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We are given the differential equation:

\[
2y \, dy - (y^2 + 1) \cos x \, dx = 0
\]

Our goal is to solve for \(y(x)\).

### Step 1: Separate Variables

Let's rearrange the equation to separate variables:

\[
2y \, dy = (y^2 + 1) \cos x \, dx
\]

Now, divide both sides by \((y^2 + 1)\):

\[
\frac{2y \, dy}{y^2 + 1} = \cos x \, dx
\]

### Step 2: Integrate Both Sides

We'll now integrate both sides.

- The left-hand side is:

\[
\int \frac{2y \, dy}{y^2 + 1}
\]

This integral can be solved using a substitution. Let \( u = y^2 + 1 \), so \( du = 2y \, dy \), and the integral becomes:

\[
\int \frac{du}{u} = \ln |u| = \ln (y^2 + 1)
\]

- The right-hand side is:

\[
\int \cos x \, dx = \sin x
\]

Thus, after integrating both sides, we have:

\[
\ln (y^2 + 1) = \sin x + C
\]

### Step 3: Solve for \(y\)

Exponentiate both sides to eliminate the logarithm:

\[
y^2 + 1 = e^{\sin x + C} = e^{\sin x} \cdot e^C
\]

Since \( e^C \) is a constant, we can replace it with \( A \), where \( A = e^C \):

\[
y^2 + 1 = A e^{\sin x}
\]

Now, solve for \(y^2\):

\[
y^2 = A e^{\sin x} - 1
\]

Finally, take the square root of both sides:

\[
y = \pm \sqrt{A e^{\sin x} - 1}
\]

### Step 4: Compare to Given Options

The solution that matches the form from the given options is:

\[
y = \sqrt{A e^{\sin x} - 1}
\]

Thus, the correct answer is the one displayed in the **yellow box**.

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Would you like more details on any part of the solution? 

Here are 5 related questions you might find helpful:
1. How can we identify separable differential equations?
2. What is the significance of the constant \( C \) in differential equations?
3. How do we handle initial conditions in solving differential equations?
4. What substitution methods are useful for solving integrals like this?
5. How does exponentiation help in solving logarithmic equations?

**Tip**: Always check if a differential equation can be separated into functions of \(x\) and \(y\), as this makes integration simpler.

Noah is studying the motion of a pendulum and comes across a differential equation that models the motion: 2y dy - (y^2 + 1) cos x dx = 0. Find the general solution of this differential equation.

Learn how to solve the differential equation 2y dy - (y^2 + 1) cos x dx = 0. This problem covers the use of separable equations, integration, and solving for the general solution. Suitable for students in advanced high school or undergraduate calculus.

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