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

a hyperbola

The correct answer is option (1) : a hyperbola

We have,

$\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}$

$⇒2xy\, dy = (x^2 + y^2 +1) dx$

$⇒2xy \, dy  - y^2 dx= (x^2 + 1) dx$

$⇒xd(y^2)-y^2dx = (x^2+1) dx$

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

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

On integrating, we get

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

$⇒y^2=x^2-1+C\, x $

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

Clearly, it represents a hyperbola.