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Thread: ???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!
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Feb 2008
    Yuma, AZ, USA
    $\displaystyle mg-k\frac{dx}{dt}=m\frac{d^2x}{dt^2}$

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

    so we get

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

    This can be solved by using an integrating factor.

    see how it goes from here
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