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

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 \(x = 2\):

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

Separate variables: \(\frac{dy}{1-y^2} = \frac{dx}{x}\).

Express \(\frac{1}{1-y^2}\) as partial fractions: \(\frac{1}{1-y^2} = \frac{1}{2(1-y)} + \frac{1}{2(1+y)}\).

Integrate both sides: \(-\frac{1}{2} \ln(1-y) + \frac{1}{2} \ln(1+y) = \ln x + C\).

Combine logarithms: \(\frac{1}{2} \ln \left( \frac{1+y}{1-y} \right) = \ln x + C\).

Use initial condition \(x = 2, y = 0\) to find \(C\):

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

Substitute back: \(\frac{1}{2} \ln \left( \frac{1+y}{1-y} \right) = \ln x - \ln 2\).

Exponentiate to solve for \(y\):

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

Cross-multiply and solve for \(y\):

\(4(1+y) = x^2(1-y)\).

\(4 + 4y = x^2 - x^2y\).

\(4y + x^2y = x^2 - 4\).

\(y(4 + x^2) = x^2 - 4\).

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

(a) Start by separating variables: \(e^{-y} dy = x e^{-x} dx\).

Integrate both sides: \(\int e^{-y} dy = \int x e^{-x} dx\).

The left side integrates to \(-e^{-y}\).

For the right side, use integration by parts: let \(u = x\) and \(dv = e^{-x} dx\), then \(du = dx\) and \(v = -e^{-x}\).

Integration by parts gives: \(\int x e^{-x} dx = -xe^{-x} - \int -e^{-x} dx = -xe^{-x} - e^{-x} + C\).

Thus, \(-e^{-y} = -xe^{-x} - e^{-x} + C\).

Using the initial condition \(y = 0\) when \(x = 0\), we find \(-1 = -1 + C\), so \(C = 0\).

Therefore, \(-e^{-y} = -xe^{-x} - e^{-x}\).

Solve for \(y\): \(e^{-y} = (x+1)e^{-x}\), so \(y = -\ln((x+1)e^{-x})\).

(b) Substitute \(x = 1\) into the expression for \(y\):

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

Thus, \(y = 1 - \ln 2\).

First, separate the variables:

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

Integrate both sides:

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

The left side integrates to:

\(-\frac{1}{y}\).

For the right side, use integration by parts:

Let \(u = x\) and \(dv = 4e^{-2x} dx\).

Then \(du = dx\) and \(v = -2e^{-2x}\).

So, \(\int 4xe^{-2x} dx = -2xe^{-2x} - \int -2e^{-2x} dx\).

\(= -2xe^{-2x} + e^{-2x}\).

Thus, the integrated equation is:

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

Using the initial condition \(y = 1\) when \(x = 0\):

\(-1 = -2(0)e^{0} + e^{0} + C\).

\(-1 = 1 + C\).

\(C = -2\).

Substitute back:

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

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

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

Multiply numerator and denominator by \(e^{2x}\):

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

(i) Separate variables: \(\frac{dy}{dx} = xe^{x+y}\) becomes \(e^{-y} dy = x e^x dx\).

Integrate both sides: \(\int e^{-y} dy = -e^{-y}\) and \(\int x e^x dx\) by parts gives \(x e^x - e^x\).

Combine: \(-e^{-y} = x e^x - e^x + C\).

Use initial condition \(y = 0\) when \(x = 0\): \(-1 = 0 - 1 + C\), so \(C = 0\).

Final solution: \(-e^{-y} = e^x(1-x)\), thus \(y = -\ln(e^x(1-x))\).

(ii) For \(y\) to be real, \(e^x(1-x) > 0\). Since \(e^x > 0\), it follows that \(1-x > 0\), so \(x < 1\).

(i) Separate variables and integrate: \(2y^{\frac{1}{2}} = \ldots\)

Integrate by parts for \(x \sin \left( \frac{1}{3}x \right)\):

\(-3x \cos \left( \frac{1}{3}x \right) + 9 \sin \left( \frac{1}{3}x \right)\)

General solution: \(y = \left( -\frac{3}{10}x \cos \left( \frac{1}{3}x \right) + \frac{9}{10} \sin \left( \frac{1}{3}x \right) + c \right)^2\)

(ii) Use \(x = 0\) and \(y = 100\) to find constant:

Substitute \(x = 25\) to find \(y\):

\(y = 203\)

Separate the variables:

\(y^2 \, dy = 6x e^{3x} \, dx\)

Integrate both sides:

\(\int y^2 \, dy = \int 6x e^{3x} \, dx\)

The left-hand side becomes:

\(\frac{1}{3} y^3\)

For the right-hand side, use integration by parts:

Let \(u = x\) and \(dv = 6e^{3x} \, dx\).

Then \(du = dx\) and \(v = 2e^{3x}\).

\(\int 6x e^{3x} \, dx = 2xe^{3x} - \int 2e^{3x} \, dx\)

\(= 2xe^{3x} - \frac{2}{3}e^{3x} + C\)

Equating both sides:

\(\frac{1}{3} y^3 = 2xe^{3x} - \frac{2}{3}e^{3x} + C\)

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

\(\frac{1}{3} (2)^3 = 2(0)e^{0} - \frac{2}{3}e^{0} + C\)

\(\frac{8}{3} = 0 - \frac{2}{3} + C\)

\(C = \frac{10}{3}\)

Thus, the equation becomes:

\(\frac{1}{3} y^3 = 2xe^{3x} - \frac{2}{3}e^{3x} + \frac{10}{3}\)

Substitute \(x = 0.5\) to find \(y\):

\(\frac{1}{3} y^3 = 2(0.5)e^{1.5} - \frac{2}{3}e^{1.5} + \frac{10}{3}\)

Calculate \(y\) to 2 decimal places:

\(y \approx 2.44\)

First, separate the variables:

\(e^{4x} \frac{dy}{dx} = \cos^2 3y\)

\(\frac{dy}{\cos^2 3y} = e^{-4x} dx\)

Integrate both sides:

\(\int \sec^2 3y \, dy = \int e^{-4x} \, dx\)

\(\frac{1}{3} \tan 3y = -\frac{1}{4} e^{-4x} + C\)

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

\(\frac{1}{3} \tan(0) = -\frac{1}{4} e^{-8} + C\)

\(0 = -\frac{1}{4} e^{-8} + C\)

\(C = \frac{1}{4} e^{-8}\)

Substitute back to find \(y\):

\(\frac{1}{3} \tan 3y = -\frac{1}{4} e^{-4x} + \frac{1}{4} e^{-8}\)

\(\frac{1}{3} \tan 3y = \frac{1}{4} e^{-8} - \frac{1}{4} e^{-4x}\)

\(\tan 3y = \frac{3}{4} e^{-8} - \frac{3}{4} e^{-4x}\)

\(y = \frac{1}{3} \arctan \left( \frac{3}{4} e^{-8} - \frac{3}{4} e^{-4x} \right)\)

(a) To solve the differential equation \(\frac{dy}{dx} = \frac{1 + 4y^2}{e^x}\), we first separate variables:

\(e^x \, dy = (1 + 4y^2) \, dx\).

Integrate both sides:

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

The left side integrates to \(\frac{1}{2} \arctan(2y)\) and the right side to \(-e^{-x} + C\).

Thus, \(\frac{1}{2} \arctan(2y) = -e^{-x} + C\).

Using the initial condition \(y = 0\) when \(x = 1\), we find:

\(\frac{1}{2} \arctan(0) = -e^{-1} + C\).

Since \(\arctan(0) = 0\), we have \(0 = -e^{-1} + C\), so \(C = e^{-1}\).

Substitute back to get:

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

Solving for \(y\), we get:

\(y = \frac{1}{2} \tan \left( 2e^{-1} - 2e^{-x} \right)\).

(b) As \(x \to \infty\), \(e^{-x} \to 0\), so:

\(y \to \frac{1}{2} \tan \left( 2e^{-1} \right)\).

Separate variables:

\(\int \frac{1}{x+2} \, dx = \int \cot \frac{1}{2} \theta \, d\theta\)

Integrate both sides:

\(\ln(x+2) = 2 \ln \sin \frac{1}{2} \theta + C\)

Use the initial condition \(x = 1\) when \(\theta = \frac{1}{3} \pi\):

\(\ln(1+2) = 2 \ln \sin \frac{1}{6} \pi + C\)

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

\(C = \ln 3 - 2 \ln \frac{1}{2} = \ln 12\)

Substitute back:

\(\ln(x+2) = 2 \ln \sin \frac{1}{2} \theta + \ln 12\)

\(x+2 = 12 \sin^2 \frac{1}{2} \theta\)

Use the double angle identity \(\sin^2 \frac{1}{2} \theta = \frac{1 - \cos \theta}{2}\):

\(x+2 = 12 \cdot \frac{1 - \cos \theta}{2}\)

\(x+2 = 6(1 - \cos \theta)\)

\(x = 6 - 6 \cos \theta - 2\)

\(x = 4 - 6 \cos \theta\)

(i) To differentiate \(\frac{1}{\sin^2 \theta}\), rewrite it as \(\csc^2 \theta\). The derivative of \(\csc \theta\) is \(-\csc \theta \cot \theta\). Using the chain rule, the derivative of \(\csc^2 \theta\) is:

\(-2 \csc \theta \cot \theta \cdot \csc \theta = -2 \csc^2 \theta \cot \theta\).

(ii) Start with the differential equation:

\(x \tan \theta \frac{dx}{d\theta} + \csc^2 \theta = 0\).

Rearrange to separate variables:

\(x \tan \theta \frac{dx}{d\theta} = -\csc^2 \theta\).

\(x \frac{dx}{d\theta} = -\frac{\csc^2 \theta}{\tan \theta}\).

Integrate both sides:

\(\int x \, dx = \int -\csc^2 \theta \cot \theta \, d\theta\).

\(\frac{1}{2}x^2 = \frac{1}{\sin^2 \theta} + C\).

Use the initial condition \(x = 4\) when \(\theta = \frac{1}{6}\pi\):

\(\frac{1}{2}(4)^2 = \frac{1}{\sin^2 \frac{1}{6}\pi} + C\).

\(8 = 4 + C\).

\(C = 4\).

Thus, the solution is:

\(\frac{1}{2}x^2 = \frac{1}{\sin^2 \theta} + 4\).

\(x^2 = \frac{2}{\sin^2 \theta} + 8\).

\(x = \sqrt{\csc^2 \theta + 12}\).

First, separate the variables:

\(\frac{dy}{y^3} = ke^{-x} dx\)

Integrate both sides:

\(\int y^{-3} \, dy = \int ke^{-x} \, dx\)

This gives:

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

Using the initial condition \(y = 1\) when \(x = 0\):

\(-\frac{1}{2(1)^2} = -k + c\)

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

Using the second condition \(y = \sqrt{e}\) when \(x = 1\):

\(-\frac{1}{2(\sqrt{e})^2} = -ke^{-1} + c\)

\(-\frac{1}{2e} = -\frac{k}{e} + c\)

Solving these equations gives \(k = \frac{1}{2}\) and \(c = 0\).

Substitute back to find \(y\):

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

\(y^2 = e^x\)

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

(i) To show that \(\frac{d}{d\theta}(\tan^2 \theta) = \frac{2 \tan \theta}{\cos^2 \theta}\), we start by using the chain rule:

\(\frac{d}{d\theta}(\tan^2 \theta) = 2 \tan \theta \cdot \frac{d}{d\theta}(\tan \theta).\)

Since \(\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta = \frac{1}{\cos^2 \theta}\), we have:

\(\frac{d}{d\theta}(\tan^2 \theta) = 2 \tan \theta \cdot \frac{1}{\cos^2 \theta} = \frac{2 \tan \theta}{\cos^2 \theta}.\)

(ii) To solve the differential equation \(x \cos^2 \theta \frac{dx}{d\theta} = 2 \tan \theta + 1\), we separate variables:

\(x \frac{dx}{d\theta} = \frac{2 \tan \theta + 1}{\cos^2 \theta}.\)

Integrating both sides, we have:

\(\int x \, dx = \int \left( \frac{2 \tan \theta}{\cos^2 \theta} + \frac{1}{\cos^2 \theta} \right) d\theta.\)

The left side integrates to \(\frac{1}{2}x^2\).

The right side integrates to \(\tan^2 \theta + \tan \theta\).

Thus, \(\frac{1}{2}x^2 = \tan^2 \theta + \tan \theta + C\).

Using the initial condition \(x = 1\) when \(\theta = \frac{1}{4}\pi\), we find:

\(\frac{1}{2}(1)^2 = \tan^2 \left(\frac{1}{4}\pi\right) + \tan \left(\frac{1}{4}\pi\right) + C.\)

\(\frac{1}{2} = 1 + 1 + C \Rightarrow C = -\frac{3}{2}.\)

Thus, \(\frac{1}{2}x^2 = \tan^2 \theta + \tan \theta - \frac{3}{2}\).

Substituting \(\theta = \frac{1}{3}\pi\), we find:

\(\frac{1}{2}x^2 = \tan^2 \left(\frac{1}{3}\pi\right) + \tan \left(\frac{1}{3}\pi\right) - \frac{3}{2}.\)

\(\frac{1}{2}x^2 = \left(\sqrt{3}\right)^2 + \sqrt{3} - \frac{3}{2} = 3 + \sqrt{3} - \frac{3}{2}.\)

\(\frac{1}{2}x^2 = \frac{3}{2} + \sqrt{3}.\)

\(x^2 = 3 + 2\sqrt{3}.\)

\(x = \sqrt{3 + 2\sqrt{3}} \approx 2.54.\)

Separate the variables: \(\frac{1}{\cos^2 y} dy = 4 \tan x \, dx\).

Integrate both sides: \(\int \sec^2 y \, dy = \int 4 \tan x \, dx\).

The integrals are: \(\tan y = 4 \ln |\sec x| + C\).

Use the initial condition \(x = 0\) and \(y = \frac{1}{4}\pi\):

\(\tan \left( \frac{1}{4}\pi \right) = 4 \ln |\sec 0| + C\).

\(1 = 4 \cdot 0 + C \Rightarrow C = 1\).

The solution is \(\tan y = 4 \ln |\sec x| + 1\).

Substitute \(y = \frac{1}{3}\pi\):

\(\tan \left( \frac{1}{3}\pi \right) = 4 \ln |\sec x| + 1\).

\(\sqrt{3} = 4 \ln |\sec x| + 1\).

\(4 \ln |\sec x| = \sqrt{3} - 1\).

\(\ln |\sec x| = \frac{\sqrt{3} - 1}{4}\).

\(|\sec x| = e^{\frac{\sqrt{3} - 1}{4}}\).

\(\sec x = e^{\frac{\sqrt{3} - 1}{4}}\).

\(x = \arccos \left( \frac{1}{e^{\frac{\sqrt{3} - 1}{4}}} \right)\).

\(x \approx 0.587\).

To solve the differential equation \(\frac{dy}{dx} = e^{-2y} \tan^2 x\), we first separate variables:

\(e^{2y} \, dy = \tan^2 x \, dx\).

Integrate both sides:

\(\int e^{2y} \, dy = \int \tan^2 x \, dx\).

The left side integrates to \(\frac{1}{2} e^{2y}\), and the right side integrates to \(\tan x - x + C\), where \(C\) is the constant of integration.

Thus, \(\frac{1}{2} e^{2y} = \tan x - x + C\).

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

\(\frac{1}{2} e^{0} = \tan 0 - 0 + C\).

This gives \(\frac{1}{2} = C\).

Substitute \(C = \frac{1}{2}\) back into the equation:

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

Now, solve for \(y\) when \(x = \frac{1}{4}\pi\):

\(\frac{1}{2} e^{2y} = \tan \frac{1}{4}\pi - \frac{1}{4}\pi + \frac{1}{2}\).

Calculate \(\tan \frac{1}{4}\pi = 1\), so:

\(\frac{1}{2} e^{2y} = 1 - \frac{1}{4}\pi + \frac{1}{2}\).

\(\frac{1}{2} e^{2y} = \frac{3}{2} - \frac{1}{4}\pi\).

\(e^{2y} = 3 - \frac{1}{2}\pi\).

\(2y = \ln(3 - \frac{1}{2}\pi)\).

\(y = \frac{1}{2} \ln(3 - \frac{1}{2}\pi)\).

Calculate \(y \approx 0.179\).

(i) Separate variables: \(\frac{1}{x} dx = \frac{\sin 2\theta}{3 + \cos 2\theta} d\theta\).

Integrate both sides: \(\int \frac{1}{x} dx = \int \frac{\sin 2\theta}{3 + \cos 2\theta} d\theta\).

This gives \(\ln x = -\frac{1}{2} \ln(3 + \cos 2\theta) + C\).

Use initial condition \(x = 3\) when \(\theta = \frac{1}{4}\pi\) to find \(C\):

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

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

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

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

Rearrange to find \(x\):

\(x = \sqrt{\frac{27}{3 + \cos 2\theta}}\)

(ii) The least value of \(x\) occurs when \(\cos 2\theta = 1\), giving:

\(x = \sqrt{\frac{27}{3 + 1}} = \frac{3\sqrt{3}}{2}\)

To solve the differential equation \(\frac{dx}{dθ} = (x + 2) \sin^2 2θ\), we first separate variables:

\(\frac{1}{x + 2} dx = \sin^2 2θ \, dθ\).

Integrating both sides, we have:

\(\int \frac{1}{x + 2} \, dx = \int \sin^2 2θ \, dθ\).

The left side integrates to \(\ln(x + 2)\).

For the right side, use the identity \(\sin^2 2θ = \frac{1 - \cos 4θ}{2}\):

\(\int \sin^2 2θ \, dθ = \int \frac{1 - \cos 4θ}{2} \, dθ = \frac{1}{2} \int (1 - \cos 4θ) \, dθ\).

This integrates to \(\frac{1}{2} \left( θ - \frac{1}{4} \sin 4θ \right)\).

Thus, we have:

\(\ln(x + 2) = \frac{1}{2} θ - \frac{1}{8} \sin 4θ + C\).

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

\(\ln(2) = C\).

So the solution is:

\(\ln(x + 2) = \frac{1}{2} θ - \frac{1}{8} \sin 4θ + \ln 2\).

To find \(x\) when \(θ = \frac{1}{4}π\):

\(\ln(x + 2) = \frac{1}{2} \left( \frac{1}{4}π \right) - \frac{1}{8} \sin \left( π \right) + \ln 2\).

\(\ln(x + 2) = \frac{π}{8} + \ln 2\).

Solving for \(x\):

\(x + 2 = e^{\frac{π}{8} + \ln 2} = 2e^{\frac{π}{8}}\).

\(x = 2e^{\frac{π}{8}} - 2\).

Calculating this gives \(x \approx 0.962\) to 3 significant figures.

First, separate the variables:

\(2 \cos^2 \theta \frac{dx}{d\theta} = \sqrt{2x + 1}\)

\(\Rightarrow \frac{dx}{\sqrt{2x + 1}} = \frac{1}{2 \cos^2 \theta} d\theta\)

Express \(\frac{1}{\cos^2 \theta}\) as \(\sec^2 \theta\):

\(\Rightarrow \frac{dx}{\sqrt{2x + 1}} = \frac{1}{2} \sec^2 \theta d\theta\)

Integrate both sides:

\(\int \frac{dx}{\sqrt{2x + 1}} = \frac{1}{2} \int \sec^2 \theta d\theta\)

The left side integrates to \(\sqrt{2x + 1}\), and the right side integrates to \(\frac{1}{2} \tan \theta + C\):

\(\sqrt{2x + 1} = \frac{1}{2} \tan \theta + C\)

Use the initial condition \(x = 0\) when \(\theta = \frac{1}{4}\pi\):

\(\sqrt{2(0) + 1} = \frac{1}{2} \tan \left(\frac{1}{4}\pi\right) + C\)

\(1 = \frac{1}{2} \cdot 1 + C\)

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

Substitute \(C\) back into the equation:

\(\sqrt{2x + 1} = \frac{1}{2} \tan \theta + \frac{1}{2}\)

Square both sides:

\(2x + 1 = \left(\frac{1}{2} \tan \theta + \frac{1}{2}\right)^2\)

\(2x + 1 = \frac{1}{4}(\tan \theta + 1)^2\)

\(2x = \frac{1}{4}(\tan \theta + 1)^2 - 1\)

\(x = \frac{1}{8}(\tan \theta + 1)^2 - \frac{1}{2}\)

First, separate the variables:

\(\int \tan \theta \, \mathrm{d}\theta = \int \frac{x}{x^2 + 3} \, \mathrm{d}x.\)

Integrate both sides:

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

Use the initial condition \(x = 1\) when \(\theta = 0\) to find \(C\):

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

\(0 = \frac{1}{2} \ln(4) + C\)

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

Substitute back to get:

\(-\ln(\cos \theta) = \frac{1}{2} \ln(x^2 + 3) - \ln(2)\)

Rearrange to solve for \(x^2\):

\(\ln(\cos \theta) + \ln(2) = \frac{1}{2} \ln(x^2 + 3)\)

\(\ln(2 \cos \theta) = \frac{1}{2} \ln(x^2 + 3)\)

\(2 \ln(2 \cos \theta) = \ln(x^2 + 3)\)

\(x^2 + 3 = 4 \cos^2 \theta\)

\(x^2 = \frac{4}{\cos^2 \theta} - 3\)

First, separate the variables:

\(\frac{dy}{y} = \frac{6e^{3x}}{2 + e^{3x}} \, dx\).

Integrate both sides:

\(\int \frac{1}{y} \, dy = \int \frac{6e^{3x}}{2 + e^{3x}} \, dx\).

The left side integrates to \(\ln y\).

For the right side, use substitution: let \(u = 2 + e^{3x}\), then \(du = 3e^{3x} \, dx\), so \(dx = \frac{du}{3e^{3x}}\).

Thus, \(\int \frac{6e^{3x}}{u} \cdot \frac{du}{3e^{3x}} = 2 \int \frac{1}{u} \, du = 2 \ln |u| + C\).

Substitute back \(u = 2 + e^{3x}\):

\(\ln y = 2 \ln |2 + e^{3x}| + C\).

Exponentiate both sides:

\(y = e^{C} (2 + e^{3x})^2\).

Given \(y = 36\) when \(x = 0\):

\(36 = e^{C} (2 + e^{0})^2 = e^{C} \cdot 3^2 = 9e^{C}\).

Thus, \(e^{C} = 4\).

Therefore, \(y = 4(2 + e^{3x})^2\).

First, separate the variables:

\(\frac{dy}{dx} = e^{2x+y} \Rightarrow e^{-y} dy = e^{2x} dx\).

Integrate both sides:

\(\int e^{-y} \, dy = \int e^{2x} \, dx\).

This gives:

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

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

\(-e^{0} = \frac{1}{2} e^{0} + C \Rightarrow -1 = \frac{1}{2} + C \Rightarrow C = -\frac{3}{2}\).

Substitute back:

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

Rearrange to solve for \(y\):

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

\(-y = \ln\left(\frac{2}{3 - e^{2x}}\right)\).

Thus, \(y = \ln\left(\frac{2}{3 - e^{2x}}\right)\).

Separate the variables:

\(\frac{dx}{x+1} = \frac{\cos 2θ}{\sin 2θ} \, dθ\)

Integrate both sides:

\(\int \frac{dx}{x+1} = \int \frac{\cos 2θ}{\sin 2θ} \, dθ\)

\(\ln |x+1| = \frac{1}{2} \ln |\sin 2θ| + C\)

Use the initial condition \(θ = \frac{1}{12}π, x = 0\):

\(\ln |0+1| = \frac{1}{2} \ln |\sin \frac{1}{6}π| + C\)

\(0 = \frac{1}{2} \ln \frac{1}{2} + C\)

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

Substitute back:

\(\ln |x+1| = \frac{1}{2} \ln |\sin 2θ| - \frac{1}{2} \ln \frac{1}{2}\)

\(\ln |x+1| = \frac{1}{2} \ln \left( \frac{\sin 2θ}{\frac{1}{2}} \right)\)

\(\ln |x+1| = \frac{1}{2} \ln (2 \sin 2θ)\)

\(x+1 = \sqrt{2 \sin 2θ}\)

\(x = \sqrt{2 \sin 2θ} - 1\)

(i) Separate variables: \(\frac{dx}{\cos^2 x} = e^{-2t} dt\).

Integrate both sides: \(\int \sec^2 x \, dx = \int e^{-2t} \, dt\).

This gives \(\tan x = -\frac{1}{2} e^{-2t} + C\).

Use initial condition \(t = 0, x = 0\): \(\tan 0 = 0 = -\frac{1}{2} e^{0} + C\) so \(C = \frac{1}{2}\).

Thus, \(\tan x = \frac{1}{2} - \frac{1}{2} e^{-2t}\).

Rearrange to find \(x\): \(x = \arctan\left(\frac{1}{2} - \frac{1}{2} e^{-2t}\right)\).

(ii) As \(t \to \infty\), \(e^{-2t} \to 0\), so \(x \to \arctan\left(\frac{1}{2}\right)\).

(iii) As \(t\) increases, \(e^{-2t}\) decreases, making \(1 - e^{-2t}\) increase, and thus \(x\) increases.

(a) Separate variables: \(\int \frac{1}{4 + 9y^2} \, dy = \int e^{-(2x+1)} \, dx\).

Integrate both sides: \(\frac{1}{6} \arctan \left( \frac{3y}{2} \right) = -\frac{1}{2} e^{-(2x+1)} + C\).

Use initial condition \(y = 0\) when \(x = 1\) to find \(C\):

\(\frac{1}{6} \arctan(0) = -\frac{1}{2} e^{-3} + C\) gives \(C = \frac{1}{2} e^{-3}\).

Substitute \(C\) back: \(\frac{1}{6} \arctan \left( \frac{3y}{2} \right) = -\frac{1}{2} e^{-(2x+1)} + \frac{1}{2} e^{-3}\).

Solve for \(y\): \(y = \frac{2}{3} \arctan (3e^{-3} - 3e^{-2x-1})\).

(b) As \(x \to \infty\), \(e^{-2x-1} \to 0\), so \(y \to \frac{2}{3} \arctan (3e^{-3})\).

First, separate the variables:

\(\sin^2 3y \, dy = 4 \sec 2x \tan 2x \, dx\).

Integrate both sides:

\(\int \sin^2 3y \, dy = \int 4 \sec 2x \tan 2x \, dx\).

The left side integrates to:

\(\frac{1}{2}y - \frac{1}{12}\sin 6y\).

The right side integrates to:

\(2 \sec 2x\).

Using the initial condition \(y = 0\) when \(x = \frac{1}{6}\pi\), find the constant of integration:

\(\frac{1}{2}(0) - \frac{1}{12}\sin(0) = 2 \sec \left(\frac{1}{3}\pi\right) - C\).

\(C = 2 \sec \left(\frac{1}{3}\pi\right)\).

Substitute back to find \(x\) when \(y = \frac{1}{6}\pi\):

\(\frac{1}{2}\left(\frac{1}{6}\pi\right) - \frac{1}{12}\sin\left(\frac{1}{2}\pi\right) = 2 \sec 2x - 2 \sec \left(\frac{1}{3}\pi\right)\).

Solve for \(x\):

\(x = 0.541\).

To solve the differential equation \(\frac{dy}{dx} = e^{3y} \sin^2 2x\), we first separate the variables:

\(\frac{1}{e^{3y}} dy = \sin^2 2x \, dx\).

Integrate both sides:

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

The left side becomes \(-\frac{1}{3} e^{-3y}\).

For the right side, use the double angle formula: \(\sin^2 2x = \frac{1}{2} (1 - \cos 4x)\).

Integrate to get \(\frac{1}{2} \left( x - \frac{1}{4} \sin 4x \right)\).

Combine the results:

\(-\frac{1}{3} e^{-3y} = \frac{1}{2} \left( x - \frac{1}{4} \sin 4x \right) + C\).

Use the initial condition \(y = 0\) when \(x = 0\) to find \(C\):

\(-\frac{1}{3} = C\).

Substitute \(x = \frac{1}{2}\) to find \(y\):

\(-\frac{1}{3} e^{-3y} = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{4} \sin 2 \right) - \frac{1}{3}\).

Simplify to find \(y = 0.175\) or \(y = -\frac{1}{3} \ln \left( \frac{1}{4} + \frac{3}{8} \sin 2 \right)\).

(a) To find \(\frac{d}{d\theta}(\cot^2 \theta)\), use the chain rule:

\(\frac{d}{d\theta}(\cot^2 \theta) = 2 \cot \theta \cdot \frac{d}{d\theta}(\cot \theta)\).

Given \(\frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta\), substitute:

\(\frac{d}{d\theta}(\cot^2 \theta) = 2 \cot \theta (-\csc^2 \theta) = -\frac{2 \cot \theta}{\sin^2 \theta}\).

(b) Start with the differential equation:

\(x \sin^2 \theta \frac{dx}{d\theta} = \tan^2 \theta - 2 \cot \theta\).

Separate variables:

\(\int x \, dx = \int \frac{\tan^2 \theta - 2 \cot \theta}{\sin^2 \theta} \, d\theta\).

Integrate both sides:

\(\frac{1}{2}x^2 = \int \frac{\tan^2 \theta}{\sin^2 \theta} \, d\theta - \int \frac{2 \cot \theta}{\sin^2 \theta} \, d\theta\).

\(\frac{1}{2}x^2 = \tan \theta + \cot^2 \theta + C\).

Use the initial condition \(x = 2\) when \(\theta = \frac{1}{4}\pi\):

\(\frac{1}{2}(2)^2 = \tan \frac{1}{4}\pi + \cot^2 \frac{1}{4}\pi + C\).

\(2 = 1 + 1 + C \Rightarrow C = 0\).

Thus, \(\frac{1}{2}x^2 = \tan \theta + \cot^2 \theta\).

Substitute \(\theta = \frac{1}{6}\pi\):

\(\frac{1}{2}x^2 = \tan \frac{1}{6}\pi + \cot^2 \frac{1}{6}\pi\).

Calculate \(x\):

\(x = \sqrt{2(\tan \frac{1}{6}\pi + \cot^2 \frac{1}{6}\pi)} \approx 2.67\).

(a) Let \(y = \ln(\ln x)\). Then \(\frac{dy}{dx} = \frac{d}{dx}(\ln(\ln x))\).

Using the chain rule, \(\frac{dy}{dx} = \frac{1}{\ln x} \cdot \frac{d}{dx}(\ln x) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}\).

(b) The differential equation is \(x \ln x + t \frac{dx}{dt} = 0\).

Rearrange to get \(t \frac{dx}{dt} = -x \ln x\).

Separate variables: \(\frac{dx}{x \ln x} = -\frac{dt}{t}\).

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

The left side integrates to \(\ln(\ln x)\) and the right side to \(-\ln t + C\).

Thus, \(\ln(\ln x) = -\ln t + C\).

Using the initial condition \(x = e\) when \(t = 2\), we find \(\ln(\ln e) = -\ln 2 + C\).

Since \(\ln e = 1\), \(C = \ln 2 + 1\).

Substitute back: \(\ln(\ln x) = -\ln t + \ln 2 + 1\).

Exponentiate to solve for \(x\): \(\ln x = e^{-\ln t + \ln 2 + 1}\).

\(x = e^{\frac{2}{t}}\).

(c) As \(t \to \infty\), \(\frac{2}{t} \to 0\), so \(x = e^{\frac{2}{t}} \to e^0 = 1\).

(a) Separate variables:

\(\frac{dy}{y} = \frac{\sin x}{1 - \cos x} dx\)

Integrate both sides:

\(\int \frac{dy}{y} = \int \frac{\sin x}{1 - \cos x} dx\)

\(\ln y = -\ln(1 - \cos x) + C\)

\(\ln y = \ln(1 - \cos x)^{-1} + C\)

\(y = e^C (1 - \cos x)^{-1}\)

Use initial condition \(y = 4\) when \(x = \pi\):

\(4 = e^C (1 - \cos \pi)^{-1}\)

\(4 = e^C (1 - (-1))^{-1}\)

\(4 = e^C \cdot \frac{1}{2}\)

\(e^C = 8\)

Thus, \(y = 8(1 - \cos x)^{-1}\)

\(y = 2(1 - \cos x)\)

(b) The graph of \(y = 2(1 - \cos x)\) is a sinusoidal curve with a maximum at \(x = \pi\) for \(0 < x < 2\pi\).

(a) To solve the differential equation \(e^{3t} \frac{dx}{dt} = \cos^2 2x\), we first separate variables:

\(\int \sec^2 2x \, dx = \int e^{-3t} \, dt\).

Integrating both sides, we get:

\(\frac{1}{2} \tan 2x = -\frac{1}{3} e^{-3t} + C\).

Using the initial condition \(x = 0\) when \(t = 0\), we find \(C = \frac{1}{3}\).

Thus, \(\frac{1}{2} \tan 2x = -\frac{1}{3} e^{-3t} + \frac{1}{3}\).

Solving for \(x\), we obtain:

\(x = \frac{1}{2} \arctan \left( \frac{2}{3} (1 - e^{-3t}) \right)\).

(b) As \(t \to \infty\), \(e^{-3t} \to 0\), so \(x \to \frac{1}{2} \arctan \left( \frac{2}{3} \right)\).