ and $f'\left(x\right)>3f\left(x\right)\ $ $∀\ x\ \ge0$ then prove that $f\left(x\right)\ge e^{3x}\ ∀\ x\ \ge0$

Q:

$f\left(x\right)$ is a continuous and differentiable function defined in $[0,\infty)$. If $f\left(0\right)=1$ and $f'\left(x\right)>3f\left(x\right)\ $ $∀\ x\ \ge0$ then prove that $f\left(x\right)\ge e^{3x}\ ∀\ x\ \ge0$

Approach:

Given $$f'\left(x\right)>3f\left(x\right)\ $$ Let $f\left(x\right)=t$, $$\to\frac{dt}{dx}>3t\ \to\frac{dt}{t}>3dx$$ since $t\ (or f(x)) >0$, equality sign doesn't remain same. $$\ln t>3x+C$$ Put $x=0\ \to f\left(0\right)=1$, We get, $$\ln1>0+C$$ $$0>0+C$$

What do I do now, how shall I proceed further, where am I wrong?

Also, $t\ (or f(x)) >0$ because $f\left(0\right)=1$ and $f'\left(x\right)>3f\left(x\right)\ $ so $f'\left(x\right)$ is always positive so function only takes positive values.

Need help

$$g\left(x\right)=\frac{f\left(x\right)}{e^{3x}}\to g\left(0\right)=\frac{f\left(0\right)}{e^{0}}\to g\left(0\right)=1$$ Now, $$g'\left(x\right)=\frac{f'\left(x\right)-3f\left(x\right)}{e^{3x}}$$ Since, $f'\left(x\right)-3f\left(x\right)>0\ $ for all $x∈[0,∞)$, $g'\left(x\right)>0$ in x $∈[0,∞)$

Thus, $$g\left(x\right)\ge1$$and so $$f\left(x\right)\ge e^{3x}\ in\ x∈[0,∞)$$

Let $p(x)\ge 0$ such that

$$ f'(x)-p(x) = 3f(x) $$

Using the Laplace transform we have

$$ (s-3)F(s) = P(s)+f(0)\Rightarrow F(s) = \frac{1}{s+3}P(s)+\frac{1}{s+3}f(0) $$

or inverting

$$ f(x) = e^{3x}\circledast p(x)+f(0)e^{3x} $$

now as $f(0) = 1$ and $e^{3x}\circledast p(x)\ge 0$ we have that

$$ f(x) \ge e^{3x} $$