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 \(x\):

\(\frac{1}{2} \ln(2^2 + 4) + 2 \arctan \frac{2}{2} = \ln x + \frac{2\pi}{3}\)

\(\frac{1}{2} \ln 8 + 2 \times \frac{\pi}{4} = \ln x + \frac{2\pi}{3}\)

\(\frac{1}{2} \ln 8 + \frac{\pi}{2} = \ln x + \frac{2\pi}{3}\)

\(\ln x = \frac{1}{2} \ln 8 + \frac{\pi}{2} - \frac{2\pi}{3}\)

\(\ln x = \frac{1}{2} \ln 8 - \frac{\pi}{6}\)

\(x = e^{\frac{1}{2} \ln 8 - \frac{\pi}{6}}\)

\(x = \frac{\sqrt{8}}{e^{\frac{\pi}{6}}}\)

Start by separating the variables:

\(\frac{1}{y} \frac{dy}{dx} = \frac{6x}{x^2 + 4}\)

Integrate both sides:

\(\int \frac{1}{y} \, dy = \int \frac{6x}{x^2 + 4} \, dx\)

This gives:

\(\ln y = 3 \ln(x^2 + 4) + C\)

Use the initial condition \(y = 32\) when \(x = 0\):

\(\ln 32 = 3 \ln 4 + C\)

Solving for \(C\), we find:

\(C = \ln 32 - 3 \ln 4\)

Substitute back to find \(y\):

\(\ln y = 3 \ln(x^2 + 4) + \ln 32 - 3 \ln 4\)

\(\ln y = \ln((x^2 + 4)^3) + \ln \frac{32}{4^3}\)

\(\ln y = \ln((x^2 + 4)^3 \times \frac{1}{2})\)

\(y = \frac{1}{2}(x^2 + 4)\)

Start by separating variables:

\(\frac{dx}{dt} = \frac{1}{x} - \frac{x}{4}\)

Rearrange to:

\(x \frac{dx}{dt} = 1 - \frac{x^2}{4}\)

Separate variables:

\(\frac{x}{1 - \frac{x^2}{4}} dx = dt\)

Integrate both sides:

\(\int \frac{x}{1 - \frac{x^2}{4}} dx = \int dt\)

Let \(u = 1 - \frac{x^2}{4}\), then \(du = -\frac{x}{2} dx\).

Substitute and integrate:

\(-2 \int \frac{1}{u} du = t + C\)

\(-2 \ln|u| = t + C\)

Substitute back for \(u\):

\(-2 \ln|1 - \frac{x^2}{4}| = t + C\)

Use initial condition \(x = 1\) when \(t = 0\):

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

\(-2 \ln\left(\frac{3}{4}\right) = C\)

\(C = 2 \ln 3\)

Substitute \(C\) back:

\(-2 \ln|1 - \frac{x^2}{4}| = t + 2 \ln 3\)

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

\(\ln|1 - \frac{x^2}{4}| = -\frac{1}{2}t - \ln 3\)

\(|1 - \frac{x^2}{4}| = 3\exp\left(-\frac{1}{2}t\right)\)

\(1 - \frac{x^2}{4} = 3\exp\left(-\frac{1}{2}t\right)\)

\(x^2 = 4 - 3\exp\left(-\frac{1}{2}t\right)\)

Start by separating variables: \(xy \frac{dy}{dx} = y^2 + 4\) becomes \(y \frac{dy}{y^2 + 4} = \frac{dx}{x}\).

Integrate both sides: \(\int \frac{y \, dy}{y^2 + 4} = \int \frac{dx}{x}\).

The left side integrates to \(\frac{1}{2} \ln(y^2 + 4)\) and the right side to \(\ln x + C\).

Thus, \(\frac{1}{2} \ln(y^2 + 4) = \ln x + C\).

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

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

Substitute back: \(\frac{1}{2} \ln(y^2 + 4) = \ln x + \frac{1}{2} \ln 4\).

Rearrange to find \(y^2\):

\(\ln(y^2 + 4) = 2 \ln x + \ln 4\).

\(\ln(y^2 + 4) = \ln(4x^2)\).

\(y^2 + 4 = 4x^2\).

\(y^2 = 4x^2 - 4\).

Thus, \(y^2 = 4(x^2 - 1)\).

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 \(y^2\):

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

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

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