Separate variables:

\(\frac{dy}{y-1} = \frac{dx}{(x+1)(x+3)}\)

Integrate both sides:

\(\int \frac{dy}{y-1} = \int \frac{dx}{(x+1)(x+3)}\)

\(\ln|y-1| = \int \left( \frac{A}{x+1} + \frac{B}{x+3} \right) dx\)

Find \(A\) and \(B\) such that:

\(\frac{1}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}\)

Solving gives \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\).

Integrate:

\(\ln|y-1| = \frac{1}{2} \ln|x+1| - \frac{1}{2} \ln|x+3| + C\)

\(\ln|y-1| = \frac{1}{2} \ln\left(\frac{x+1}{x+3}\right) + C\)

Exponentiate both sides:

\(|y-1| = e^C \sqrt{\frac{x+1}{x+3}}\)

Use initial condition \(y = 2\) when \(x = 0\):

\(1 = e^C \sqrt{\frac{1}{3}}\)

\(e^C = \sqrt{3}\)

Thus, \(y - 1 = \sqrt{3} \sqrt{\frac{x+1}{x+3}}\)

\(y = 1 + \sqrt{\frac{3x+3}{x+3}}\)