The solution of `(y(1+x^(-1))+siny)dx +(x+log x +x cos y)dy=0` is

Options

A. `(1 + y^(-1) sin y) + x^(-1) log x = c`
B. `(y+sin y)+xy log x =C`
C. `xy + y log x + x sin y = C`
D. None of these

Explanation

To solve the differential equation

\[ y(1 + x^{-1})dx + (x + \log x + x \cos y)dy = 0, \]

we will follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the terms in the equation:

\[ y(1 + \frac{1}{x})dx + (x + \log x + x \cos y)dy = 0. \]

### Step 2: Expressing in a Differential Form
We can express this equation in a more manageable form:

\[ y(1 + \frac{1}{x})dx + (x + \log x + x \cos y)dy = 0. \]

This can be rewritten as:

\[ y(1 + \frac{1}{x})dx + (x + \log x)dy + x \cos y dy = 0. \]

### Step 3: Identifying the Total Differential
We can identify the total differential of a function \(F(x, y)\):

\[ F(x, y) = y(x + \log x) + x \sin y. \]

### Step 4: Finding the Partial Derivatives
To confirm that \(F\) is indeed a potential function, we will compute the partial derivatives:

1. \( \frac{\partial F}{\partial x} = y(1 + \frac{1}{x}) + \sin y \)
2. \( \frac{\partial F}{\partial y} = x \cos y + \log x \)

### Step 5: Setting Up the Equation
Now, we can set up the equation:

\[ dF = 0 \Rightarrow F(x, y) = C. \]

### Step 6: Integrating
Integrating gives us:

\[ y(x + \log x) + x \sin y = C. \]

### Step 7: Final Form
Rearranging the equation, we have:

\[ xy + y \log x + x \sin y = C. \]

Thus, the solution to the differential equation is:

\[ xy + y \log x + x \sin y = C. \]

### Conclusion
The correct option from the given choices is option (C):

\[ xy + y \log x + x \sin y = C. \]