Unique Solution for ODE y' = y^2/(x^2+y^2) | Region of Uniqueness

[aaronfue]

Homework Statement

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

y' = \frac{y^2}{x^2+y^2}

The Attempt at a Solution

\frac{\partial f}{\partial y} = \frac{2x^y}{(x^2+y^2)^2}

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!?

 

[haruspex]

\frac{\partial f}{\partial y} = \frac{2x^y}{(x^2+y^2)^2}

I think you mean \frac{\partial f}{\partial y} = \frac{2x^2y}{(x^2+y^2)^2}
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)?

 

[aaronfue]

I think you mean \frac{\partial f}{\partial y} = \frac{2x^2y}{(x^2+y^2)^2}
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\frac{dy}{dx} = f(x,y) with y(x) = y_o, if f(x,y) and \frac{\partial f}{\partial y} are continuous on an interval containing the initial point (x_o, y_o) then there is a unique function, y that satisfies the IVP.

 

[haruspex]

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)?