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Qualitative Differential equations

  1. Dec 19, 2008 #1
    Problem:
    Let x(t), y(t) e a solution of

    dx/dt=y+x^2
    dydt=x+y^2

    with x(t0) NOT = y(t0)

    Show that x(t) NOT = y(t) for all t

    Attempt:

    I feel like the easiest way to show this would be to show that x=y is an orbit of the system and then simply use the fact that orbits may not cross due to the uniqueness of IVP's at every point in the solution space?

    So if I set x=y

    dx/dt = y+y^2
    dy/dt = y+y^2

    This implies
    dy/dx = 1

    So y=x+c for all t.

    Is this a reasonable solution?
    Is there anything that needs clearing up?

    Thanks guys.
     
  2. jcsd
  3. Dec 20, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    you started by assuming that y= x and showed that y= x+ c. How does that show that y= x is an orbit of the system? It might be better to determine a differential equation for y- x.
     
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