Implication of $f(t)\le2e^{5t},t\in[0,\infty)$

Question

Let $f:[0,\infty)\to \mathbb R$ be a $C^1$-function that fulfills: $$f'(t)\le 5f(t),\ t\in[0,\infty), \ and \ \ f(0)=2 $$ Prove that $f(t)\le2e^{5t},t\in[0,\infty)$. Can somebody give me some tips, pleasE?

Answer 1

By product rule we have $(e^{-5t} f(t))'=e^{-5t} (f'(t)-5f(t)) \leq 0$ so $e^{-5t} f(t)$ is decreasing. This gives $e^{-5t} f(t) \leq e^{-(5)(0)}f(0)=2$. Mutiply by $e^{5t}$ to finish the proof.

Answer 2

Let $g(t)=e^{-5t}f(t).$ Show that $g'(t) \le 0$ for $t \ge 0$. Hence $g$ is decreasing. Thus

$$g(t) \le 2$$

for all $t \ge 0.$

Answer 3

Let $f$ be a $\mathcal{C}^1$ function on $\mathbb{R}_+$ such that $ \forall t\geqslant 0,~f'(t) \leqslant 5f(t)$ and $f(0)=2$. Remark that \begin{align} \dfrac{\mathrm{d}}{\mathrm{d}t}\left(f(t)e^{-5t} \right) = \left(f'(t)-5f(t)\right)e^{-5t} \leqslant 0 \end{align} by assumptions, so that $f(t)e^{-5t}$ is non-increasing. Thus its value is less that $f(0) e^{-0}=2$, and $f(t)e^{-5t}\leqslant 2$.