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

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

we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the given equation in a more manageable form:

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

This can be expressed as:

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

### Step 2: Group terms
Next, we can group the terms involving \(dx\) and \(dy\):

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

This can be rearranged as:

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

### Step 3: Factor out common terms
We can factor out common terms from the equation:

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

### Step 4: Identify exactness
To check if the equation is exact, we denote:

\[ M = y(1 + \frac{1}{x}) \quad \text{and} \quad N = x + \log x + x \cos y. \]

We compute the partial derivatives:

\[ \frac{\partial M}{\partial y} = 1 + \frac{1}{x}, \quad \frac{\partial N}{\partial x} = 1 + \frac{1}{x} + \cos y. \]

Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.

### Step 5: Find an integrating factor
To solve the non-exact equation, we look for an integrating factor. In this case, we can try using an integrating factor of the form \(\mu(x)\).

### Step 6: Multiply through by the integrating factor
Assuming we find an appropriate integrating factor, we multiply the entire equation by \(\mu(x)\) to make it exact.

### Step 7: Solve the exact equation
Once we have an exact equation, we can find a potential function \(F(x,y)\) such that:

\[ \frac{\partial F}{\partial x} = M \quad \text{and} \quad \frac{\partial F}{\partial y} = N. \]

### Step 8: Integrate to find the solution
Integrate \(M\) with respect to \(x\) and \(N\) with respect to \(y\) to find the function \(F(x,y)\).

### Step 9: Set the function equal to a constant
Finally, we set \(F(x,y) = C\), where \(C\) is a constant, to obtain the implicit solution of the differential equation.

### Final Solution
The final solution will be of the form:

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