Results 1 to 2 of 2

Math Help - ???homogenous, partial! don't know teachers a joke!

  1. #1
    Member i_zz_y_ill's Avatar
    Mar 2008

    ???homogenous, partial! don't know teachers a joke!

    mg - k(dx/dt)^1 = m(d^2x/dt2) x=displacement initial condition
    t=time x(o)=0
    m=mass dx/dt=0 i.e x'(o)=0
    object(particle) falls from statonary position, only resistance is air resistance(unknown) . Was told i can 'solve this second order D.E three ways
    1. using an integrating factor,standard form.
    2.homogenously? auxiliary equation, complementary function assumption of some kind of y=Ae^(b(1)x) + Ce^(b(2)x) depending on roots of a.e.
    3. re-arrange into a function v(t) = v then into t(v) = t and you'll get there!

    this is where im supposed to get x= (gm^2/k^2)e^(-kt/m) - gm^2/k^2 + gmt/k which gives in the end
    x=gm/k((m/k)e^(-kt/m)-(m/k) + t)

    So i've had practice with all three methods, and my teacher can't be bothered to explain because everybody else got it. Having got it the first time then lost my notes and some genius in class showed me impossible way of doing it! if someone could spare time thnx!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Feb 2008
    Yuma, AZ, USA

    try the substitution \frac{dx}{dt}=v \mbox{ and } \frac{d^2x}{dt^2}=\frac{dv}{dt}

    so we get

    mg-kv=m\frac{dv}{dt} \iff \frac{dv}{dt}+\frac{k}{m}v=g

    This can be solved by using an integrating factor.

    see how it goes from here
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Math Joke
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: October 7th 2007, 12:15 AM
  2. Teachers and students
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 12th 2005, 05:19 PM

Search Tags

/mathhelpforum @mathhelpforum