Qualitative Differential equations

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SUMMARY

The discussion revolves around the qualitative analysis of the differential equations dx/dt = y + x² and dy/dt = x + y². The main objective is to demonstrate that if x(t0) ≠ y(t0), then x(t) ≠ y(t) for all t. Participants suggest that assuming x = y leads to the conclusion y = x + c, indicating that the solutions do not intersect, thus confirming the uniqueness of initial value problems (IVPs) in this context. The need for a clearer approach using the differential equation for y - x is also highlighted.

PREREQUISITES
  • Understanding of qualitative analysis of differential equations
  • Familiarity with initial value problems (IVPs)
  • Knowledge of orbit theory in dynamical systems
  • Basic skills in manipulating differential equations
NEXT STEPS
  • Explore the uniqueness theorem for initial value problems in differential equations
  • Study the concept of orbits in dynamical systems
  • Learn how to derive and analyze differential equations for differences, such as y - x
  • Investigate the implications of non-intersecting solutions in systems of differential equations
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Mathematicians, students studying differential equations, and researchers interested in dynamical systems and qualitative analysis.

dismo
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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.
 
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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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