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..
 A: 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
$$0 \leq \lim_{t\to\infty}y(t) \leq \lim_{t\to\infty}Mt e^{-ct^2/2} = 0$$
A: here is way to do this. we will set the lower arbitrarily to $t_0.$ you can change it to anything more convenient if you like. suppose 
$$A = exp\left(\int_{t_0}^t a(s)\,ds\right), \dfrac{1}{A}= exp\left(-\int_{t_0}^t a(s)\,ds\right)$$ is the unique solution of $$ \frac{dA}{dt} - aA = 0, \, A(t_0) = 1.$$  $A$ is called the integrating factor because if you multiply $\frac{dy}{dt} + ay = f$ by $A$ you get $$\dfrac{d}{dt}(Ay)  = Af$$ which can be integrated to give a particular solution 
$$y_P =  \dfrac{1}{A(t)}\left(\int_{t_0}^t A(s)f(s)\,ds\right) = \int_{t_0}^t exp\left(-\int_s^t a(u)\,du\right)f(s)\,ds$$ 
we will show that $y_P(t) \to 0$ as $t \to \infty.$
$$ |y_P| \le \int_{t_0}^t exp\left(-\int_s^t a(u)\,du\right)|f(s)|\,ds 
\le \int_{t_0}^t exp\left(-\int_s^t c\,du\right)|f(s)| \,ds$$
picking $t_0$ large enough to make $|f|$ small so that the limit of $y_P$ is zero as $t \to \infty.$
there must be a cleaner way to show that the limit is zero than the one i have here. i will think about it.
A: After your last step:
$$e^{\frac{1}{2}a^2(t)}y=\int e^{\frac{1}{2}a^2(t)}[f(t)]dt$$
You want to divide both sides by $e^{\frac{1}{2}a^2(t)}$ to solve for $y$
Giving us:
$$y=e^{-\frac{1}{2}a^2(t)}\int e^{\frac{1}{2}a^2(t)}[f(t)]dt$$
