Start by separating variables: \(\frac{dy}{y} = \frac{(2 - x^2)}{x} dx\).

Integrate both sides: \(\int \frac{dy}{y} = \int \left(\frac{2}{x} - x\right) dx\).

This gives \(\ln y = 2 \ln x - \frac{1}{2} x^2 + C\).

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

\(\ln 1 = 2 \ln 1 - \frac{1}{2} \cdot 1^2 + C\)

\(0 = 0 - \frac{1}{2} + C\)

\(C = \frac{1}{2}\)

Substitute \(C\) back into the equation: \(\ln y = 2 \ln x - \frac{1}{2} x^2 + \frac{1}{2}\).

Rearrange to solve for \(y\):

\(y = e^{2 \ln x - \frac{1}{2} x^2 + \frac{1}{2}}\)

\(y = e^{2 \ln x} \cdot e^{-\frac{1}{2} x^2} \cdot e^{\frac{1}{2}}\)

\(y = x^2 \cdot e^{-\frac{1}{2} x^2} \cdot e^{\frac{1}{2}}\)

\(y = x^{-2} \cdot e^{\frac{1}{2} - \frac{1}{2} x^2}\)