Let y(x) be a solution of `xdy+ydx+y^(2)(xdy-ydx)=0` satisfying y(1)=1.
Statement -1 : The range of y(x) has exactly two points.
Statement-2 : The constant of integration is zero.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given differential equation \( x dy + y dx + y^2 (x dy - y dx) = 0 \) with the initial condition \( y(1) = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x dy + y dx + y^2 (x dy - y dx) = 0 \]
This can be rewritten as:
\[ x dy + y dx + y^2 x dy - y^3 dx = 0 \]
Grouping the terms gives:
\[ (x + y^2 x) dy + (y - y^3) dx = 0 \]
Factoring out common terms:
\[ x(1 + y^2) dy + y(1 - y^2) dx = 0 \]

### Step 2: Separate variables
Rearranging the equation, we can separate the variables:
\[ \frac{dy}{dx} = -\frac{y(1 - y^2)}{x(1 + y^2)} \]

### Step 3: Integrate both sides
We can now integrate both sides. We can rewrite this as:
\[ \frac{1 + y^2}{y(1 - y^2)} dy = -\frac{1}{x} dx \]

### Step 4: Solve the left-hand side
The left-hand side can be simplified:
\[ \frac{1 + y^2}{y(1 - y^2)} = \frac{1}{y(1 - y^2)} + \frac{y}{1 - y^2} \]
This can be integrated separately:
1. \(\int \frac{1}{y(1 - y^2)} dy\)
2. \(\int \frac{y}{1 - y^2} dy\)

Using partial fractions, we can integrate:
\[ \int \left( \frac{1}{2y} + \frac{1}{2(1-y)} + \frac{1}{2(1+y)} \right) dy \]
This gives:
\[ \frac{1}{2} \ln |y| - \frac{1}{2} \ln |1 - y^2| + C_1 \]

### Step 5: Solve the right-hand side
The right-hand side integrates to:
\[ -\ln |x| + C_2 \]

### Step 6: Combine results
Combining both integrals, we have:
\[ \frac{1}{2} \ln |y| - \frac{1}{2} \ln |1 - y^2| = -\ln |x| + C \]

### Step 7: Solve for the constant using the initial condition
Using the initial condition \( y(1) = 1 \):
\[ \frac{1}{2} \ln |1| - \frac{1}{2} \ln |1 - 1^2| = -\ln |1| + C \]
This simplifies to \( C = 0 \).

### Step 8: Final equation
Substituting \( C \) back into our equation gives:
\[ \frac{1}{2} \ln |y| - \frac{1}{2} \ln |1 - y^2| = -\ln |x| \]
Rearranging leads to:
\[ y^2 = 1 \quad \text{or} \quad y = \pm 1 \]

### Conclusion
The solution \( y(x) \) has exactly two values: \( y = 1 \) and \( y = -1 \). Therefore, the range of \( y(x) \) has exactly two points.