Ball Thrown Upwards Subject to Air Resistance Proportional to Speed

Suppose the air resistance on a particle of mass
$m$
is proportional to the velocity, so that
$R=kv$
/ A particle thrown up will then be subject to two forces slowing the particle - the force of gravity and air resistance. Obviously when
$t=0, \; y=0$
. Suppose the particle is thrown up with initial speed
$v_0=\frac{dy}{dt}\|_{t=0}$
. Applying
$F=ma$
to the particle while it is moving upwards gives
$ma =-mg-kv$

Using
$v= \frac{dy}{dt}, \; a= \frac{d^2 y}{dt^2}$
this equation becomes
$m \frac{d^2 y}{dt^2} = -mg - k \frac{dy}{dt} \rightarrow \frac{d^2 y}{dt^2} +\frac{k}{m}\frac{dy}{dt}=-g$
.
We can write this last equation as
$\frac{d}{dt} (\frac{dy}{dt}+\frac{k}{m}y)=-g$
.
Integrating gives
$[ \frac{dy}{dt}+\frac{k}{m}y \ ] = [ -gt \ ]^t_0$

$\frac{dy}{dt}-v_0 + \frac{k}{m}y =-gt$
.
Write this as
$\frac{dy}{dt}+ \frac{k}{m}y =v_0-gt$
.
Use the integrating factor method with integrating factor
$e^{\frac{k}{m}t}$
.
The equation becomes
$e^{\frac{k}{m}t} \frac{dy}{dt}+e^{\frac{k}{m}t} \frac{k}{m}y =-e^{\frac{k}{m}t}gt$
.
$\frac{d}{dt}(e^{\frac{k}{m}t}y) =-e^{\frac{k}{m}}gt$
.
Integrate again.
$[ e^{\frac{k}{m}t}y \ ]^y_0 = \int^t_0 e^{\frac{k}{m} t}(v_0-gt)dt$
.
Integrate the right hand side by parts to give
$e^{\frac{k}{m}t}y = \int^t_0 e^{\frac{k}{m} t}(v_0-gt)dt=\frac{k}{m}((v_o-gt)e^{\frac{k}{m}t}-1)+ \frac{k^2g}{m^2}(e^{\frac{k}{m}t}-1)$
.
Hence
$y=\frac{k}{m}((v_o-gt)- e^{-\frac{k}{m}t})+ \frac{k^2g}{m^2}(1- e^{- \frac{k}{m}t})$