$y'+a(t)y=f(t)$, $\lim_{t\to \infty} f(t)= 0 \implies \lim_{t\to \infty} y(t) = 0 $ I've come across the following ODE problem (1.2.19 Braun,4th edition, Differential Equations and Their Applications):

given $a(t)\geq a_{min} > 0$ and $\lim_{t\to \infty}f(t) = 0$, with $a$ and $f$ continuous on $(-\infty, \infty)$, every solution $y$ of $y'+a(t)y=f(t)$ vanishes as t tends to infinity.

The integrating factor for this ODE is $\mu (t) = e^{\int a(t)} \geq e^{a_{min} \int dt} = e^{a_{min}t}$.
Hence we have $$\int (\mu (t)y)'dt = \int \mu(t)f(t)dt\\ e^{\int a(t) dt}y(t)= \int e^{\int a(t)}f(t)dt \\ y(t) = e^{-\int a(t) dt} \int e^{\int a(t)}f(t)dt \\ \ . $$
This is where I get stuck.
I tried using the information on $f$ by rewriting it as a product of a bounded part $b(t)$ and an unbounded part $d(t) > 0$, with $d(t) \to \infty $ as $t \to \infty$: $$f(t) = \frac{b(t)}{d(t)}\ . $$
Now, I guess, I need to show that $e^{-\int a(t) dt} $ vanishes faster than $\int e^{\int a(t)}f(t)dt$ grows as $t\to \infty$. I don't see how.
 A: In fact, the solution of the equation is
$$ y(t) = e^{-\int_0^t a(s) ds}\bigg[y_0+ \int_0^t e^{\int_0^s a(\tau)d\tau}f(s)ds\bigg]. $$
Since $\lim_{t\to\infty}f(t)=0$, for any $\epsilon>0$, there is $N_1>0$ such that for $t>N_1$, $|f(t)|<\frac12 a_{min}\epsilon.$
Thus for $t>N_1$,
\begin{eqnarray}
|y(t)| &\le& e^{-\int_0^t a(s) ds}\bigg[|y_0|+ \int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds+\int_{N_1}^t e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]\\
&\le& e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12 a_{min} \epsilon\int_{N_1}^te^{-\int_s^t a(\tau) d\tau}\\
&\le&e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12 a_{min}\epsilon \int_{N_1}^te^{-\int_s^t a_{min} d\tau}\\
&\le&e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12 a_{min}\epsilon \int_{N_1}^te^{-a_{min}(t-s)}\\
&\le&e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12(1-e^{-a_{min}(t-{N_1})})\epsilon.
\end{eqnarray}
Since
$$\lim_{t\to\infty}e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]=0, $$
there is $N_2>0$ such that when $t>N_2$,
$$ e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]<\frac12\epsilon. $$
Let $N=\max\{N_1,N_2\}$ and then when $t>N$,
$$ |y(t)|<\frac12\epsilon+\frac12\epsilon=\epsilon. $$
So
$$ \lim_{t\to\infty}y(t)=0. $$
A: Let's let $Y=\lim_{t\to\infty}y(t)$. First, let's say that $Y$ is positive, or that $Y=\infty$. Then for some real $\delta_1$, and some fixed positive real $r$, whenever $t>\delta_1$, we have $r<y(t)$. (Which positive real $r$, you may ask? If $Y$ is positive and finite, then $\frac Y2$ works. If $Y$ is infinite, pick your favorite number.)
So then for $t>\delta_1$ we have $y'=f(t)-a(t)y<f(t)-a_{min}r$. Now, we are given that $f$ approaches $0$, so for some real $\delta_2$, whenever $t>\delta_2$ we have that $f(t)<\frac{a_{min}r}2$. So, whenever $t>\max{(\delta_1,\delta_2)}$, we have that $y'(t)<-\frac 12a_{min}r$.
The mean value theorem guarantees that if $\max(\delta_1,\delta_2)\leq t_1<t_2$ then $y(t_2)-y(t_1)\leq-\frac12a_{min}r(t_2-t_1)$. So, if $t_1>\max(\delta_1,\delta_2)$ then $y(t)\leq y(t_1)-\frac12a_{min}r(t-t_1)$, where the right hand side of that final inequality approaches $-\infty$, and so $Y=-\infty$.
So, if $Y$ is positive (or if the limit approaches positive infinity) then $Y=-\infty$. In much the same way, if $Y$ is negative (or negative infinity) then $Y=\infty$. Either way, we get a contradiction, and so the only conclusion left is that $Y=0$.
