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 differential equation and find the value of \( f(\pi, \pi) \), we will follow these steps:

### Step 1: Rewrite the differential equation
The given differential equation is:
\[ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} \]

### Step 2: Integrate both sides
We can integrate both sides of the equation. The left-hand side can be integrated as follows:
\[ \int x \, dx + \int y \, dy = \frac{x^2}{2} + \frac{y^2}{2} \]

### Step 3: Recognize the right-hand side
The right-hand side can be recognized as the derivative of the arctangent function:
\[ \frac{xdy - ydx}{x^2 + y^2} = d(\tan^{-1}(\frac{y}{x})) \]
Thus, integrating the right-hand side gives:
\[ \tan^{-1} \left(\frac{y}{x}\right) + C \]

### Step 4: Set up the equation
Now we equate both sides:
\[ \frac{x^2}{2} + \frac{y^2}{2} = \tan^{-1} \left(\frac{y}{x}\right) + C \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \frac{x^2}{2} + \frac{y^2}{2} - C = \tan^{-1} \left(\frac{y}{x}\right) \]

### Step 6: Apply the tangent function
Taking the tangent of both sides, we have:
\[ \tan\left(\frac{x^2 + y^2}{2} - C\right) = \frac{y}{x} \]

### Step 7: Identify the function \( f(x, y) \)
From the equation, we can identify:
\[ f(x, y) = \frac{x^2 + y^2}{2} \]

### Step 8: Use the condition \( f(1, 1) = 1 \)
We are given that \( f(1, 1) = 1 \):
\[ f(1, 1) = \frac{1^2 + 1^2}{2} = \frac{2}{2} = 1 \]
This condition is satisfied.

### Step 9: Calculate \( f(\pi, \pi) \)
Now we need to find \( f(\pi, \pi) \):
\[ f(\pi, \pi) = \frac{\pi^2 + \pi^2}{2} = \frac{2\pi^2}{2} = \pi^2 \]

### Final Answer
Thus, the value of \( f(\pi, \pi) \) is:
\[ \boxed{\pi^2} \]