Yh dat will mk it undefined but then we needed de values of x instead dat will make it ubdefined

The solution is y=1/(x^2-1), and the interval is(0,infinity).

Hi Dr Qassem, The equation is dy/y^2 = - 2xdx (i.e., y' + 2xy^2 = 0), the solution is y = 1/ (x^2 - 1) , The initial condition means c = 1.

Hi Dr. Lakhdar

Thanks, that is exactly what I said.

Good work

Hii Adrien........ The largest interval over which solution is defined is (-infinity,-1) U (-1,1) U (1,infinity). As the solution of This IVP is y=1/(x^2 -1), so the only points where it is not defined are x=-1 and x= 1. So the interval is whole real line except these two points.

Hii, I would like to modify my answer... I read ur question again.. u have asked the largest interval, not largest set... so the actual answer should be (1,infinity) and it is justified as it should also contain the point where initial condition is given i.e. at x=2.

yeah after solving it as Dr Qassem said i came to largerst interval...thanks you all again