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Consider the Cauchy problem$$(*)\begin{cases}y'=\frac{y^2-1}{x^2} \\y(1)=1.\end{cases}$$ We note that $y'=h(x)k(y)$ with $h(x)$ continuous on $\Bbb R\setminus\{0\}$ and $k(y)$ continuously differentiable on $\Bbb R$, and since $(x_0,y_0)=(1,1)\in(0,+\infty)\times\Bbb R$, this is the set we consider.

Cauchy-Lipschitz theorem then guarantees $y(x)\equiv1$ is the unique solution of $(*)$ on $(0,+\infty)$. Since it's differentiable on all $\Bbb R$, does it make sense to say it can be prolonged on all $\Bbb R$?

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    $\begingroup$ The differential equation does not exist at $x=0$, thus you can not have any solution through that point. That you can continue the function does not matter, you can not continue the solution as solution. $\endgroup$ – LutzL Jan 15 at 13:59
  • $\begingroup$ @LutzL Many thanks! $\endgroup$ – Learner Jan 15 at 14:06
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As per @LutzL's comment: no. That you can continue the function does not matter, you can not continue the solution as solution.

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