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I'm confused regarding the following;

If we have a linear second order ODE, for example: $$y'' -\frac{8}{x}y' + \frac{8}{x^2}y = 0$$subject to the initial conditions $y(1)=1$ and $y'(1) = 4$, then there exists a unique solution in an interval $I$ where the coefficients $-\frac{8}{x}$ and $\frac{8}{x^2}$ are continuous.

My questions is this, does there exist a solution in $\mathbb{R}$, with the exception of zero, or is there a unique solution in $\mathbb{R^+},$ because that's where the initial condition lies, or can I not say this?

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  • $\begingroup$ This is the thing about existence and uniqueness theorems, they can't really specify the biggest interval of existence, but you can definitely say that a solution exists in an interval containing $x=1$. That is, it's at least existent in $(0,\alpha)$ for some $\alpha >1$. $\endgroup$ – DaveNine Feb 24 '16 at 0:33
  • $\begingroup$ To add on, if you go into the negative reals, you no longer have an interval, but two disjoint intervals separated by the bad point at 0 $\endgroup$ – DaveNine Feb 24 '16 at 0:35

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