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.

This page provides a detailed solution to the separable differential equation (x^2 + 4)y'cos(3y) = 2 + sin(3y). It covers key concepts such as separation of variables, integration by substitution, and the use of inverse trigonometric functions. Suitable for undergraduate students studying calculus or differential equations.

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