I hope this is a suited question for this site since it contains a mix of physics and mathematics. In case I should post this on the physics stackexchange site, please let me know.
A spherical raindrop is falling from the sky. Because of the humid atmosphere the raindrop will gain mass during his fall. The increase in mass per time is proportional to the current surface area.
- Find an equation for the radius of the drop as a function of time ($r(t)$). ($r(0) =r_0$)
- Find and solve the equation of motion for the raindrop. The equation of motion should depend on $r_0$
My work so far:
Mass: $m(t)=\frac{4}{3} \pi r(t)^3 \rho $
Sufrace Area: $A(t)=4\pi r(t)^2$
$$\begin{aligned} & \implies \frac{dm(t)}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r(t)^3 \rho)=4 \pi r(t)^2 r'(t)\rho\\ & \iff 4 \pi \rho r(t)^2 r'(t)=\lambda 4 \pi r(t) ^2 \iff r'(t)=\frac{\lambda}{\rho} \iff \frac{dr}{dt}=\frac{\lambda}{\rho} \\ & \iff \int dr = \frac{\lambda}{\rho} \int dt \iff r(t)=\frac{\lambda}{\rho}t+c \\ & \color{blue}{\implies r(t)=\alpha t+r_0} \space \space \space \space \space \space \space \space \space \space \space \text{where}\space \space\alpha=\frac{\lambda}{\rho} \end{aligned}$$
Equation of motion:
Gravitational Force: $mg$
Air Friction: $kv(t)$
$$\begin{aligned} &F=mg-kv \iff \frac{d}{dt}(mv)=m'v+v'm=mg-kv \\ & \end{aligned}$$
I tried substituting $m'$ and $m$ into my equation and I am left with:
$$\begin{aligned}& mv'+\frac{3m \alpha}{r}=mg-kv\end{aligned}$$
I tried solving this in wolfram alpha and I get the following result: LINK
It looks super messy and I don't think that is the solution I am supposed to get. Am I doing something wrong? Is there some substitution I can make to make it easier to solve?
If you assume additionally that the friction is proportional to surface area and to velocity, then you can run the following logic to get a pretty simple ODE for $v$:
\begin{align} r & =c_1t + c_2 \\ m & =c_3 r^3 \\ F & =c_4 m - c_5 r^2 v = c_6 r^3 - c_5 r^2 v \\ F & =m'v+mv' \\ & = 3c_3 r^2 r' v + c_3r^3 v' \\ & = 3c_3 c_1 r^2 v + c_3 r^3 v' \\ & = c_6 r^3 - c_5 r^2 v \\ & \Rightarrow 3 c_3 c_1 v + c_3 r v' = c_6 r - c_5 v. \end{align}
In normal form for first order linear ODEs:
$$v' + \frac{3 c_3 c_1 + c_5}{c_3 r} v = \frac{c_6}{c_3}.$$
Actually plugging in what all these constants are in your notation:
$$v'+\frac{4 \pi \rho \alpha + k}{\frac{4}{3} \pi \rho (\alpha t + r_0)} v = g.$$
This is quite tractable using the method of integrating factors.