Start by separating the variables:

\(\frac{1}{y} dy = \frac{1 - 2x^2}{x} dx\)

Integrate both sides:

\(\int \frac{1}{y} dy = \int \frac{1 - 2x^2}{x} dx\)

This gives:

\(\ln y = \ln x - x^2 + C\)

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

\(\ln 1 = \ln 1 - 1^2 + C\)

\(0 = 0 - 1 + C\)

\(C = 1\)

Substitute \(C\) back into the equation:

\(\ln y = \ln x - x^2 + 1\)

Exponentiate both sides to solve for \(y\):

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

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

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