Can we say that the graph is symmetric about origin. Because replacing $x$, $y$ with $-x$, $-y$ does not change the equation
Also the slope becomes larger as we move away from origin.
Anything else that you guys can conclude?
Thanks
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The slope is always positive? There's an extrema at the origin whose nature you can find by substituting $(0,0)$ into the second derivative. – Zain Patel Jul 20 '15 at 12:02
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This is a standard example for a Riccati equation. See Riccati D.E., vertical asymptotes and the related topics in the right sidebar. – Lutz Lehmann Sep 28 '19 at 22:34
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The equation of a circle about the origin is $x^2 + y^2 = r^2$, for radius $r$. So we see that the rate $y'$ is constant along such circles. That's just one observation I can think of.
Colm Bhandal
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1) Solutions are increasing.
2) The graph is not in general symmetric around the origin, it's only symmetric (odd) if $y(0)=0$.
3) The slope of the entire family is constant on concentric circles.
4) The solution (starting at $x=0$) grows faster than $y(0)+x^3/3$ and also faster than $y(0)/(1-xy(0))$. From the second property, we know it has poles.
orion
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