To solve the given differential equation:

$(x^2 + 4) y' \cos(3y) = 2 + \sin(3y)$

Step 1: Isolate $y'$

First, we want to express $y'$ (which is the derivative of $y$ with respect to $x$) explicitly. To do that, divide both sides of the equation by $(x^2 + 4) \cos(3y)$:

$y' = \frac{2 + \sin(3y)}{(x^2 + 4) \cos(3y)}$

Step 2: Separate variables

This equation is separable, meaning we can separate the variables $y$ and $x$. Rewrite it as:

$\frac{\cos(3y)}{2 + \sin(3y)} \, dy = \frac{1}{x^2 + 4} \, dx$

Now, we can integrate both sides.

Step 3: Integrate both sides

For the left-hand side (in terms of $y$):

Let $u = 2 + \sin(3y)$. Then, $du = 3 \cos(3y) \, dy$, or $\frac{du}{3} = \cos(3y) \, dy$.

Thus, the left-hand side becomes:

$\int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \ln |u| = \frac{1}{3} \ln |2 + \sin(3y)|$

For the right-hand side (in terms of $x$):

The integral of $\frac{1}{x^2 + 4}$ is a standard result:

$\int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right)$

Step 4: Combine the results

After integrating both sides, we have:

$\frac{1}{3} \ln |2 + \sin(3y)| = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C$

Step 5: Solve for $y$ (optional)

You can solve for $y$ if needed by exponentiating both sides and simplifying:

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

At this point, the equation involves $y$, but solving explicitly for $y$ might require further manipulation depending on the form desired.

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Would you like further clarification or a deeper explanation of any part of the process?

Here are five related questions:

1. What are separable differential equations, and how do we solve them?
2. Can all first-order differential equations be separated into variables?
3. How do you compute integrals involving trigonometric functions?
4. What are the common methods for solving first-order differential equations?
5. What are inverse trigonometric functions, and where do they commonly appear in integration?

Tip: Always check whether an equation is separable when solving first-order differential equations—it can simplify the process significantly.