To solve the given differential equation:

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

Step 1: Rearranging the equation

First, isolate $y'$ (the derivative of $y$ with respect to $x$).

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

Step 2: Separate the variables

The next goal is to separate the variables $x$ and $y$. We'll write the equation as:

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

Now, separate $y$-terms on one side and $x$-terms on the other side:

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

Step 3: Integrate both sides

Integrating both sides with respect to their respective variables:

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

Left-hand side:

Let $u = 2 + \sin(3y)$, then $du = 3 \cos(3y) \, dy$. So, the left-hand side becomes:

$\frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln|u| + C_1 = \frac{1}{3} \ln|2 + \sin(3y)| + C_1$

Right-hand side:

The integral of $\frac{1}{x^2 + 4}$ is $\frac{1}{2} \arctan\left(\frac{x}{2}\right)$, so:

$\int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C_2$

Step 4: Combine the results

Now, combining the results from both sides:

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

Where $C = C_2 - C_1$ is the constant of integration.

Final form:

To simplify the expression, you can multiply both sides by 3:

$\ln|2 + \sin(3y)| = \frac{3}{2} \arctan\left(\frac{x}{2}\right) + 3C$

Exponentiating both sides:

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

Or more simply:

$2 + \sin(3y) = A e^{\frac{3}{2} \arctan\left(\frac{x}{2}\right)}$

where $A = e^{3C}$.

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Would you like further clarification or details on this solution?

Related Questions:

1. How can you solve differential equations using substitution methods?
2. What are the common techniques for separating variables in differential equations?
3. How do you integrate trigonometric expressions involving sine and cosine functions?
4. Can you solve first-order differential equations using integrating factors?
5. How does the arctangent function behave, and where is it commonly used in integration?

Tip:

When dealing with trigonometric differential equations, look for substitution possibilities to simplify integrals!