Given the differential equation $\frac{dy}{dt}+a(t)y=f(t)$ , show that every solution tends to 0 as t approaches infinity. Given the differential equation $\frac{dy}{dt}+a(t)y=f(t)$ 
with a(t) and f(t) continuous for:
$-\infty<t<\infty$
$a(t) \ge c>0$
$lim_{t_->0}f(t)=0$
Show that every solution tends to 0 as t approaches infinity.
$$\frac{dy}{dt}+a(t)y=f(t)$$
$$μ(t)=e^{\int a(t)dt}$$
$$μ(t)=e^{\frac{1}{2}a^2(t)}$$
Multiplying both sides of the equation by $μ(t)$:
$$\frac{dy}{dt}+a(t)y=f(t)$$ 
$$μ(t)[\frac{dy}{dt}+a(t)y]=μ(t)[f(t)]$$ 
$$e^{\frac{1}{2}a^2(t)}[\frac{dy}{dt}+a(t)y]=e^{\frac{1}{2}a^2(t)}[f(t)]$$ 
$$\frac{d}{dt}e^{\frac{1}{2}a^2(t)}y=e^{\frac{1}{2}a^2(t)}[f(t)]$$ 
$$e^{\frac{1}{2}a^2(t)}y=\int e^{\frac{1}{2}a^2(t)}[f(t)]dt$$ 
Did I make a mistake in the integrating factor step? I am not sure how to proceed..

I will give a solution for the case $f \geq 0$. First we note that the solution can be written (since $y'+ay = f \implies (ye^{\mu})' = fe^{\mu}$ and $y(0)=0$)
$$y(t) = \int_0^t f(t')e^{\mu(t')-\mu(t)} dt'$$
where $\mu(t) = \int_0^t a(t')dt'$. This shows that $y(t)\geq 0$ for all $t\geq 0$. Now using $a(t)\geq c > 0$ we find
$$\mu(t) =  \int_{0}^t a(t')dt' \geq ct$$
Using this we can write the ODE as
$$y' = f - ay \leq f - ct y$$
and apply Gronwall's inequality to get
$$y(t) \leq \left(\int_0^t f(t')dt'\right)e^{\int_0^t (-ct')dt'} = \left(\int_0^t f(t')dt'\right) e^{-ct^2/2}$$
Since $f$ is continuous and $\lim_{t\to \infty} f(t) = 0$ we have that $f$ is bounded (by $M$) on $\mathbb{R}$ and it follows that
$$0 \leq \lim_{t\to\infty}y(t) \leq \lim_{t\to\infty}Mt e^{-ct^2/2} = 0$$
