(i) Given that the rate of formation of X is proportional to the mass of Y, we have \(\frac{dx}{dt} = k(100 - x)\). Using the initial condition \(\frac{dx}{dt} = 1.9\) when \(x = 5\), we find \(1.9 = k(100 - 5)\), giving \(k = 0.02\). Thus, \(\frac{dx}{dt} = 0.02(100 - x)\).

(ii) To solve \(\frac{dx}{dt} = 0.02(100 - x)\), separate variables: \(\frac{1}{100 - x} dx = 0.02 dt\). Integrate both sides: \(-\ln|100 - x| = 0.02t + C\). Solve for x: \(x = 100 - Ce^{-0.02t}\). Use the initial condition \(x = 5\) when \(t = 0\) to find \(C = 95\). Thus, \(x = 100 - 95e^{-0.02t}\).

(iii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(x \to 100\).

(i) The rate of growth of the infected area is proportional to the product of the infected area \(a\) and the uninfected area \(10 - a\). Therefore, \(\frac{da}{dt} = k a (10 - a)\). Given that initially \(a = 5\) and \(\frac{da}{dt} = 0.1\), we have:

\(0.1 = k \times 5 \times (10 - 5)\)

\(0.1 = 25k\)

\(k = 0.004\)

Thus, \(\frac{da}{dt} = 0.004a(10 - a)\).

(ii) Express \(\frac{1}{a(10-a)}\) in partial fractions:

\(\frac{1}{a(10-a)} = \frac{A}{a} + \frac{B}{10-a}\)

Solving, we find \(A = \frac{1}{10}\) and \(B = \frac{1}{10}\).

Separate variables and integrate:

\(\int \frac{da}{a(10-a)} = \int 0.004 \, dt\)

\(\frac{1}{10} \ln a - \frac{1}{10} \ln (10-a) = 0.004t + C\)

\(\ln \left( \frac{a}{10-a} \right) = 0.04t + C\)

Using initial conditions \(t = 0, a = 5\), solve for \(C\):

\(\ln 1 = C \Rightarrow C = 0\)

Thus, \(\ln \left( \frac{a}{10-a} \right) = 0.04t\)

\(t = 25 \ln \left( \frac{a}{10-a} \right)\)

(iii) For 90% infection, \(a = 9\):

\(t = 25 \ln \left( \frac{9}{1} \right)\)

\(t = 25 \ln 9 \approx 54.9\)

(a) The rate of change of volume is given by \(\frac{dV}{dt} = -k\sqrt{h}\). The volume of the hemisphere is \(V = \frac{1}{3} \pi (3rh^2 - h^3)\). Differentiating with respect to h, \(\frac{dV}{dh} = 2\pi rh - \pi h^2\).

Using the chain rule, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\), we have \(-k\sqrt{h} = (2\pi rh - \pi h^2) \cdot \frac{dh}{dt}\).

Rearranging gives \(\frac{dh}{dt} = -\frac{k\sqrt{h}}{2\pi rh - \pi h^2}\). Let \(B = k\pi\), then \(\frac{dh}{dt} = -\frac{B}{2rh^2 - h^3}\).

(b) Separate variables: \((2rh^2 - h^3) dh = -B dt\). Integrate both sides: \(\int (2rh^2 - h^3) dh = -\int B dt\).

The left side integrates to \(\frac{2}{3}rh^3 - \frac{1}{4}h^4\) and the right side to \(-Bt + C\).

At \(t = 0, h = r\), so \(\frac{2}{3}r^4 - \frac{1}{4}r^4 = C\).

At \(t = 14, h = 0\), so \(0 = -14B + C\).

Solving these gives \(C = \frac{5}{12}r^4\) and \(B = \frac{1}{15}r^2\).

Substitute back to find \(t\): \(t = 14 - 20\left(\frac{h}{r}\right)^{\frac{3}{2}} + 6\left(\frac{h}{r}\right)^{\frac{5}{2}}\).

(i) The rate of increase of m is proportional to \((50 - m)^2\), so \(\frac{dm}{dt} = k(50 - m)^2\). Given \(\frac{dm}{dt} = 5\) when \(m = 0\), we have \(5 = k(50)^2\). Solving for \(k\), we find \(k = \frac{5}{2500} = 0.002\). Thus, \(\frac{dm}{dt} = 0.002(50 - m)^2\).

(ii) Separate variables: \(\frac{1}{(50 - m)^2} dm = 0.002 dt\). Integrate both sides: \(-\frac{1}{50 - m} = 0.002t + C\). Using \(t = 0\), \(m = 0\), we find \(C = -\frac{1}{50}\). Thus, \(-\frac{1}{50 - m} = 0.002t - \frac{1}{50}\). Solving for m, we get \(m = 50 - \frac{500}{t + 10}\).

(iii) Substitute \(t = 10\) into the solution: \(m = 50 - \frac{500}{10 + 10} = 25\). For \(m = 45\), solve \(45 = 50 - \frac{500}{t + 10}\), giving \(t = 90\).

(iv) As \(t\) becomes very large, \(\frac{500}{t + 10}\) approaches 0, so \(m\) approaches 50.

(a) The differential equation is \(\frac{dy}{dx} = k \frac{y}{x\sqrt{x}}\). Separate variables: \(\frac{dy}{y} = k \frac{dx}{x\sqrt{x}}\). Integrate both sides: \(\ln y = -2k \frac{1}{\sqrt{x}} + C\). Use the point \((1, 1)\) to find \(C\): \(\ln 1 = -2k \cdot 1 + C\), so \(C = 2k\). Use the point \((4, e)\) to find \(k\): \(\ln e = -2k \cdot \frac{1}{2} + 2k\), solving gives \(k = 1\). Thus, \(y = \exp\left(-\frac{2}{\sqrt{x}} + 2\right)\).

(b) As \(x\) tends to infinity, \(-\frac{2}{\sqrt{x}}\) tends to 0, so \(y\) approaches \(e^2\).