Solve the differential equation : `x "cos" y dy=e^(x)(x "log"x+1)dx`.

Answer

To solve the differential equation \( x \cos y \, dy = e^x (x \log x + 1) \, dx \), we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to separate the variables \( y \) and \( x \):

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

This simplifies to:

\[ \cos y \, dy = e^x \left( \log x + \frac{1}{x} \right) \, dx \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side integrates to:

\[ \int \cos y \, dy = \sin y + C_1 \]

For the right side, we need to integrate:

\[ \int e^x \left( \log x + \frac{1}{x} \right) \, dx \]

This can be split into two separate integrals:

\[ \int e^x \log x \, dx + \int e^x \frac{1}{x} \, dx \]

### Step 3: Using Integration by Parts
For the first integral \( \int e^x \log x \, dx \), we will use integration by parts. Let:

- \( u = \log x \) → \( du = \frac{1}{x} \, dx \)
- \( dv = e^x \, dx \) → \( v = e^x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ \int e^x \log x \, dx = e^x \log x - \int e^x \frac{1}{x} \, dx \]

### Step 4: Combining the Integrals
Now substituting back into our equation, we have:

\[ \sin y + C_1 = e^x \log x - \int e^x \frac{1}{x} \, dx + \int e^x \frac{1}{x} \, dx \]

The integrals of \( e^x \frac{1}{x} \) cancel out, leading to:

\[ \sin y + C_1 = e^x \log x + C_2 \]

### Step 5: Final Rearrangement
We can combine the constants \( C_1 \) and \( C_2 \) into a single constant \( C \):

\[ \sin y = e^x \log x + C \]

### Final Solution
Thus, the solution to the differential equation is:

\[ e^x \log x = \sin y + C \]