The velocity $v$ of a freefalling skydiver is modeled by the differential equation
$$ m\frac{dv}{dt} = mg - kv^2,$$
where $m$ is the mass of the skydiver, $g$ is the gravitational constant, and $k$ is the drag coefficient determined by the position of the diver during the dive. Find the general solution of the differential equation.
So is it my job to solve for velocity here ($v$)? or am I missing something?
On
As Xander said, $v(t)$ is what they are asking for here. Differentiation shouldn't be on the mind when you solve this problem however. The differential equation $$m\frac{dv}{dt} = mg - kv^2$$ can be classified as a first order non-linear equation in $v.$ The equation is separable which means we can solve it using integration. Observe that if we write the system as $$\frac{dv}{dt} = g - \frac{k}{m}v^2\ \ \ \ \ \ \text{and then}$$
$$\frac{dv}{g - \frac{k}{m}v^2} = dt$$ then we can simply integrate each side from $t=t_0$ to $t=t_f $ as
$$\int _{v(t_0)}^{v(t_f)} \frac{dv}{g - \frac{k}{m}v^2} = \int_{t_0}^{t_f}dt . $$
This is an ugly integral for which I recommend finding a good integral table. The result is that $$\frac{ \tanh^{-1}\bigg(v\sqrt{\frac{k}{mg}}\bigg) }{\sqrt{\frac{gk}{m}}} \bigg|_{v=v(t_0)}^{v=v(t_f)} = t_f - t_0 $$ which I also recommend verifying for yourself. From here you just evaluate the lefthand expression over the bounds and solve for $v(t_f).$ That's just algebra so I'll let you hash that out. Something to note is that you should probably just write $t_f$ as $t$ and use $t_0 = 0$ in your final answer. Hope this helps.
$$ m\frac{dv}{dt} = mg - kv^2,$$
$$\implies \frac{m}{mg-kv^2}{dv}=dt\implies\int_{v_o}^v\frac{m}{mg-kv^2}{dv}=\int_{t_o}^{t} dt $$ Since m,g,k are constants, $$\frac{m}{k}\int_{v_o}^v\frac{1}{\frac{mg}{k}-v^2}{dv}=t-t_o$$ Solving the integrals, $$\frac{m}{k}\left[\frac{1}{2\sqrt{\frac{mg}{k}}}\ln\left|\frac{\sqrt{\frac{mg}{k}}+v}{\sqrt{\frac{mg}{k}}-v}\right|\right]_{v_o}^v=t-t_o$$ Simplify the answer further, it will the general equation.