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Recently, I have learned the maximal existence interval of ODEs. I met a differential equation$$\frac{dy}{dx}=\frac{1}{x^2+y^2}.$$ The answer of the maximal existence interval of the ODE is$(0,+\infty)$ or $(-\infty,0)$ or $(-\infty,+\infty)$. But I don't know how to get $(0,+\infty)$ or $(-\infty,0)$. I tried to solve this ODE. But it was difficult.

R-CH2OH
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    Hint: Consider the IVP $\frac{dx}{dy}=x^2+y^2$ with $x(0)=0$, the solution $x=\varphi(y)$ can be written as $y=\psi(x)$ (prove it!) and $y=\psi(x), x\neq0$ is the solution of $\frac{dy}{dx}=\frac1{x^2+y^2}$. Then try to find a solution with maximal existence interval $(0,\infty)$. – Feng Oct 20 '22 at 03:10
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    Similar question in https://math.stackexchange.com/questions/2835521/limit-of-function-as-x-to-infty-when-fx-is-given – Lutz Lehmann Oct 20 '22 at 03:20

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