The solution of the differential equation ` x dx + y dy+ (x dy - y dx)/(x^(2)+y^(2))=0` is

Options

A. ` y = x tan ((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 the above

Explanation

To solve the differential equation \( x \, dx + y \, dy + \frac{x \, dy - y \, dx}{x^2 + y^2} = 0 \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ x \, dx + y \, dy + \frac{x \, dy - y \, dx}{x^2 + y^2} = 0 \]

### Step 2: Combine Terms
We can rewrite the equation by combining the terms:
\[ x \, dx + y \, dy + \frac{x \, dy}{x^2 + y^2} - \frac{y \, dx}{x^2 + y^2} = 0 \]

### Step 3: Multiply by \( x^2 + y^2 \)
To eliminate the fraction, we multiply the entire equation by \( x^2 + y^2 \):
\[ (x^2 + y^2)(x \, dx + y \, dy) + x \, dy - y \, dx = 0 \]

### Step 4: Expand the Terms
Expanding the terms gives us:
\[ x^3 \, dx + x^2 y \, dy + y^2 x \, dy + y^3 \, dy + x \, dy - y \, dx = 0 \]

### Step 5: Group the Terms
Now, we group the \( dx \) and \( dy \) terms:
\[ (x^3 - y) \, dx + (x^2 y + y^2 x + y^3 + x) \, dy = 0 \]

### Step 6: Identify the Total Differential
We can recognize that the left-hand side can be expressed as a total differential:
\[ d\left(\frac{1}{2}(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right)\right) = 0 \]

### Step 7: Integrate
Integrating both sides gives:
\[ \frac{1}{2}(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right) = C \]
where \( C \) is a constant of integration.

### Step 8: Final Form
Thus, the solution to the differential equation is:
\[ \frac{1}{2}(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right) = C \]