Question:

\subsection*{Question 3: Solution} We are given \[ \frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}. \] Multiply both sides by \(2y\): \[ 2y\frac{dy}{dx} =\frac{x^2+y^2+1}{x}. \] Put \[ u=y^2. \] Then \[ \frac{du}{dx}=2y\frac{dy}{dx}. \] So \[ \frac{du}{dx} =\frac{x^2+u+1}{x} =x+\frac{u}{x}+\frac{1}{x}. \] Hence \[ \frac{du}{dx}-\frac{u}{x} =x+\frac{1}{x}. \] This is a linear differential equation. Its integrating factor is \[ e^{\int -1/x\,dx}=\frac{1}{x}. \] Multiplying by \(1/x\), \[ \frac{d}{dx}\left(\frac{u}{x}\right) =1+\frac{1}{x^2}. \] Integrating, \[ \frac{u}{x}=x-\frac{1}{x}+C. \] Thus \[ u=x^2-1+Cx. \] Since \(u=y^2\), \[ y^2=x^2-1+Cx. \] Using \(y(1)=0\), \[ 0=1-1+C, \] so \[ C=0. \] Therefore \[ y^2=x^2-1, \] or \[ x^2-y^2=1. \] This represents a hyperbola. \[ \boxed{\text{hyperbola}} \]