## 2nd and last post on DEs, I promise :p

Posted: April 3, 2013 in Mathematics

Now I am neither an engineer nor a physics major, so I am not so sure if my solution is correct.  Don’t trust me on this, I don’t know physics 🙂

My guess is

$m\cdot\dfrac{dv}{dt}=mg-kv$

here m, g, and k are fixed constant.  Then if you trust me on this, read on

Since mass is non-zero,

$\dfrac{dv}{dt}=g-\dfrac{kv}{m}$

$\dfrac{dv}{dt}+\dfrac{k}{m}\cdot v=g$

Just like the last post, the integrating factor is

$e^{\int k/m \, dt} = e^{kt/m}$

Then

$\dfrac{dv}{dt}e^{kt/m}+\dfrac{k}{m}\cdot ve^{kt/m}=ge^{kt/m}$

$\dfrac{d}{dt}ve^{kt/m}=ge^{kt/m}$

$ve^{kt/m}=\displaystyle \int ge^{kt/m}\,dt$

$ve^{kt/m}=\dfrac{gm}{k}e^{kt/m} + C$

$v=\dfrac{gm}{k}+Ce^{-kt/m}$

Since when $t=0, v=v_0$, then we have

$v_0=\dfrac{gm}{k}+C$

$\therefore C=v_0-\dfrac{gm}{k}$

Finally,

$v=\dfrac{gm}{k}+\left(v_0-\dfrac{gm}{k}\right)e^{-kt/m}$

Now, I have no idea if the initial velocity is zero or not coz I don’t really fully understand the problem, I ain’t expert in physics, so if

$v_0=0$

then

$v=\dfrac{gm}{k}-\dfrac{gm}{k}\cdot e^{-kt/m}$

otherwise,

$v=\dfrac{gm}{k}+\left(v_0-\dfrac{gm}{k}\right)e^{-kt/m}$