(i) Separate variables: \(\frac{1}{x} dx = -kt dt\).

Integrate both sides: \(\int \frac{1}{x} dx = \int -kt dt\).

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

Use initial condition \(t = 0, x = 100\) to find \(C\):

\(\ln 100 = C\).

Thus, \(\ln x = -\frac{1}{2}kt^2 + \ln 100\).

(ii) Use \(t = 20, x = 90\) to find \(k\):

\(\ln 90 = -\frac{1}{2}k(20)^2 + \ln 100\).

Solve for \(k\):

\(k = \frac{2(\ln 100 - \ln 90)}{400}\).

Substitute \(x = 50\) to find \(t\):

\(\ln 50 = -\frac{1}{2}k t^2 + \ln 100\).

\(t^2 = \frac{400(\ln 100 - \ln 50)}{\ln 100 - \ln 90}\).

Calculate \(t\):

\(t = 51.3\).

(i) Separate variables and integrate:

\\(\int \frac{1}{\sqrt{P-A}} \, dP = \int -k \, dt \\)

Integrating gives:

\\(2\sqrt{P-A} = -kt + c \\)

(ii) Use initial conditions to find constants:

When \\(t = 0, P = 5A \\):

\\(2\sqrt{5A-A} = c \\)

\\(c = 4\sqrt{A} \\)

When \\(t = 2, P = 2A \\):

\\(2\sqrt{2A-A} = -2k + 4\sqrt{A} \\)

Solve for \\(k \\):

\\(2\sqrt{A} = -2k + 4\sqrt{A} \\)

\\(k = \sqrt{A} \\)

(iii) Substitute \\(P = A \\) and solve for \\(t \\):

\\(2\sqrt{A-A} = -kt + 4\sqrt{A} \\)

\\(0 = -kt + 4\sqrt{A} \\)

\\(t = 4 \\)

(iv) Rearrange to express \\(P \\) in terms of \\(A \\) and \\(t \\):

\\(2\sqrt{P-A} = -kt + 4\sqrt{A} \\)

Square both sides:

\\(4(P-A) = (4\sqrt{A} - kt)^2 \\)

\\(P = \frac{1}{4}A(4 + (4-t)^2) \\)

(a) Separate variables: \(\int \frac{1}{x} \, dx = \int ke^{-0.1t} \, dt\).

Integrate both sides: \(\ln x = -10ke^{-0.1t} + C\).

Use initial condition \(x = 20\) when \(t = 0\): \(\ln 20 = C\).

Thus, \(\ln x = 10k(1 - e^{-0.1t}) + \ln 20\).

(b) Given \(x = 40\) when \(t = 10\), substitute into the equation:

\(\ln 40 = 10k(1 - e^{-1}) + \ln 20\).

Solve for \(10k\): \(10k = \frac{\ln 2}{1 - e^{-1}} = 1.09654\).

As \(t \to \infty\), \(e^{-0.1t} \to 0\), so \(x \to 20e^{10k}\).

Substitute \(10k = 1.09654\): \(x \to 59.9\).

(a) Separate variables: \(\frac{dN}{N^{\frac{3}{2}}} = k \cos 0.02t \, dt\).

Integrate both sides: \(-\frac{2}{\sqrt{N}} = 50k \sin 0.02t - 0.2\).

Use initial condition \(t = 0, N = 100\) to find constant: \(-\frac{2}{10} = -0.2\).

(b) Substitute \(N = 625\) and \(t = 50\) into the expression: \(-\frac{2}{25} = 50k \sin 1 - 0.2\).

Solve for \(k\): \(k = 0.00285\).

(c) Rearrange to find \(N\) in terms of \(t\): \(N = 4(0.2 - 0.142(607) \sin 0.02t)^{-2}\).

Find the greatest value of \(N\) by evaluating the expression at critical points.

(i) Separate variables: \(\frac{dx}{(20-x)(40-x)} = \frac{1}{5} dt\).

Use partial fractions: \(\frac{1}{(20-x)(40-x)} = \frac{A}{20-x} + \frac{B}{40-x}\).

Find \(A\) and \(B\): \(A = \frac{1}{20}\), \(B = -\frac{1}{20}\).

Integrate both sides: \(\int \left( \frac{1}{20} \ln(20-x) - \frac{1}{20} \ln(40-x) \right) dx = \int \frac{1}{5} dt\).

Integrate: \(-\frac{1}{20} \ln(20-x) + \frac{1}{20} \ln(40-x) = \frac{1}{5} t + C\).

Use initial condition \(x = 10\) when \(t = 0\) to find \(C\).

Solve for \(x\): \(x = \frac{60e^{4t} - 40}{3e^{4t} - 1}\).

(ii) As \(t\) becomes large, \(x\) approaches 20.