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?

$F = mg - mkv = ma \rightarrow g - kv = a \rightarrow g - kv = dv/dt$ Integrate and we will get the graph second case we do something similar.
Now for the second equation we have $F = mg - mkv^2 = ma$
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: $$v_{\text{terminal}}=\lim_{t\to\infty}v(t)=\sqrt{\dfrac{g}{k}}$$