Start by separating variables: \(xy \frac{dy}{dx} = y^2 + 4\) becomes \(y \frac{dy}{y^2 + 4} = \frac{dx}{x}\).

Integrate both sides: \(\int \frac{y \, dy}{y^2 + 4} = \int \frac{dx}{x}\).

The left side integrates to \(\frac{1}{2} \ln(y^2 + 4)\) and the right side to \(\ln x + C\).

Thus, \(\frac{1}{2} \ln(y^2 + 4) = \ln x + C\).

Use the initial condition \(y = 0\) when \(x = 1\) to find \(C\):

\(\frac{1}{2} \ln(0^2 + 4) = \ln 1 + C\) gives \(\frac{1}{2} \ln 4 = C\).

Substitute back: \(\frac{1}{2} \ln(y^2 + 4) = \ln x + \frac{1}{2} \ln 4\).

Rearrange to find \(y^2\):

\(\ln(y^2 + 4) = 2 \ln x + \ln 4\).

\(\ln(y^2 + 4) = \ln(4x^2)\).

\(y^2 + 4 = 4x^2\).

\(y^2 = 4x^2 - 4\).

Thus, \(y^2 = 4(x^2 - 1)\).