(a) Start by separating variables:

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

Express \(\frac{1}{y^2 + y}\) in partial fractions:

\(\frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}\)

Integrate both sides:

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

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

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

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

\(-\ln 2 = 1 + C\)

\(C = -1 - \ln 2\)

Substitute back to find \(y\):

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

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

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

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

(b) As \(x\) tends to infinity, \(\frac{1}{x}\) tends to 0, so:

\(y = \frac{e^{0 - 1}}{2 - e^{0 - 1}} = \frac{e^{-1}}{2 - e^{-1}} = \frac{1}{2e - 1}\)

(i) To express \(\frac{100}{x^2(10-x)}\) in partial fractions, assume the form:

\(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{10-x}\)

Multiply through by the denominator \(x^2(10-x)\) to get:

\(100 = A x (10-x) + B (10-x) + C x^2\)

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

\(A = 1, B = 10, C = 1\)

Thus, \(\frac{100}{x^2(10-x)} = \frac{1}{x} + \frac{10}{x^2} + \frac{1}{10-x}\).

(ii) Separate variables and integrate:

\(\int \frac{1}{x^2(10-x)} \, dx = \int \frac{1}{100} \, dt\)

Integrate the left side using partial fractions:

\(\int \left( \frac{1}{x} + \frac{10}{x^2} + \frac{1}{10-x} \right) \, dx\)

\(= \ln |x| - \frac{10}{x} - \ln |10-x|\)

Integrate the right side:

\(\frac{t}{100} + C\)

Apply initial conditions \(x = 1, t = 0\):

\(\ln 1 - \frac{10}{1} - \ln 9 = 0 + C\)

\(C = -10 - \ln 9\)

Substitute back to find \(t\):

\(\ln |x| - \frac{10}{x} - \ln |10-x| = \frac{t}{100} - 10 - \ln 9\)

Simplify to obtain:

\(t = \ln \left( \frac{9x}{10-x} \right) - \frac{10}{x} + 10\)

(i) Express \(\frac{1}{y(4-y)}\) in partial fractions: \(\frac{A}{y} + \frac{B}{4-y}\).

Solving for \(A\) and \(B\), we get \(A = \frac{1}{4}\) and \(B = -\frac{1}{4}\).

Thus, \(\frac{1}{y(4-y)} = \frac{1}{4} \left( \frac{1}{y} - \frac{1}{4-y} \right)\).

Integrate: \(\int \frac{1}{4} \left( \frac{1}{y} - \frac{1}{4-y} \right) \, dy = \frac{1}{4} \ln y - \frac{1}{4} \ln (4-y) + C\).

(ii) Separate variables: \(\frac{dy}{y(4-y)} = dx\).

Integrate both sides: \(\int \left( \frac{1}{y} + \frac{1}{4-y} \right) \, dy = \int dx\).

Using initial condition \(y = 1\) when \(x = 0\), solve for constant.

Final expression: \(y = \frac{4}{3e^{-4x} + 1}\).

(iii) As \(x \to \infty\), \(e^{-4x} \to 0\), so \(y \to 4\).

First, separate the variables:

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

Integrate both sides:

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

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

Using partial fraction decomposition, find \(A\) and \(B\):

\(A = -\frac{1}{2}, \quad B = \frac{3}{2}\)

Integrate:

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

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

\(\ln 1 = -\frac{1}{2} \ln 2 + \frac{1}{2} \ln 4 + C\)

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

Substitute \(x = 3\) into the equation:

\(\ln |y| = -\frac{1}{2} \ln 4 + \frac{1}{2} \ln 10 + \frac{1}{2} \ln 2\)

\(\ln |y| = \frac{1}{2} \ln 5\)

\(|y| = e^{\frac{1}{2} \ln 5} = \sqrt{5}\)

Thus, \(y = \frac{1}{2} \sqrt{5}\).

1. Start with the differential equation \(\frac{dx}{dt} = x^2(1 + 2x)\).

2. Separate variables: \(\frac{1}{x^2(1 + 2x)} dx = dt\).

3. Use partial fractions to express \(\frac{1}{x^2(1 + 2x)}\) as \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{1+2x}\).

4. Solve for constants: \(A = -2\), \(B = 1\), \(C = 4\).

5. Integrate both sides: \(\int \left( \frac{-2}{x} + \frac{1}{x^2} + \frac{4}{1+2x} \right) dx = \int dt\).

6. Integrate: \(-2 \ln x - \frac{1}{x} + 2 \ln(1+2x) = t + C\).

7. Use initial condition \(x = 1\) when \(t = 0\) to find \(C\): \(C = 1\).

8. Solve for \(t\): \(t = -\frac{1}{x} + 2 \ln \left( \frac{1+2x}{3x} \right) + 1\).