I am given the solution y=yo/(1-yo x^2). If yo>0, the solution is valid only on ( -1/sqrt(yo), 1/sqrt(yo) ). How is this determined? I need a proof like answer and I need it programmed in Mathematica. Can you help? Thank you .

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    $\begingroup$ Asking for help suggests that you've made a start on your problem yourself and need some assistance to finish it and to remove the rough edges. Show us what you've got so far. $\endgroup$ – High Performance Mark Feb 21 at 10:26
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    $\begingroup$ Perhaps you're looking for Reduce[]. I'm not sure what a "proof-like answer" would look like. Unless it's FindEquationalProof[]. $\endgroup$ – Michael E2 Feb 22 at 1:54

Setting the denominator equal to zero gives the points where your function goes to infinity. You can use Mathematica (or algebra) to find those points where 1-y0 x^2 is zero:

g[x_,y0_]:=y0/(1-y0 x^2);
Assuming[{y0>0}, Solve[1/g[x,y0]==0,x]

This gives

{{x -> -1/Sqrt[y0]},{x -> 1/Sqrt[y0]}}

The solution is valid between these two points.

Here is an example with y0=9. The solution between -1/3 and 1/3 is finite, and approaches infinity at both ends of the interval. The function also exists to the left and to the right of the interval. Perhaps you could say why you believe y does not exist there also.

 y0 = 9; Plot[g[x, y0], {x, -1, 1}, PlotRange -> All, Frame -> True, Axes -> False]

The function y=y(x).


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