Let $x(t)\ge 0$ obey the following differential equation:
$$ \dot{x}(t) = -\alpha(t)x(t) + b\mathrm{e}^{-\lambda t}, $$
where $b>0$, $\lambda>0$, $\alpha(t)\in\mathbb{R}$ is both lower- and upper-bounded, and
$$ \int_0^{+\infty} \alpha(\tau)\ \mathrm{d}\tau = +\infty. $$
Simulations seem to show that $x(t)\to 0$.
(i) Do you see a formal proof?
(ii) If not, what further hypotheses do we need on $\alpha(t)$ to get $x(t)\to 0$?
Possible approach
The Laplace solution reads
$$ x(t) = \mathrm{e}^{-A(t,0)}x(0) + b \int_0^t \mathrm{e}^{-A(t,\tau)} e^{-\lambda \tau} \mathrm{d}\tau, $$
where
$$ A(t,\tau) = \int_\tau^t \alpha(\theta) \mathrm{d}\theta. $$
By hypothesis, $A(t,0)\to+\infty$, so the addend $\mathrm{e}^{-A(t,0)}x(0)$ in $x(t)$ vanishes. Therefore we only need to show that the second addend also vanishes, i.e., that
$$ \lim_{t\to+\infty} \int_0^t \mathrm{e}^{-A(t,\tau)} e^{-\lambda \tau} \mathrm{d}\tau = 0. $$
We can also observe that $A(t,\tau)=A(t,0)-A(\tau,0)$, which substituted in the limit yields
$$ \lim_{t\to+\infty} \mathrm{e}^{-A(t,0)}\int_0^t \mathrm{e}^{A(\tau,0)-\lambda \tau}\mathrm{d}\tau.$$