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[tex]y'=2x/(1+2y) , y(2)=0[/tex] and after solving, we get the equation [tex] y= -0.5 + \sqrt{x^{2} - 15/4}[/tex](discarded the negative squareroot since governed by initial condition).

Therefore, the interval is found to be [tex] x > or = \sqrt{15/4}[/tex], but my question is, why can't the interval also be defined at [tex] x < \sqrt{15/4}[/tex]? How does it violate the initial condition? The I.C. only points out that when x=2, y=0.

I think i'm missing some basic knowledge for initial value problems...

Thanks in advance