Prove that a differential function is bounded by 1/2

[DeadOriginal]

Homework Statement

Suppose $\phi(x)$ is a function with a continuous derivative on $0\leq x<\infty$ such that $\phi'(x)+2\phi(x)\leq 1$ for all such $x$ and $\phi(0)=0$. Show that $\phi(x)<\frac{1}{2}$ for $x\geq 0$.

The Attempt at a Solution

I tried to solve this like I would any other first order differential equation.
$$
\phi'(x)+2\phi(x)\leq 1\Leftrightarrow e^{2x}(\phi'(x)+2\phi(x))\leq e^{2x}\Leftrightarrow e^{2x}\phi(x)\leq e^{2x}
$$
so
$$
\phi(x)\leq e^{-2x}\int\limits_{x_{0}}^{x}e^{2t}dt + ce^{-2x}=\frac{1}{2}+ce^{-2x}
$$
but that was as far as I could get. Any help would be greatly appreciated.

 

[LCKurtz]

Homework Statement

Suppose $\phi(x)$ is a function with a continuous derivative on $0\leq x<\infty$ such that $\phi'(x)+2\phi(x)\leq 1$ for all such $x$ and $\phi(0)=0$. Show that $\phi(x)<\frac{1}{2}$ for $x\geq 0$.

The Attempt at a Solution

I tried to solve this like I would any other first order differential equation.
$$
\phi'(x)+2\phi(x)\leq 1\Leftrightarrow e^{2x}(\phi'(x)+2\phi(x))\leq e^{2x}\Leftrightarrow e^{2x}\phi(x)\leq e^{2x}
$$
so
$$
\phi(x)\leq e^{-2x}\int\limits_{x_{0}}^{x}e^{2t}dt + ce^{-2x}=\frac{1}{2}+ce^{-2x}
$$
but that was as far as I could get. Any help would be greatly appreciated.

You almost have it. After you multiply by your integrating factor you have$$
(e^{2x}\phi(x))'\le e^{2x}$$Instead of doing an indefinite integral, integrate this from $0$ to $x$:$$
\int_0^x(e^{2t}\phi(t))'~dt\le \int_0^x e^{2t}~dt$$and see what happens.

 

[DeadOriginal]

Ahh! I see it! Thanks!