Why can a slope field include points outside the domain of the original function?

Question

I'm looking specifically at the slope field for $y'=\frac{2x}{y}$, which is the derivative of the function $2x^2-y^2=1$ (one of the solutions). But for no "family of functions" is a point, say $(0,1)$ valid for this solution, because if I solve for the family of functions: $y=\pm\sqrt{2x^2-1} + C$ ......then $(0,1)$ is still not a valid point, so why should it be part of the slope field? Or part of the derivative for that matter?

Answer 1

You were given the differential equation $$y'={2x\over y}\quad (y\ne0)\tag{1}$$ and found the solution $$2x^2-y^2=1\ .\tag{2}$$ You are right that there is an infinite family of solutions. But this family cannot be derived from the random particular solution you have found without recurse to $(1)$. You just solved $(2)$ for $y$ and claimed that $y(x)=\pm\sqrt{2x^2-1}+C$ is the general solution of $(1)$. Things don't work this way.

In fact you can write $(1)$ in the form $2yy'-4x=0$, or $$\bigl(y^2-2x^2)'=0\ .$$ This says that for any solution $x\mapsto y(x)$ the quantity $$y^2(x)-2x^2={\rm const.}$$ It follows that all solutions are parts of curves $$y^2-2x^2=C\ ,$$ all of them hyperbolas with asymptotes $y=\pm\sqrt{2}\,x$.