Solve : `(dy)/(dx)=(x e^(x)log x +e^(x))/(x cos y)`.

Answer

To solve the differential equation

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

we can start by separating the variables.

### Step 1: Separate the variables

Rearranging the equation, we can write it as:

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

This can be simplified to:

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

### Step 2: Integrate both sides

Now, we will integrate both sides:

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

The left side integrates to:

\[ \sin y + C_1, \]

where \(C_1\) is a constant of integration.

### Step 3: Solve the right side integral

For the right side, we can split the integral:

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

The integral \(\int \frac{e^x}{x} \, dx\) does not have a simple closed form, but we can denote it as \(Ei(x)\) (the Exponential Integral function).

For the first integral, we can use integration by parts:

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

Using integration by parts:

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

Thus, we can write:

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

### Step 4: Combine the results

Now we can combine the results of both integrals:

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

where \(C\) is a constant that combines \(C_1\) and the constant from the integral.

### Final Solution

Thus, the solution to the differential equation is:

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