(a) Separate variables: \(\int \frac{1}{4 + 9y^2} \, dy = \int e^{-(2x+1)} \, dx\).

Integrate both sides: \(\frac{1}{6} \arctan \left( \frac{3y}{2} \right) = -\frac{1}{2} e^{-(2x+1)} + C\).

Use initial condition \(y = 0\) when \(x = 1\) to find \(C\):

\(\frac{1}{6} \arctan(0) = -\frac{1}{2} e^{-3} + C\) gives \(C = \frac{1}{2} e^{-3}\).

Substitute \(C\) back: \(\frac{1}{6} \arctan \left( \frac{3y}{2} \right) = -\frac{1}{2} e^{-(2x+1)} + \frac{1}{2} e^{-3}\).

Solve for \(y\): \(y = \frac{2}{3} \arctan (3e^{-3} - 3e^{-2x-1})\).

(b) As \(x \to \infty\), \(e^{-2x-1} \to 0\), so \(y \to \frac{2}{3} \arctan (3e^{-3})\).