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DE of the form y'' + by' = a

  1. Jul 5, 2008 #1
    Hi, I'm trying to solve the following equation

    y'' + by' = a

    But my answer doesn't make sense:

    The question:
    an object is flying through space, with velocity could be approximated as:
    v_next = v_current + a*dt - damp*v*dt

    dt - time increment taken repetitively
    a - acceleration
    damp - a constant

    For large dt the approximation is inappropriate, find an equation that will do for large dt.

    My go:

    it looks like the above equation is "similar" to
    x'' = a - damp*x'
    x'' + damp*x' = a

    part 1: x_c
    [tex]x'' + damp*x = 0 => r*r + damp*r = 0; r = 0, r = \frac{-1}{damp}[/tex]
    [tex]x_c = c_1 + c_2e^{\frac{-t}{damp}}[/tex]

    part 2: x_p
    x(t) = k*t
    x'' + damp*k = a
    (k*t)'' = 0
    [tex]k = \frac{a}{damp}[/tex]
    [tex]x_p = \frac{a*t}{damp}[/tex]

    part 3:
    [tex]x = x_c + x_p = c_1 + c_2e^{\frac{-t}{damp}} + \frac{a*t}{damp}[/tex]
    we want x' approximation so
    [tex]x' = \frac{-c_2}{damp}e^{\frac{-t}{damp}} + \frac{a}{damp}[/tex]

    what doesn't make sense is lets say damp -> 0 then x' should be a streight line but it doesn't look like it?

    Where may I have gone wrong?

    Thank you
    Last edited: Jul 5, 2008
  2. jcsd
  3. Jul 5, 2008 #2
    Your solution for nonzero damp looks fine to me: the system starts with some initial speed and then approaches asymptotically the so called "terminal speed", v(terminal) = a/damp, at which the pulling force balances the friction exactly: m*a = m*damp*v(terminal).

    Your solution doesn't apply to damp=0 case because you assumed you got two distinct roots of the characteristic equation and hence the full general solution of the homogeneous equation in part 1. That assumption breaks down when damp =0.

    When damp=0, the homogeneous diff. equation in part 1 is x" = 0.
    You get a double zero root of the characteristic equation hence only one exponent which can't capture the full general solution required to depend on two arbitrary constants not one. In such cases the general prescription tells you to look for solution of other types not exp(kt). In this case the solution is a linear function xc = c1 + c2*t. The double zero root in the exponent produces only the c1 term.
    Last edited: Jul 5, 2008
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