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ODE - First encounter, not understanding them

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    First time I've had to deal with ODEs, an I'm pretty confused.

    This SHOULD be a simple ODE for finding air resistance, that is only dealing with the y vector (up and down in this case)

    [itex]m\frac{dv_{y}}{dt}=mg-kv_{y}[/itex]

    2. Relevant equations
    F=ma
    f=-kV

    3. The attempt at a solution
    First order differential equation, so we only need to integrate once.

    [tex]m\frac{dv_{y}}{dt}=mg-kv_{y}=g-\frac{kv}{m}[/tex]

    [tex]\int\frac{dv}{dt}=gt-\frac{k}{m}\int v dt[/tex]

    I have no idea where to go from here. The answer in the book gives:

    [tex]v=\frac{mg}{k}(1-e^{-\frac{kt}{m}})[/tex]
     
  2. jcsd
  3. Jan 28, 2013 #2

    Mark44

    Staff: Mentor

    This (above) won't work.
    Divide both sides by m to get
    dv/dt = g - (k/m)v

    This equation is separable, so get everything involving v and dv on one side, and everything involving t and dt on the other. (There is nothing that involves t, though.)
     
  4. Jan 28, 2013 #3

    rude man

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    The book answer does not include a possible initial velocity .

    I'm sure there's a "classical" way to solve this equation. As an EE I would of course use the Laplace transform method, which sadly is not taught in most diff. eq. courses but should be.

    The book answer is correct if initial velocity = 0.

    (If you're interested in the Laplace method let me know.)

    EDIT: Mark44 has the way to go. I tried but could not figure out how to separate the variables.
     
  5. Jan 28, 2013 #4
    I'll be honest, I had to look up what a separable ODE was.

    So, after doing that, I got:

    [tex]\int\frac{k}{m}v-g dv=\int dt[/tex]

    so this equals: [itex]\frac{k}{2m}v^{2}-gv=t[/itex]

    And the next step is to just rearrange?
     
  6. Jan 28, 2013 #5

    Mark44

    Staff: Mentor

    This isn't right, due to an algebra error.

    Starting from dv/dt = g - (k/m)v,

    we have dv/(g - (k/m)v) = dt

    Now the equation is separated, and you can integrate both sides.
     
  7. Jan 28, 2013 #6

    Mark44

    Staff: Mentor

    I taught diff. eqns. for a lot of years, and Laplace transforms were one of the techniques presented. They weren't taught at the beginning of the course, though, and that's where the OP is, I believe.
     
  8. Jan 28, 2013 #7

    rude man

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    I had two texts on diff. eq's and neither mentioned Laplace. One was by MIT people.
     
  9. Jan 29, 2013 #8

    HallsofIvy

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    Science Advisor

    Personally, I dislike the "Laplace transform method". It loved by engineers because it gives a very "mechanical" way of solving differential equations. You just "look up" the inverse transfors in a table. But I have never met a differentia equation, solvable by Laplace transform, that was not, in my opinion, more easily solved by direct methods.
     
  10. Jan 29, 2013 #9

    rude man

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    I'm not sure that "guessing" a solution to a diff. eq., as is done classically, is more insightful than looking up an inverse in a table. And I like the convenience of not having to do two solutions (homogeneous & inhomogeneous) and also including the initial conditions automatically.

    Chaqu'un a son gout.
     
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