First, separate the variables:
\(\frac{y + 4}{y^2 + 4} \, dy = \frac{1}{x} \, dx\)
Integrate both sides:
\(\int \frac{y + 4}{y^2 + 4} \, dy = \int \frac{1}{x} \, dx\)
This gives:
\(\frac{1}{2} \ln(y^2 + 4) + 2 \arctan \frac{y}{2} = \ln x + C\)
Use the initial condition \(x = 4\) when \(y = 2\sqrt{3}\) to find \(C\):
\(\frac{1}{2} \ln((2\sqrt{3})^2 + 4) + 2 \arctan \frac{2\sqrt{3}}{2} = \ln 4 + C\)
\(\frac{1}{2} \ln(16) + 2 \times \frac{\pi}{3} = \ln 4 + C\)
\(\ln 4 + \frac{2\pi}{3} = \ln 4 + C\)
\(C = \frac{2\pi}{3}\)
Substitute \(y = 2\) to find \(x\):
\(\frac{1}{2} \ln(2^2 + 4) + 2 \arctan \frac{2}{2} = \ln x + \frac{2\pi}{3}\)
\(\frac{1}{2} \ln 8 + 2 \times \frac{\pi}{4} = \ln x + \frac{2\pi}{3}\)
\(\frac{1}{2} \ln 8 + \frac{\pi}{2} = \ln x + \frac{2\pi}{3}\)
\(\ln x = \frac{1}{2} \ln 8 + \frac{\pi}{2} - \frac{2\pi}{3}\)
\(\ln x = \frac{1}{2} \ln 8 - \frac{\pi}{6}\)
\(x = e^{\frac{1}{2} \ln 8 - \frac{\pi}{6}}\)
\(x = \frac{\sqrt{8}}{e^{\frac{\pi}{6}}}\)
