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

Options

A. `x^(2) - y^(2) = Cx`
B. `x^(2) + y^(2) = Cx`
C. ` 2(x^(2) - y^(2)) = Cx`
D. None of these

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ x^2 + y^2 - 2xy \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 2xy \frac{dy}{dx} = x^2 + y^2 \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \]

### Step 2: Substituting \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting \(y = vx\) into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2 + v^2 x^2}{2vx^2} = \frac{1 + v^2}{2v} \]

### Step 3: Setting the Two Expressions Equal
Now we set the two expressions for \(\frac{dy}{dx}\) equal to each other:
\[ v + x \frac{dv}{dx} = \frac{1 + v^2}{2v} \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v \]
Simplifying the right-hand side:
\[ x \frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v} \]

### Step 5: Separating Variables
Now we separate the variables:
\[ \frac{2v}{1 - v^2} dv = \frac{dx}{x} \]

### Step 6: Integrating Both Sides
Integrating both sides:
\[ \int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x} \]
The left-hand side can be integrated using substitution:
Let \(u = 1 - v^2\), then \(du = -2v dv\), which gives:
\[ -\ln |1 - v^2| = \ln |x| + C \]

### Step 7: Exponentiating Both Sides
Exponentiating both sides results in:
\[ |1 - v^2| = \frac{K}{x} \quad \text{(where \(K = e^{-C}\))} \]
Substituting back \(v = \frac{y}{x}\):
\[ |1 - \left(\frac{y}{x}\right)^2| = \frac{K}{x} \]

### Step 8: Final Rearrangement
This leads to:
\[ 1 - \frac{y^2}{x^2} = \frac{K}{x} \]
Multiplying through by \(x^2\):
\[ x^2 - y^2 = Kx \]
Letting \(C = -K\) gives us the final form:
\[ x^2 - y^2 = Cx \]

### Conclusion
Thus, the solution of the differential equation is:
\[ x^2 - y^2 = Cx \]