First, separate 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 \(y = 2\) when \(x = 1\) to find \(C\):

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

\(C = \ln 2 + 1\)

Substitute back to get:

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

Exponentiate to solve for \(y\):

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

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