Differential Equations Challenge!

Let us write the above DE as the following:

y' = (x^2 + y^2 + 1)/2xy;

or x*(2yy') = x^2 + y^2 + 1;

or x*(2yy') - y^2 = x^2 + 1.

If we now divide both sides by x^2, we arrive at:

[x*(2yy') - y^2] / x^2 = (x^2 + 1) / x^2;

or (y^2 / x)' = 1 + 1/x^2 (i).

Integrating both sides of (i) with respect to x gives: y^2 / x = x - 1/x + C, and the boundary condition y(1) = 0 yields C = 0. We ultimately end up with the equation:

y^2 / x = x - 1/x => y^2 = x^2 - 1 => x^2 - y^2 = 1

or a hyperbola.

Set $y'=0$ to arrive at $x^2+y^2+1=0$ . Which is true for no pair of real $x,y$ . We can discard the first two options with this since both have zero slope at some point. Consider the scalar function given by $y'(x,y) = \frac{x^2+y^2+1}{2xy}$ . And now consider $y'(1,y)=\frac{1}{y} + \frac{y}{2}$ We notice that for positive "y" this function gives positives values, and the same thing happens for negative y. As y approaches 0, y' tends to infinity; which suggests that we're dealing with an hyperbola with one vertex at (0,1) (since at such point it has a vertical slope) and extending along the positive x-axis. Propose $(x/a)^2-(y/b)^2=1$ . Plug it in the ode to obtain: $b^2-1=x^2 (1-(b/a)^2)$ . Alude to the linear independence of polynomials to get that both a and b are +-1. Hence the desired solution.