Can the unique, stationary solution of a Cauchy problem be prolonged on all $\Bbb R$ even if the differential equation is undefined at one point?

Question

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 $\Bbb R$?

Answer 1

As per @LutzL's comment: no. That you can continue the function does not matter, you can not continue the solution as solution.