The solution curve of `(dy)/(dx) =(y^(2)-2xy-x^(2))/(y^(2)+2xy-x^(2)), y(-1)=1` is

Options

A. St. line
B. Circle
C. Parabola
D. Ellipse

Answer

To solve the differential equation given by

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

with the initial condition \( y(-1) = 1 \), we will follow these steps:

### Step 1: Simplify the Differential Equation
We start by simplifying the right-hand side of the equation. We can factor out \( x^2 \) from both the numerator and the denominator:

\[ \frac{dy}{dx} = \frac{y^2/x^2 - 2y/x - 1}{y^2/x^2 + 2y/x - 1} \]

Let \( v = \frac{y}{x} \), then \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).

### Step 2: Substitute and Rearrange
Substituting \( y = vx \) into the equation gives:

\[ v + x \frac{dv}{dx} = \frac{(vx)^2 - 2(vx)x - x^2}{(vx)^2 + 2(vx)x - x^2} \]

This simplifies to:

\[ v + x \frac{dv}{dx} = \frac{v^2 - 2v - 1}{v^2 + 2v - 1} \]

### Step 3: Separate Variables
Rearranging gives:

\[ x \frac{dv}{dx} = \frac{v^2 - 2v - 1}{v^2 + 2v - 1} - v \]

This can be simplified further to:

\[ x \frac{dv}{dx} = \frac{(v^2 - 2v - 1) - v(v^2 + 2v - 1)}{v^2 + 2v - 1} \]

### Step 4: Simplify the Right-Hand Side
After simplifying the numerator, we find:

\[ x \frac{dv}{dx} = \frac{-v^3 - v^2 - 2v - 1}{v^2 + 2v - 1} \]

### Step 5: Separate Variables Again
Now we can separate the variables:

\[ \frac{v^2 + 2v - 1}{-v^3 - v^2 - 2v - 1} dv = \frac{dx}{x} \]

### Step 6: Integrate Both Sides
Integrating both sides, we have:

\[ \int \frac{v^2 + 2v - 1}{-v^3 - v^2 - 2v - 1} dv = \int \frac{dx}{x} \]

### Step 7: Solve the Integrals
The left-hand side can be solved using partial fractions, and the right-hand side integrates to \( \ln |x| + C \).

### Step 8: Apply Initial Condition
Using the initial condition \( y(-1) = 1 \) (which means \( v = -1 \) when \( x = -1 \)), we can find the constant \( C \).

### Step 9: Final Equation
After solving for \( C \), we can express the final solution in terms of \( y \) and \( x \).

### Conclusion
The solution curve is a straight line represented by the equation:

\[ x + y = 0 \]