The solution of the differential equation $\frac{d y}{d x}=\frac{x^2+y^2+1}{2 x y}$ satisfying $y(1)=1$, is

a hyperbola

We have,

$\frac{d y}{d x}=\frac{x^2+y^2+1}{2 x y}$

$\Rightarrow 2 x y d y=\left(x^2+y^2+1\right) d x$

$\Rightarrow 2 x y d y-y^2 d x=\left(x^2+1\right) d x$

$\Rightarrow x d\left(y^2\right)-y^2 d x=\left(x^2+1\right) d x$

$\Rightarrow \frac{x d\left(y^2\right)-y^2 d x}{x^2}=\left(1+\frac{1}{x^2}\right) d x$

$\Rightarrow d\left(\frac{y^2}{x}\right)=d\left(x-\frac{1}{x}\right)$

On integrating, we get

$\frac{y^2}{x} =x-\frac{1}{x}+C$

$\Rightarrow y^2=x^2-1+C x \Rightarrow y^2=\left(x+\frac{C}{2}\right)^2-1-\frac{C^2}{4}$

Clearly, it represents a hyperbola.