Question

If $\varphi$ is a differentiable function with $\varphi \left(0\right)=0$ and ${\varphi }^{\prime }\left(x\right)+2\varphi \left(x\right)\le 1$ then maximum value of $\varphi \left(x\right)$ is

Answer 1

${\varphi }^{\prime }\left(x\right)+2\varphi \left(x\right)\le 1;$ Put $y=\varphi \left(x\right)$
$\frac{dy}{dx}+2y\le 1$
$e^{2x}\frac{dy}{dx}+2y{e}^{2x}\le e^{2x}$
$\frac{d}{dx}\left(y{e}^{2x}\right)\le e^{2x}$
$\therefore {\int }_0^1\frac{d}{dx}\left(y{e}^{2x}\right)dx\le {\int }_0^1{e}^{2x}dx$
$\therefore y{e}^{2x}{|}_0^t\le \frac{e^{2t}}{2}-\frac{1}{2}$
$\therefore e^{2t}\varphi \left(t\right)-\varphi \left(0\right)\le \frac{e^{2t}}{2}-\frac{1}{2}$
$\therefore \varphi \left(t\right)\le \frac{1}{2}\forall t$