$\frac{dx}{dt} + \alpha x = 1$ asymptotic behaviour and solution 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 $\frac{1}{\alpha}$. 
 A: This is an inhomogeneous first-order linear ordinary differential equation. The standard way to solve such an equation is to find all solutions of the corresponding homogeneous equation and then add any particular solution of the inhomogeneous equation.
The inhomogeneous equation is solved by a constant:
$$x(t)=x_0 \rightarrow \alpha x_0=1 \rightarrow x_0=\frac{1}{\alpha}\;.$$
The standard way to find all solutions of the homogeneous equation
$$\frac{\mathrm dx}{\mathrm dt}+\alpha x=0$$
is through the ansatz
$$x(t)=c\mathrm e^{\lambda x}\;,$$
which leads to the characteristic equation
$$\lambda+\alpha=0\;,$$
and hence $\lambda=-\alpha$ and
$$x(t)=c\mathrm e^{-\alpha t}\;.$$
So the general solution of the inhomogeneous equation is
$$x(t)=c\mathrm e^{-\alpha t}+\frac{1}{\alpha}\;.$$
Substituting $x(0)=2$ yields
$$c+\frac{1}{\alpha}=2\;,$$
$$c=2-\frac{1}{\alpha}\;,$$
and thus
$$x(t)=\left(2-\frac{1}{\alpha}\right)e^{-\alpha t}+\frac{1}{\alpha}\;.$$
A: joriki's approach is fine. Alternatively, the equation is "variables separable" and can be solved by rewriting as $${1\over1-\alpha x}\,dx=dt$$ and then integrating; $$\int{1\over1-\alpha x}\,dx=\int\,dt,\qquad -{1\over\alpha}\log|1-\alpha x|=t+C$$ stick in $t=0$ to get $$C=-{1\over\alpha}\log|1-2\alpha|,\qquad -{1\over\alpha}\log|1-\alpha x|=t-{1\over\alpha}\log|1-2\alpha|$$ and now solve for $x$. 
