To solve the differential equation \(3y^2 \frac{dy}{dx} = 4(y^3 + 1) \cos^2 x\), we first use the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\) to rewrite the equation as:

\(3y^2 \frac{dy}{dx} = 2(y^3 + 1)(1 + \cos 2x)\).

Separate variables and integrate:

\(\int \frac{3y^2}{y^3 + 1} \, dy = \int 2(1 + \cos 2x) \, dx\).

The left side integrates to \(\ln(y^3 + 1)\), and the right side integrates to \(2x + \sin 2x + C\).

Thus, \(\ln(y^3 + 1) = 2x + \sin 2x + C\).

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

\(\ln(2^3 + 1) = \ln 9 = C\).

So, the equation becomes:

\(\ln(y^3 + 1) = 2x + \sin 2x + \ln 9\).

At stationary points, \(\frac{dy}{dx} = 0\), which implies \(\cos^2 x = 0\). Therefore, \(x = \frac{\pi}{2}\) or \(x = \frac{3\pi}{2}\).

For \(x = \frac{\pi}{2}\):

\(\ln(y^3 + 1) = 2\left(\frac{\pi}{2}\right) + \sin \pi + \ln 9 = \pi + \ln 9\).

\(y^3 + 1 = e^{\pi + \ln 9} = 9e^{\pi}\).

\(y^3 = 9e^{\pi} - 1\).

\(y = (9e^{\pi} - 1)^{1/3} \approx 5.9\).

For \(x = \frac{3\pi}{2}\):

\(\ln(y^3 + 1) = 2\left(\frac{3\pi}{2}\right) + \sin 3\pi + \ln 9 = 3\pi + \ln 9\).

\(y^3 + 1 = e^{3\pi + \ln 9} = 9e^{3\pi}\).

\(y^3 = 9e^{3\pi} - 1\).

\(y = (9e^{3\pi} - 1)^{1/3} \approx 48.1\).

(i) The volume of the cone is given by \(V = \frac{1}{3} \pi r^2 h\). Given the semi-vertical angle is 60°, \(r = h \tan(60^{\circ}) = \sqrt{3}h\). Thus, \(V = \pi h^3\). The rate of change of volume is \(\frac{dV}{dt} = -k\sqrt{h}\). Using \(V = \pi h^3\), \(\frac{dV}{dh} = 3\pi h^2\). Therefore, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 3\pi h^2 \cdot \frac{dh}{dt}\). Equating gives \(3\pi h^2 \cdot \frac{dh}{dt} = -k\sqrt{h}\), leading to \(\frac{dh}{dt} = -\frac{k}{3\pi} h^{-\frac{3}{2}}\). Let \(A = \frac{k}{3\pi}\), then \(\frac{dh}{dt} = -Ah^{-\frac{3}{2}}\).

(ii) Separate variables: \(\int h^{\frac{3}{2}} \, dh = -A \int \, dt\). Integrating gives \(\frac{2}{5} h^{\frac{5}{2}} = -At + C\). At \(t = 0, h = H\), so \(C = \frac{2}{5} H^{\frac{5}{2}}\). At \(t = 60, h = 0\), so \(0 = -60A + \frac{2}{5} H^{\frac{5}{2}}\), giving \(A = \frac{1}{150} H^{\frac{5}{2}}\). Substitute back to find \(t = 60 \left( 1 - \left( \frac{h}{H} \right)^{\frac{5}{2}} \right)\).

(iii) Substitute \(h = \frac{1}{2}H\) into the expression for t: \(t = 60 \left( 1 - \left( \frac{1}{2} \right)^{\frac{5}{2}} \right) = 49.4\).

(i) The rate of change of volume \(\frac{dV}{dt}\) is given by the inflow minus the outflow: \(\frac{dV}{dt} = 80 - kV\).

Separate variables: \(\frac{dV}{80 - kV} = dt\).

Integrate both sides: \(-\frac{1}{k} \ln(80 - kV) = t + C\).

Using \(t = 0\) and \(V = 0\), find \(C\): \(-\frac{1}{k} \ln(80) = C\).

Substitute back: \(-\frac{1}{k} \ln(80 - kV) = t - \frac{1}{k} \ln(80)\).

Simplify to find \(V\): \(V = \frac{1}{k}(80 - 80e^{-kt})\).

(ii) Use the iterative formula \(k_{n+1} = \frac{4 - 4e^{-15k_n}}{25}\) with \(k_0 = 0.1\).

Perform iterations: \(k_1 = 0.1368\), \(k_2 = 0.1399\), \(k_3 = 0.1400\), \(k_4 = 0.1400\).

Final value of \(k\) is 0.14 to 2 significant figures.

(iii) Substitute \(t = 20\) into \(V = \frac{1}{k}(80 - 80e^{-kt})\) with \(k = 0.14\).

Calculate \(V \approx 535\) cm³.

As \(t \to \infty\), \(V \to \frac{80}{k} \approx 569\) cm³.

(i) Substitute \(x = 5e^{-3t}\) into the differential equation \(\frac{dy}{dt} = -0.6xy\) to get \(\frac{dy}{dt} = -0.6(5e^{-3t})y = -3e^{-3t}y\).

Separate variables: \(\frac{1}{y} dy = -3e^{-3t} dt\).

Integrate both sides: \(\ln y = e^{-3t} + C\).

Use the initial condition \(t = 0, y = 70\) to find \(C\): \(\ln 70 = 1 + C\), so \(C = \ln 70 - 1\).

Thus, \(\ln y = e^{-3t} + \ln 70 - 1\).

Rearrange to find \(y\): \(y = 70(e^{-3t} - 1)\).

(ii) As \(t \to \infty\), \(e^{-3t} \to 0\), so \(y \to 70(-1) = -70\).

The percentage of the initial mass of B remaining is \(p = \frac{y}{70} \times 100\).

Thus, \(p \to \frac{100}{e}\) as \(t \to \infty\).

(i) The rate of formation of X is proportional to the product of the masses of Y and Z, which are \(10-x\) and \(20-x\) respectively. Therefore, \(\frac{dx}{dt} = k(10-x)(20-x)\). Given \(\frac{dx}{dt} = 2\) when t = 0 and x = 0, we have \(2 = k(10)(20)\), so \(k = 0.01\). Thus, \(\frac{dx}{dt} = 0.01(10-x)(20-x)\).

(ii) Separate variables: \(\frac{1}{(10-x)(20-x)} dx = 0.01 dt\).

Use partial fractions: \(\frac{1}{(10-x)(20-x)} = \frac{A}{10-x} + \frac{B}{20-x}\).

Solving gives \(A = \frac{1}{10}\) and \(B = -\frac{1}{10}\).

Integrate: \(-\frac{1}{10} \ln(10-x) + \frac{1}{10} \ln(20-x) = 0.01t + C\).

At t = 0, x = 0: \(C = \frac{1}{10} \ln 2\).

Rearrange: \(-\ln(10-x) + \ln(20-x) = 0.1t + \ln 2\).

\(\ln\left(\frac{20-x}{10-x}\right) = 0.1t + \ln 2\).

\(\frac{20-x}{10-x} = 2\exp(0.1t)\).

\(20-x = 2(10-x)\exp(0.1t)\).

\(x = \frac{20(\exp(0.1t)-1)}{2\exp(0.1t)-1}\).

(iii) As t becomes large, \(\exp(0.1t)\) becomes very large, so \(x\) approaches 10.