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