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Solving a first-order nonlinear differential equation.

  1. Dec 19, 2007 #1
    [Differential equation at the end; All the intermediary stuff is the problem behind it.]

    I was curious about finding the velocity function for a free-falling object using solely newton's equations. Using the force diagram, I've deduced that

    g-gravitational constant
    f_net - net force
    A - cross-sectional area
    v - velocity

    f_net = mg-1/4Av^2
    ma = mg - (Av^2)/4
    v'=g-(Av^2)/4m and g, A and m are constants. In this case, I simplified by assuming A=1, m=1 so the equation becomes the nonlinear first-order differential equation

    v'=9.8-.25v^2. How do you... solve this kind of equation?
  2. jcsd
  3. Dec 19, 2007 #2


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    We might tidy up your problem as follows:
    Set [tex]\epsilon^{2}=\frac{A}{4mg},w(t)=\epsilon{v}(t)[/tex]
    Multiplying your diff.eq with [itex]\epsilon[/itex], we get:
    Agreed so far?
    Last edited: Dec 19, 2007
  4. Dec 19, 2007 #3


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    Latex was a bit grumpy, it should be okay now..
  5. Dec 19, 2007 #4


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    Looks to me like an easily separable equation: get all "v"s on one side, all "t"s on the other and integrate.
  6. Dec 19, 2007 #5
    Took quite some time to mentally process(This format is extremely unfamiliar) and sure. I'm not really sure where you go from here(My experience with differential equations is practically nothing, beyond several glimpses on wikipedia.) but I guess I can shoot off into several random directions and hope one hits the target.
  7. Dec 19, 2007 #6


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    Have you done separable diff. eqs yet?
  8. Dec 19, 2007 #7
    So I see how this might solve a problem that I *think* I was having earlier.



    dw/(1-w^2) = eg dt

    Then the left side would be integrated in respect to initial and final "w", which then could be converted into v terms post-integration(or would be it better to do it pre-integration? I'm not immediately seeing how that would simplify things since the current form seems to be an immediately integrable form, I just need to look up the charts.), and the right side could be integrated in respect to initial and final t, and then velocity can be straightforwardly solved for algebraically? Is this a fruitful direction?

    (Where do I learn how to use LaTeX(Wrong capitalization sequence? baha) on the forums, by the way?)

    Ok. So I think I've found a definite way to solve it.

    w = ev
    dw = e dv

    dw/(1-w^2) = e dv/(1-e^2v^2) (So pre-integration conversion seems to be better).

    Looking at the table of integrals, it appears that the last form integrates to an inverse hyperbolic tangent which then the function v can be solved for.
    Last edited: Dec 19, 2007
  9. Dec 19, 2007 #8
    Implicit differentiation about 3 years ago in Highschool is about the closest(That I know of). I didn't actually do "Calculus 2"(Differential equations included, looking at the book) but I did complete a system of differential equations sometime ago in a Calculus 3 test. But I think that was simply an initial value differential equations problem and did I struggle trying to complete it having only seen examples of it in Stewart's calculus book. But I solved it. :D
    Last edited: Dec 19, 2007
  10. Dec 19, 2007 #9


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    Quite right.
    You should get, starting from rest:
    [tex]w(t)=Tanh(egt)[/tex], where a positive velocity means a downwards velocity, due to the sign choice you made at the start.

    Tanh is the hyperbolic tangent function.
  11. Dec 19, 2007 #10

    So w(t) = tanh(egt)?
    Then ev(t) = tanh(egt)
    v(t) = tanh(egt)/e

    When I tried solving it by pre-converting w in terms of v before integrating, I ended up

    v(t) = tanh(e^2gt)/e. I'm thinking that I superfluously inserted an e somewhere, but I don't know where.


    w = ev

    dw/dt = e dv/dt (Right?)
    dw = e dv(right? Imagine another "Right?" to each step hereafter.)

    dw/(1-w^2) =
    e dv/(1-e^2v^2)
    = e tanh....

    Hehe... nevermind... I forgot to multiply the e in front of the integral by the e in the denominator of the evaluated integral(Which cancels out all tanh^(-1)-external e's on the left side). So, yeah, pre-integration works as well as post-integration. Sweet.

    Even cooler; The graph ACTUALLY looks realistic!

    (I just realized another method to solve this problem could be done by using an energy approach. It seems like it wouldn't have to involve differential equations, either.)

    Anyways, thanks! This has been extremely informative! So... with differential equations like this, just basically restate the equation so that one side could be completely expressed as products and the other side as the derivative and.... divide and conquer.
    (Perhaps there's a quicker more "standard" way, but this seems to get the job done.)
  12. Dec 19, 2007 #11


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    Note that 1/e is the "steady-state" velocity as time goes to infinity.

    We ASSUMED that we let the object from rest, falling freely.
    But, we may ask:
    If we throw the stone upwards, how will that affect the motion?

    In this case, BOTH gravity and air resistance work against the motion, we will get a (trigonometric) tangent relationship between velocity and time, rather than a hyperbolic one.
    Try it out! (assume some initial velocity V>0)
  13. Sep 14, 2008 #12

    I have encountered a nonlinear first-order differential equation in my research. It appears rather simple yet I have been unable to solve it. Any clues/tips would be highly appreciated. First I state it in the original form:

    y'[x] - 1/y[x] = - Exp[-1/x] / x^2

    Next I state it written in the form of an Abel equation, obtained by setting q[x]=1/y[x]:

    q'[x] + q[x]^3 = q[x]^2 * Exp[-1/x] / x^2

    Thanks in advance for any thoughts!

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