Behaviour of solutions of an ODE Show that any solution of the ODE given below as:
$$x'+a(t)x=f(t)$$
where 
$a(t)\geq c>0$ and
$f(t) \to 0 $ when $t \to +\infty$
goes to $0$ when $t \to +\infty$
 A: Look for a solution
$$
x(t) = C(t) \exp\left[ -A(t)\right]
\\
A(t) = \int_0^t a(s) ds
$$ one gets
$$
C'(t) = f(t) \exp\left[ A(t)\right]\\
C(t) = \int_0^t f(s) \exp\left[ A(s)\right]ds + C(0)\\
x(t) = \left[
\int_0^t f(s) \exp\left[ A(s)\right]ds + x(0)\right] \exp\left[ -A(t)\right]
\\
=\int_0^t f(s) 
\exp\left[ A(s) -A(t)\right]ds + x(0) \exp\left[ -A(t)\right]
$$
Now as $a(t) \ge c$, for $t\ge s$
$$A(s) - A(t) = -\int_s^t a(s) ds \le -c(t-s)
\\
|x(t)| \le \int_0^t |f(s)| 
\exp\left[ A(s) -A(t)\right]ds + |x(0)| \exp\left[ -A(t)\right]
\\
\le
\int_0^t |f(s)| 
\exp\left[ -c(t-s)\right]ds + |x(0)| \exp\left[ -ct\right]
$$
The second term goes to 0. Now let us take care of the first one, using the fact that $|f(t)|\to 0$, there is an $A$ such as $x>A\implies |f(x)|<\frac {rc}2$, and on $[0,A]$ $f$ is bounded (assuming it is continuous) by a certain $M$:
$$
\int_0^t |f(s)| 
\exp\left[-c(t-s)\right]ds \le
M \int_0^A  
\exp\left[-c(t-s)\right]ds
+
r\int_A^t 
\exp\left[-c(t-s)\right]ds\\
\le \frac 1c\left[
\frac{rc}2(1 - \exp(-c(t-A)) + M(\exp(-c(t-A)) -\exp(-ct))
\right]\\
\le
\frac r2 + \frac Mc\exp(ca) \exp(-ct) \le r
$$
for $t$ large enough.
So $x(t)\to 0$ when $t\to\infty$.
