How would the following graphs differ in shape? This is a mechanics question but is pretty much mathematical so I figured I should post it here.

If I had a particle dropped from rest and it had resistance $mkv$ where mass is $m$, $v$ is velocity, and $k$ is a constant, how would the velocity-time graph change if the resistance was given by $mkv^2$ instead?

I know both would approach terminal velocity but how would the shape change?
 A: First of all by newtons second law we have choosing the coordinate axis as down positive and up as negative we have the following equation for the first one.
$F = mg - mkv = ma \rightarrow g - kv = a \rightarrow g - kv = dv/dt$
Integrate and we will get the graph second case we do something similar.
Now for the second equation we have
$F = mg - mkv^2 = ma$
A: First case:
From Newton's Second Law, you have:
$$F=ma=mg-mkv\Longrightarrow\dfrac{dv}{dt}=g-kv$$
We can perform an integration using separation of variables:
$$\int_0^v\dfrac{dv}{g-kv}=\int_0^t dt\Longleftrightarrow -\dfrac{1}{k}\ln{|g-kv|}\Bigg\vert_0^v=t\Bigg\vert_0^t$$
Rearranging:
$$v(t)=\dfrac{g}{k}\left(1-e^{-tk}\right)$$
And you can see that:
$$v_{\text{terminal}}=\lim_{t\to\infty}v(t)=\dfrac{g}{k}$$

Second case: From Newton's Second Law, you have:
$$F=ma=mg-mkv^2\Longrightarrow \dfrac{dv}{dt}=g-kv^2$$
Using separation of variables again:
$$\int_0^v\dfrac{dv}{g\left(1-\dfrac{kv^2}{g}\right)}=\int_0^t dt$$
Making the substitution $u=\dfrac{kv^2}{g}$, the integral becomes:
$$\dfrac{1}{\sqrt{gk}}\int_0^v\dfrac{du}{1-u^2}=\int_0^t dt\Longleftrightarrow\dfrac{\tanh^{-1}\left(v\sqrt{\frac{k}{g}}\right)}{\sqrt{gk}}\Bigg\vert_0^v=t\Bigg\vert_0^t$$
Rearranging:
$$v(t)=\sqrt{\dfrac{g}{k}}\tanh\left(\sqrt{gk}t\right)$$
And you can see that:
$$v_{\text{terminal}}=\lim_{t\to\infty}v(t)=\sqrt{\dfrac{g}{k}}$$
