I am trying to solve $\frac{dx}{dt} + \alpha x = 1$, $x(0) = 2$, $\alpha > 0$ where $\alpha$ is a constant.
[some very badly done mathematics deleted]
Continuing with Gerry's suggestion:
$\log|1-\alpha x | = -t\alpha + \log|1-2\alpha|$
$1-\alpha x = e^{-t\alpha}(1-2\alpha)$
$x(t) = \frac{1 - e^{-t\alpha}(1-2\alpha)}{\alpha}$
Then the asymptotic behaviour of $x(t)$ as $t$ goes infinity would be $e^{-t\alpha}$ approaching zero, therefore the overall $x(t)$ would approach $\frac{1}{\alpha}$.