The solution of the differential equation
` (dy)/(dx) = (x(2 log x+1))/(sin y +y cos y)` is

Options

A. `y sin y = x^(2) log x +C`
B. `y = x^(2) +logx+C`
C. `y sin y = x^(2) +C`
D. None of these

Explanation

To solve the differential equation

\[ \frac{dy}{dx} = \frac{x(2 \log x + 1)}{\sin y + y \cos y}, \]

we will use the method of separation of variables. Here are the steps to find the solution:

### Step 1: Separate the variables

We can rearrange the equation to separate the variables \(y\) and \(x\):

\[ (\sin y + y \cos y) \, dy = x(2 \log x + 1) \, dx. \]

### Step 2: Integrate both sides

Now we will integrate both sides:

\[ \int (\sin y + y \cos y) \, dy = \int x(2 \log x + 1) \, dx. \]

### Step 3: Solve the left-hand side integral

To solve the left-hand side, we can break it into two parts:

1. \(\int \sin y \, dy = -\cos y + C_1\)
2. For \(\int y \cos y \, dy\), we will use integration by parts. Let \(u = y\) and \(dv = \cos y \, dy\).

Using integration by parts:

- \(du = dy\)
- \(v = \sin y\)

So,

\[ \int y \cos y \, dy = y \sin y - \int \sin y \, dy = y \sin y + \cos y + C_2. \]

Combining these results, we have:

\[ \int (\sin y + y \cos y) \, dy = -\cos y + (y \sin y + \cos y) = y \sin y + C. \]

### Step 4: Solve the right-hand side integral

Now, we will solve the right-hand side:

\[ \int x(2 \log x + 1) \, dx. \]

We can split this into two integrals:

1. \(\int 2x \log x \, dx\)
2. \(\int x \, dx\)

For \(\int 2x \log x \, dx\), we again use integration by parts:

- Let \(u = \log x\) and \(dv = 2x \, dx\).
- Then, \(du = \frac{1}{x} \, dx\) and \(v = x^2\).

This gives us:

\[ \int 2x \log x \, dx = x^2 \log x - \int x^2 \cdot \frac{1}{x} \, dx = x^2 \log x - \int x \, dx = x^2 \log x - \frac{x^2}{2} + C_3. \]

Now, combining both parts:

\[ \int x(2 \log x + 1) \, dx = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C = x^2 \log x + C. \]

### Step 5: Equate both sides

Now we have:

\[ y \sin y = x^2 \log x + C. \]

### Final Solution

Thus, the solution of the given differential equation is:

\[ y \sin y = x^2 \log x + C. \]