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

Posted: April 3, 2013 in Mathematics

ODEs

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}

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