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System of Implicit Non-Linear First Order ODEs

  1. Oct 7, 2015 #1
    I have an extremely messy system of differential equations. Can anyone offer any ideas for a general solution?


    p(t) is a function of t, and A is a constant.
  2. jcsd
  3. Oct 7, 2015 #2
    An idea would be to rewrite ##\cos \arctan \frac{\dot{y}}{\dot{x}}## as ##\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}##, and similar for the other equation. Then try to simplify it a bit.
  4. Oct 7, 2015 #3
    Didn't think of that, thanks! I'll keep working on it.
  5. Oct 8, 2015 #4


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    Homework Helper

    You may be best served by switching to intrinsic coordinates [itex](s, \psi)[/itex] where [tex]
    \dot x = \dot s \cos\psi, \\
    \dot y = \dot s \sin \psi.[/tex] Your system is then [tex]
    (\dot s - p) \cos \psi = 0 \\
    (\dot s - p) \sin \psi = -At.[/tex] Now either [itex]\dot s = p[/itex] or [itex]\cos \psi = 0[/itex]. The first is impossible unless [itex]A = 0[/itex]. The second requires that [itex]\dot x = 0[/itex] and [itex]\dot y = \pm p - At[/itex].

    If [itex]A = 0[/itex] then [itex]\dot s = p[/itex] and [itex]\psi[/itex] can be an arbitrary function of [itex]t[/itex].
  6. Oct 8, 2015 #5
    Sorry, but it looks like something went wacky with your latex code. Could you retype it? It's hard to understand what your saying.
  7. Oct 8, 2015 #6
    It takes time before equations render correctly but eventually they do. What are you seeing? Try a different browser or device.
    Basically he says to substitute ##\dot x = \dot s \cos\psi##, ##\dot y = \dot s \sin \psi##. Then it's very simple to arrive at a solution.

    BTW, perhaps the first solution should have been ##\dot x=0##, ##\dot y=p\pm At##.
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