TITLE: Show that the Limit of ODE Solutions Tends to $0$
QUESTION [6 upvotes]: I am revising some exercises from a course on ODE's that I have not managed to do. This exercise asks the reader to show that any solution to an ODE of the following form goes to $0$ as $t\rightarrow\infty$: $$y'+a(t)y=f(t)$$
with $a(t)$ and $f(t)$ continuous, $a(t) \geq c>0$ and $\lim\limits_{t\rightarrow\infty}f(t)=0$.
We can find an integrating factor for this ODE, namely $\mu(t)=e^{\int a(t)}$, so that we find that any solution looks like $$y = e^{-\int a(t)}\int f(t)e^{\int a(t)}$$ and from the property that $a(t) \geq c>0$ we can conclude that $$|y| \leq e^{-ct}\int f(t)e^{\int a(t)}$$
however I don't know how to continue from here. Any help would be appreciated.

REPLY [3 votes]: Let $0 < \varepsilon \le 2$, let $T > 0$ be such that $|f(t)| \le \varepsilon$ for $t \ge T$, and set $z(t) = |y(t)|^2$. Then for $t \ge T$
$$
\frac{d}{dt} z(t) = - 2a(t) z(t) + |f(t)y(t)| \le -2cz(t) + \varepsilon \sqrt{z(t)}
\le - 2 c z(t) +\varepsilon (\frac{1}{2c} + \frac{c}{2} z(t)) 
$$
and therefore 
$$
\frac{d}{dt} z(t) \le - cz(t) + \frac{\varepsilon}{2c} \, 
$$
Now set $w(t) = e^{ct}z(t)$, then $\frac{d}{dt} w(t) \le \frac{\varepsilon e^{ct}}{2c}$.
Integrating this inequality from $T$ to $t > T$, it follows that 
$$
w(t) \le w(T) + \frac{\varepsilon}{2c^2}(e^{c(t-T)}-1)
$$ 
and therefore
$$
z(t) \le e^{-ct}\cdot const. + \frac{\varepsilon}{2c^2} \,
$$
Consequently $\limsup_{t \to \infty} z(t) \le \frac{\varepsilon}{2c^2}$.
Since $\varepsilon$ was arbitrary, it follows that $y(t) \to 0$.
This argument also works for vector valued differential equations, even nonlinear ones. An integrating factor for the equation is not needed.