First, separate the variables:

\(\frac{dy}{1 + y^2} = dx.\)

Integrate both sides:

\(\int \frac{dy}{1 + y^2} = \int dx.\)

This gives:

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

Use the initial condition \(y = 2\) when \(x = 0\) to find \(C\):

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

\(\frac{1}{2} \ln 5 = C.\)

Substitute \(C\) back into the equation:

\(\frac{1}{2} \ln(1 + y^2) = x + \frac{1}{2} \ln 5.\)

Rearrange to solve for \(y^2\):

\(\ln(1 + y^2) = 2x + \ln 5\)

\(1 + y^2 = 5e^{2x}\)

\(y^2 = 5e^{2x} - 1.\)