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 solve the equation:
### 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 Side Integral
For the left side, we can break it down:
1. \(\int \sin y \, dy = -\cos y\)
2. For \(\int y \cos y \, dy\), we will use integration by parts:
- Let \(u = y\) and \(dv = \cos y \, dy\).
- Then \(du = dy\) and \(v = \sin y\).
- Using integration by parts: \(\int u \, dv = uv - \int v \, du\):
\[
\int y \cos y \, dy = y \sin y - \int \sin y \, dy = y \sin y + \cos y
\]
Combining these results, we have:
\[
\int (\sin y + y \cos y) dy = -\cos y + (y \sin y + \cos y) = y \sin y
\]
### Step 4: Solve the Right Side Integral
Now, we will solve the right side integral:
\[
\int x(2 \log x + 1) dx
\]
We can split this into two parts:
1. \(\int x \cdot 2 \log x \, dx\) and
2. \(\int x \, dx\).
For \(\int x \cdot 2 \log x \, dx\), we will again use integration by parts:
- Let \(u = \log x\) and \(dv = 2x \, dx\).
- Then \(du = \frac{1}{x} \, dx\) and \(v = x^2\).
Using integration by parts:
\[
\int 2x \log x \, dx = 2 \left( x^2 \log x - \int x^2 \cdot \frac{1}{x} \, dx \right) = 2 \left( x^2 \log x - \frac{x^3}{3} \right)
\]
Now, for the second part:
\[
\int x \, dx = \frac{x^2}{2}
\]
Combining these results, we have:
\[
\int x(2 \log x + 1) \, dx = 2 \left( x^2 \log x - \frac{x^3}{3} \right) + \frac{x^2}{2}
\]
### Step 5: Combine Results
Now we combine both sides:
\[
y \sin y = 2 \left( x^2 \log x - \frac{x^3}{3} \right) + \frac{x^2}{2} + C
\]
### Final Result
Thus, the solution of the differential equation is:
\[
y \sin y = 2x^2 \log x - \frac{2x^3}{3} + \frac{x^2}{2} + C
\]
