Separate variables:

\(\frac{dy}{y-1} = \frac{dx}{(x+1)(x+3)}\)

Integrate both sides:

\(\int \frac{dy}{y-1} = \int \frac{dx}{(x+1)(x+3)}\)

\(\ln|y-1| = \int \left( \frac{A}{x+1} + \frac{B}{x+3} \right) dx\)

Find \(A\) and \(B\) such that:

\(\frac{1}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}\)

Solving gives \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\).

Integrate:

\(\ln|y-1| = \frac{1}{2} \ln|x+1| - \frac{1}{2} \ln|x+3| + C\)

\(\ln|y-1| = \frac{1}{2} \ln\left(\frac{x+1}{x+3}\right) + C\)

Exponentiate both sides:

\(|y-1| = e^C \sqrt{\frac{x+1}{x+3}}\)

Use initial condition \(y = 2\) when \(x = 0\):

\(1 = e^C \sqrt{\frac{1}{3}}\)

\(e^C = \sqrt{3}\)

Thus, \(y - 1 = \sqrt{3} \sqrt{\frac{x+1}{x+3}}\)

\(y = 1 + \sqrt{\frac{3x+3}{x+3}}\)

(i) Express \(\frac{1}{4-y^2}\) as partial fractions:

\(\frac{1}{4-y^2} = \frac{A}{2+y} + \frac{B}{2-y}\)

Multiply through by \((2+y)(2-y)\):

\(1 = A(2-y) + B(2+y)\)

Equating coefficients, solve for \(A\) and \(B\):

\(A = \frac{1}{4}, \; B = \frac{1}{4}\)

(ii) Solve the differential equation:

Separate variables:

\(\frac{dy}{4-y^2} = \frac{x}{4-y^2} dx\)

Integrate both sides:

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

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

Use initial condition \(x = 1, y = 1\):

\(\ln 3 = \frac{1}{2} + C\)

\(C = \ln 3 - \frac{1}{2}\)

Substitute back:

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

Rearrange to find \(y\):

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

(i) To express \(\frac{1}{x(2x+3)}\) in partial fractions, assume \(\frac{1}{x(2x+3)} = \frac{A}{x} + \frac{B}{2x+3}\).

Multiply through by \(x(2x+3)\) to get: \(1 = A(2x+3) + Bx\).

Equating coefficients, solve for \(A\) and \(B\):

\(A = \frac{1}{3}\) and \(B = -\frac{2}{3}\).

(ii) Separate variables: \(\frac{dy}{y} = \frac{1}{x(2x+3)} dx\).

Integrate both sides: \(\int \frac{dy}{y} = \int \frac{1}{x} dx - \int \frac{1}{2x+3} dx\).

\(\ln y = \frac{1}{3} \ln x - \frac{1}{3} \ln(2x+3) + C\).

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

\(\ln 1 = \frac{1}{3} \ln 1 - \frac{1}{3} \ln 5 + C\) gives \(C = \frac{1}{3} \ln 5\).

Thus, \(\ln y = \frac{1}{3} \ln x - \frac{1}{3} \ln(2x+3) + \frac{1}{3} \ln 5\).

When \(x = 9\), \(\ln y = \frac{1}{3} \ln 9 - \frac{1}{3} \ln 21 + \frac{1}{3} \ln 5\).

Calculate \(y\) to 3 significant figures: \(y = 1.29\).

Separate variables and factorize to obtain:

\(\frac{dy}{(3y+1)(y+3)} = 4x \, dx\)

Express the left side as partial fractions:

\(\frac{A}{3y+1} + \frac{B}{y+3}\)

Solving for \(A\) and \(B\), we find \(A = \frac{3}{8}\) and \(B = -\frac{1}{8}\).

Integrate both sides:

\(\int \left( \frac{3}{8(3y+1)} - \frac{1}{8(y+3)} \right) dy = \int 4x \, dx\)

This gives:

\(\frac{1}{8} \ln(3y+1) - \frac{1}{8} \ln(y+3) = 2x^2 + c\)

Substitute \(x = 0\) and \(y = 1\) to find \(c\):

\(\frac{1}{8} \ln(4) - \frac{1}{8} \ln(4) = c\)

\(c = 0\)

Rearrange to solve for \(y\):

\(\ln \left( \frac{3y+1}{y+3} \right) = 16x^2\)

\(\frac{3y+1}{y+3} = e^{16x^2}\)

\(3y + 1 = (y + 3)e^{16x^2}\)

\(3y + 1 = ye^{16x^2} + 3e^{16x^2}\)

\(y(3 - e^{16x^2}) = 3e^{16x^2} - 1\)

\(y = \frac{3e^{16x^2} - 1}{3 - e^{16x^2}}\)

(i) To express \(\frac{1}{x^2(2x+1)}\) in the form \(\frac{A}{x^2} + \frac{B}{x} + \frac{C}{2x+1}\), we equate:

\(\frac{1}{x^2(2x+1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{2x+1}\)

Multiply through by \(x^2(2x+1)\) to clear the denominators:

\(1 = A(2x+1) + Bx(2x+1) + Cx^2\)

Expanding and collecting terms gives:

\(1 = (2A)x + A + (2B)x^2 + Bx + Cx^2\)

Equating coefficients, we get:

\(A = 1\), \(2B + C = 0\), \(2A + B = 0\).

Solving these, we find \(A = 1\), \(B = -2\), \(C = 4\).

(ii) Given \(y = x^2(2x+1) \frac{dy}{dx}\), separate variables:

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

Integrate both sides:

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

Using the initial condition \(y = 1\) when \(x = 1\), solve for \(c\):

\(\ln 1 = -\frac{1}{2} \ln 1 + 2 \ln 3 + c\)

\(c = -2 \ln 3\)

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

Exponentiate to solve for \(y\):

\(y = \frac{(2x+1)^2}{3^2 x^{1/2}}\)

Substitute \(x = 2\):

\(y = \frac{(5)^2}{9 \sqrt{2}} = \frac{25}{36} e^{\frac{1}{2}}\)