The solution of the differential equation `xdx+ydy =(xdy-ydx)/(x^(2)+y^(2))` is `tan(f(x, y)-C)=(y)/(x)` (where, C is an arbitrary constant). If `f(1, 1)=1`, then `f(pi, pi)` is equal to

Options

A. 2
B. `pi^(2)`
C. `-1`
D. `pi`

Answer

To solve the given problem, we need to find the function \( f(x, y) \) based on the provided differential equation and the condition \( f(1, 1) = 1 \).

### Step-by-Step Solution:

1. **Start with the given differential equation:**
\[ x \, dx + y \, dy = \frac{x \, dy - y \, dx}{x^2 + y^2} \]

2. **Rearranging the equation:**
Multiply both sides by \( x^2 + y^2 \):
\[ (x^2 + y^2)(x \, dx + y \, dy) = x \, dy - y \, dx \]

3. **Integrating both sides:**
We can integrate the left side:
\[ \int (x^2 + y^2)(x \, dx + y \, dy) \]
This can be split into two integrals:
\[ \int x^3 \, dx + \int y^2 x \, dy + \int y^2 y \, dy \]

4. **Finding the integral:**
The integrals yield:
\[ \frac{x^4}{4} + \frac{y^3 x}{3} + \frac{y^3}{3} \]

5. **Integrating the right side:**
The right side can be integrated as:
\[ \int \frac{x \, dy - y \, dx}{x^2 + y^2} \]
This gives:
\[ \tan^{-1}\left(\frac{y}{x}\right) + C \]

6. **Equating both sides:**
After integration, we have:
\[ \frac{x^2 + y^2}{2} - C = \tan^{-1}\left(\frac{y}{x}\right) \]

7. **Identifying \( f(x, y) \):**
From the equation, we can express \( f(x, y) \) as:
\[ f(x, y) = \frac{x^2 + y^2}{2} \]

8. **Using the condition \( f(1, 1) = 1 \):**
Substitute \( x = 1 \) and \( y = 1 \):
\[ f(1, 1) = \frac{1^2 + 1^2}{2} = \frac{2}{2} = 1 \]
This condition is satisfied.

9. **Finding \( f(\pi, \pi) \):**
Now, substitute \( x = \pi \) and \( y = \pi \):
\[ f(\pi, \pi) = \frac{\pi^2 + \pi^2}{2} = \frac{2\pi^2}{2} = \pi^2 \]

### Final Answer:
Thus, \( f(\pi, \pi) = \pi^2 \).