The solution of the differential equation `xdx+ydy+(xdy-ydx)/(x^(2)+y^(2))=0`, is

Options

A. `y=xtan((x^(2)+y^(2)+C)/(2))`
B. `x=y tan((x^(2)+y^(2)+C)/(2))`
C. `y=x tan((C-x^(2)-y^(2))/(2))`
D. none of these

Explanation

To solve the differential equation

\[ xdx + ydy + \frac{x dy - y dx}{x^2 + y^2} = 0, \]

we will follow these steps:

### Step 1: Rewrite the Differential Equation
We can rewrite the equation as:

\[ xdx + ydy + \frac{x dy - y dx}{x^2 + y^2} = 0. \]

### Step 2: Identify the Total Differential
We recognize that the term

\[ \frac{x dy - y dx}{x^2 + y^2} \]

is the total differential of \(\tan^{-1}(\frac{y}{x})\). This can be shown by differentiating \(\tan^{-1}(\frac{y}{x})\):

\[ d(\tan^{-1}(\frac{y}{x})) = \frac{x dy - y dx}{x^2 + y^2}. \]

### Step 3: Substitute the Total Differential
Substituting this back into the equation gives us:

\[ xdx + ydy + d(\tan^{-1}(\frac{y}{x})) = 0. \]

### Step 4: Rearranging the Equation
We can rearrange this to:

\[ xdx + ydy = -d(\tan^{-1}(\frac{y}{x})). \]

### Step 5: Integrate Both Sides
Now, we integrate both sides:

1. The left side:
- \(\int xdx = \frac{x^2}{2}\)
- \(\int ydy = \frac{y^2}{2}\)

Thus, the left side becomes:

\[ \frac{x^2}{2} + \frac{y^2}{2}. \]

2. The right side:
- The integral of \(-d(\tan^{-1}(\frac{y}{x}))\) is simply \(-\tan^{-1}(\frac{y}{x})\).

### Step 6: Combine the Results
Combining both sides, we have:

\[ \frac{x^2}{2} + \frac{y^2}{2} + \tan^{-1}(\frac{y}{x}) = C, \]

where \(C\) is the constant of integration.

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

\[ \frac{x^2 + y^2}{2} + \tan^{-1}(\frac{y}{x}) = C. \]