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