Solve the following differential equations :
`(dy)/(dx)= (x e^(x) log x + e^(x))/(x cos y)`

Options

A. `siny=e^(3x)log x + c`.
B. `siny=e^(x)log x + c`.
C. `cosy=e^(x)log x + c`.
D. `cosy=e^(3x)log x + c`.

Answer

To solve the differential equation

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

we can follow these steps:

### Step 1: Separate the variables
We can rearrange the equation to separate the variables \(y\) and \(x\):

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

### Step 2: Simplify the right-hand side
The right-hand side can be simplified:

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

### Step 3: Integrate both sides
Now we integrate both sides:

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

The left-hand side integrates to:

\[ \sin y + C_1, \]

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

### Step 4: Integrate the right-hand side
For the right-hand side, we need to integrate \(e^{x} \log x\) and \(\frac{e^{x}}{x}\):

1. For \(\int e^{x} \log x \, dx\), 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:
\[ \int u \, dv = uv - \int v \, du, \]
we get:
\[ e^{x} \log x - \int e^{x} \cdot \frac{1}{x} dx. \]

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

Thus, the right-hand side becomes:
\[ e^{x} \log x - \text{Ei}(x) + C_2, \]
where \(C_2\) is another constant of integration.

### Step 5: Combine both sides
Now we can combine both results:

\[ \sin y = e^{x} \log x - \text{Ei}(x) + C, \]
where \(C = C_2 - C_1\) is a constant.

### Final Solution
The final solution to the differential equation is:

\[ \sin y = e^{x} \log x - \text{Ei}(x) + C. \]