Knowing that acceleration is time derivative of velocity, and (if you need) velocity is time derivative of position
$v = \dot x$
$a = \dot v = \ddot a$
the equation of your question can be written as a simple first-order linear ordinary differential equation,
$\dot v + k v = g$
whose solution reads
$v(t) = \dfrac{g}{k} + A e^{-kt}$.
Prodivded the initial condition $v(0) = v_0$, it's possible to determine the constant $A$,
$A = v_0 - \dfrac{g}{k}$
to get the solution of the problem as
$v(t) = \dfrac{g}{k} \left( 1 - e^{-kt}\right) + v_0 e^{-kt} $.
From this expression, you can see that starting from the velocity $v(0) = v_0$, the system exponentially approaches the limit velocity $\lim_{t\rightarrow \infty} v(t) = \frac{g}{k}$, for which the air drag equilibrates the gravity force.
Solution of first order linear ODE
You can find the solution of the non-homogeneous linear first-order linear ODE,
$\dot y + a y = f(t)$
as the sum (linear problem $\rightarrow$ principle of superposition holds) of a solution of the homogeneous equation $y_o(t)$ and a particular solution $y_1(t)$ of the non-homogeneous equation.
Solution of the homogeneous equation. The solution of the linear equation
$\dot y + k y = 0$,
using the property of the exponential $(e^{Ct})' = A e^{Ct}$. If you look for a solution with the form $y(t) = A e^{Ct}$ and put it into the homogeneous ODE, you find
$0 = ACe^{Ct} + k Ae^{Ct} = A(C+k)e^{Ct}$,
and thus $C = -k$, and $y_o(t) = A e^{-kt}$.
Particular solution of the non-homogeneous ODE. This really depends of the forcing. For a constant forcing $f(t) = g$ a particular solution is a constant solution $y_p(t) = K$ s.t. $y_p'(t) = 0$. Putting it into the non-homogeneous equation, you get
$0 + k K = g$
and thus $y_p(t) = K = \frac{g}{k}$.
Solution. The solution of the ODE thus reads
$y(t) = y_o(t) + y_1(t) = A e^{-k t} + \dfrac{g}{k}$.
Now, and only now, you can prescribe the initial condition to find the value of constant $A$.
