ODE - Unique solutions

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Homework Statement



Identify the region that the DE will have a unique solution.

y' = [itex]\frac{y^2}{x^2+y^2}[/itex]

The Attempt at a Solution



[itex]\frac{\partial f}{\partial y}[/itex] = [itex]\frac{2x^y}{(x^2+y^2)^2}[/itex]

I'm a bit rusty with my domains, but here is what I've got.

x: (-∞, -2) U (2,-∞)
y: (-∞, -2) U (2,-∞)

I think that I'm missing something else!?
 

Answers and Replies

  • #2
haruspex
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[itex]\frac{\partial f}{\partial y}[/itex] = [itex]\frac{2x^y}{(x^2+y^2)^2}[/itex]
I think you mean [itex]\frac{\partial f}{\partial y}[/itex] = [itex]\frac{2x^2y}{(x^2+y^2)^2}[/itex]
Beyond that I cannot help, as I am ignorant of this topic. Can you quote any theorems that help in determining a domain of uniqueness (as opposed to merely proving such a domain exists)?
 
  • #3
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I think you mean [itex]\frac{\partial f}{\partial y}[/itex] = [itex]\frac{2x^2y}{(x^2+y^2)^2}[/itex]
Beyond that I cannot help, as I am ignorant of this topic. Can you quote any theorems that help in determining a domain of uniqueness (as opposed to merely proving such a domain exists)?

I see the same goes for me. I've been reading this textbook with no luck or comprehension. There is a theorem, not exactly stated, but:

Given[itex]\frac{dy}{dx}[/itex] = f(x,y) with y(x) = yo, if f(x,y) and [itex]\frac{\partial f}{\partial y}[/itex] are continuous on an interval containing the initial point (xo, yo) then there is a unique function, y that satisfies the IVP.
 
  • #4
haruspex
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Yes, I think that's the theorem I found, and it only says there exists some interval around the point where the function is unique. It gives no clue as to how large that interval is.
fwiw, I believe I can solve the equation by converting to polar.
Have you tried sketching it? Looking for specific solutions (try y = ax)?
 

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