Start by separating the variables:

\(\frac{1}{y} \frac{dy}{dx} = \frac{6x}{x^2 + 4}\)

Integrate both sides:

\(\int \frac{1}{y} \, dy = \int \frac{6x}{x^2 + 4} \, dx\)

This gives:

\(\ln y = 3 \ln(x^2 + 4) + C\)

Use the initial condition \(y = 32\) when \(x = 0\):

\(\ln 32 = 3 \ln 4 + C\)

Solving for \(C\), we find:

\(C = \ln 32 - 3 \ln 4\)

Substitute back to find \(y\):

\(\ln y = 3 \ln(x^2 + 4) + \ln 32 - 3 \ln 4\)

\(\ln y = \ln((x^2 + 4)^3) + \ln \frac{32}{4^3}\)

\(\ln y = \ln((x^2 + 4)^3 \times \frac{1}{2})\)

\(y = \frac{1}{2}(x^2 + 4)\)