Show that the Limit of ODE Solutions Tends to $0$ 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.
 A: Let $A(t) = \int_0^t a(s)\, ds$. By method of integrating factors, 
$$y(t) = e^{-A(t)}y(0) + e^{-A(t)}\int_0^t e^{A(s)}f(s)\, ds$$ For $t \ge 0$, the exponential $e^{-A(t)}$ is bounded by $e^{-ct}$, so the term $e^{-At}y(0)\to 0$ as $t \to \infty$. Now $$e^{-A(t)}\int_0^t e^{A(s)}f(s)\, ds = \int_0^t f(s)e^{-\int_s^t a(u)\, du}\, ds$$ is bounded by $\int_0^t \lvert f(s)\rvert e^{-c(t-s)}\, ds$, which we'll show tends to $0$ as well. Fix a positive number $\varepsilon$. Using the condition $\lim\limits_{t\to \infty} f(t) = 0$, we may choose a positive number $M$ such that for all $t \ge M$, $\lvert f(s)\rvert < \varepsilon$. For $t > M$, $$\int_0^t \lvert f(s)\rvert e^{-c(t-s)}\, ds = \int_0^M \lvert f(s)\rvert e^{-c(t-s)}\, ds + \int_M^t \lvert f(s)\rvert e^{-c(t-s)}\, ds \le Ce^{-ct} + \frac{\varepsilon}{c}\left[1-e^{-c(t-M)}\right]$$ where $C$ is the constant $\int_0^M \lvert f(s)\rvert e^{cs}\, ds$. Taking the limit superior as $t \to \infty$ results in $$\limsup_{t\to \infty} \int_0^t \lvert f(s)\rvert e^{-c(t-s)}\, ds \le \frac{\varepsilon}{c}$$ As $\varepsilon$ was arbitrary, $\int_0^t \lvert f(s)\rvert e^{-c(t-s)}\, ds$ tends to $0$ as $t\to \infty$, as desired.
A: 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. 
