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