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 \(y^2\):

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

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

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

Start by separating variables:

\(\frac{dy}{y^2} = \left(1 + \frac{1}{y^3}\right) dx\)

Integrate both sides:

\(\int \frac{dy}{y^2} = \int \left(1 + \frac{1}{y^3}\right) dx\)

This gives:

\(-\frac{1}{y} = x + C\)

Using the initial condition \(y = 1\) when \(x = 0\), find \(C\):

\(-1 = 0 + C\)

So, \(C = -1\).

The equation becomes:

\(-\frac{1}{y} = x - 1\)

Rearrange to solve for \(y\):

\(y = -\frac{1}{x - 1}\)

Rewriting in terms of \(y^3\), we have:

\(y^3 = (2e^{3x} - 1)\)

Thus, \(y = (2e^{3x} - 1)^{\frac{1}{3}}\).

To solve the differential equation \(\frac{dy}{dx} = \frac{xy}{1+x^2}\), we first separate the variables:

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

Integrate both sides:

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

This gives:

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

Exponentiate both sides to solve for \(y\):

\(|y| = e^C \sqrt{1+x^2}\).

Using the initial condition \(y = 2\) when \(x = 0\), we find:

\(2 = e^C \sqrt{1+0^2}\), so \(e^C = 2\).

Thus, \(y = 2\sqrt{1+x^2}\).

The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y}{\sqrt{x+1}}\), where \(k\) is a constant of proportionality.

Separate variables: \(\frac{1}{y} dy = \frac{k}{\sqrt{x+1}} dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{k}{\sqrt{x+1}} dx\).

This gives \(\ln y = 2k \sqrt{x+1} + c\), where \(c\) is the constant of integration.

Using the point \((0, 1)\), substitute into the equation: \(\ln 1 = 2k \sqrt{0+1} + c\), which simplifies to \(c = -2k\).

Using the point \((3, e)\), substitute into the equation: \(\ln e = 2k \sqrt{3+1} - 2k\), which simplifies to \(1 = 4k - 2k\).

Solve for \(k\): \(k = \frac{1}{2}\).

Substitute \(k = \frac{1}{2}\) and \(c = -1\) back into the equation: \(\ln y = \sqrt{x+1} - 1\).

Exponentiate both sides to solve for \(y\): \(y = \exp(\sqrt{x+1} - 1)\).

Start by separating the variables:

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

Integrate both sides:

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

This gives:

\(\ln y = \ln x - x^2 + C\)

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

\(\ln 1 = \ln 1 - 1^2 + C\)

\(0 = 0 - 1 + C\)

\(C = 1\)

Substitute \(C\) back into the equation:

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

Exponentiate both sides to solve for \(y\):

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

\(y = e^{\ln x} \cdot e^{-x^2} \cdot e^{1}\)

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

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

Start by separating variables: \(\frac{dy}{y} = \frac{(2 - x^2)}{x} dx\).

Integrate both sides: \(\int \frac{dy}{y} = \int \left(\frac{2}{x} - x\right) dx\).

This gives \(\ln y = 2 \ln x - \frac{1}{2} x^2 + C\).

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

\(\ln 1 = 2 \ln 1 - \frac{1}{2} \cdot 1^2 + C\)

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

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

Substitute \(C\) back into the equation: \(\ln y = 2 \ln x - \frac{1}{2} x^2 + \frac{1}{2}\).

Rearrange to solve for \(y\):

\(y = e^{2 \ln x - \frac{1}{2} x^2 + \frac{1}{2}}\)

\(y = e^{2 \ln x} \cdot e^{-\frac{1}{2} x^2} \cdot e^{\frac{1}{2}}\)

\(y = x^2 \cdot e^{-\frac{1}{2} x^2} \cdot e^{\frac{1}{2}}\)

\(y = x^{-2} \cdot e^{\frac{1}{2} - \frac{1}{2} x^2}\)

Separate variables:

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

Integrate both sides:

\(\ln y = x + \ln(x+1) + C\)

Use the initial condition \(y = 2\) when \(x = 1\):

\(\ln 2 = 1 + \ln(1+1) + C\)

\(\ln 2 = 1 + \ln 2 + C\)

\(C = -1\)

Substitute back to find \(y\):

\(\ln y = x + \ln(x+1) - 1\)

\(y = e^{x + \ln(x+1) - 1}\)

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

First, separate the variables:

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

Integrate both sides:

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

This gives:

\(\ln y = \ln x - x^2 + C\)

Use the initial condition \(y = 2\) when \(x = 1\) to find \(C\):

\(\ln 2 = \ln 1 - 1^2 + C\)

\(C = \ln 2 + 1\)

Substitute back to get:

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

Exponentiate to solve for \(y\):

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

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

(i) Separate variables and integrate:

\(\int 2x^2 \, dx = \int \frac{k - x^3}{t} \, dt\)

Integrate to obtain:

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

Use initial conditions \(t = 1, x = 1\) and \(t = 4, x = 2\) to solve for \(k\):

\(\ln 4 = -\frac{2}{3} \ln(k - 8) + \frac{2}{3} \ln(k - 1)\)

Solve for \(k\) to obtain \(k = 9\).

Substitute \(k = 9\) to find \(x\):

\(x = (9 - 8t^{\frac{3}{2}})^{\frac{1}{3}}\)

(ii) As \(t \to \infty\), \(x \to 9^{\frac{1}{3}}\).