Solve the following differential equations :
`x sin y dy + (x e^(x) log x + e^(x))dx=0`.

Answer

To solve the differential equation \( x \sin y \, dy + (x e^{x} \log x + e^{x}) \, dx = 0 \), we can follow these steps:

### Step 1: Rearrange the Equation
We can rewrite the equation in the standard form:
\[ x \sin y \, dy = - (x e^{x} \log x + e^{x}) \, dx \]

### Step 2: Separate Variables
Next, we can separate the variables \( y \) and \( x \) by dividing both sides by \( x \sin y \):
\[ dy = -\frac{(x e^{x} \log x + e^{x})}{x \sin y} \, dx \]
This simplifies to:
\[ dy = -\left( e^{x} \log x + \frac{e^{x}}{x} \right) \frac{1}{\sin y} \, dx \]

### Step 3: Integrate Both Sides
Now we can integrate both sides:
\[ \int \sin y \, dy = -\int \left( e^{x} \log x + \frac{e^{x}}{x} \right) \, dx \]

The left side integrates to:
\[ -\cos y \]

For the right side, we can split the integral:
\[ -\int e^{x} \log x \, dx - \int \frac{e^{x}}{x} \, dx \]

### Step 4: Solve the Right Side Integrals
1. The integral \( \int e^{x} \log x \, dx \) can be solved using integration by parts. Let:
- \( u = \log x \) and \( dv = e^{x} \, dx \)
- Then \( du = \frac{1}{x} \, dx \) and \( v = e^{x} \)

Applying integration by parts:
\[ \int e^{x} \log x \, dx = e^{x} \log x - \int e^{x} \frac{1}{x} \, dx \]

2. The integral \( \int \frac{e^{x}}{x} \, dx \) is known as the Exponential Integral, denoted as \( \text{Ei}(x) \).

Thus, we have:
\[ -\left( e^{x} \log x - \text{Ei}(x) \right) - \text{Ei}(x) \]

### Step 5: Combine Results
Combining the results from both sides gives us:
\[ -\cos y = -e^{x} \log x + 2\text{Ei}(x) + C \]

### Final Form
Rearranging gives:
\[ \cos y = e^{x} \log x - 2\text{Ei}(x) - C \]