First, separate the variables:

\(\frac{y}{y^2 + 5} dy = \frac{1}{x + 1} dx\)

Integrate both sides:

\(\int \frac{y}{y^2 + 5} dy = \int \frac{1}{x + 1} dx\)

The left side integrates to \(\frac{1}{2} \ln(y^2 + 5)\) and the right side to \(\ln(x + 1)\).

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

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

\(\frac{1}{2} \ln(4 + 5) = \ln(1) + C\)

\(\frac{1}{2} \ln(9) = C\)

\(C = \frac{1}{2} \ln(9)\)

Substitute back to get:

\(\frac{1}{2} \ln(y^2 + 5) = \ln(x + 1) + \frac{1}{2} \ln(9)\)

Multiply through by 2:

\(\ln(y^2 + 5) = 2\ln(x + 1) + \ln(9)\)

Combine the logarithms:

\(\ln(y^2 + 5) = \ln(9(x + 1)^2)\)

Exponentiate both sides:

\(y^2 + 5 = 9(x + 1)^2\)

Thus, \(y^2 = 9(x + 1)^2 - 5\).