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:

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

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

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