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

Options

A. `tan^(-1).(x)/(y)+logy+C=0`
B. `2tan^(-1).(x)/(y)+logx+C=0`
C. `log(y+sqrt(x^(2)+y^(2)))+logy+C=0`
D. `logy=tan^(-1).(y)/(x)+C`

Explanation

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

### Step 1: Rewrite the equation
We can rewrite the given differential equation in the form:
\[ y^2 dx = - (x^2 - xy + y^2) dy \]
This allows us to express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{y^2}{x^2 - xy + y^2} \]

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

### Step 3: Substitute into the equation
Now, substitute \( y = vx \) into the equation:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} = -\frac{(vx)^2}{x^2 - x(vx) + (vx)^2} \]
This simplifies to:
\[ v + x \frac{dv}{dx} = -\frac{v^2 x^2}{x^2 - v x^2 + v^2 x^2} \]
Simplifying the denominator:
\[ x^2(1 - v + v^2) \]
Thus, we have:
\[ v + x \frac{dv}{dx} = -\frac{v^2}{1 - v + v^2} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x \frac{dv}{dx} = -\frac{v^2}{1 - v + v^2} - v \]
This can be simplified further to:
\[ x \frac{dv}{dx} = -\frac{v^2 + v(1 - v + v^2)}{1 - v + v^2} \]

### Step 5: Separate variables
Now, we can separate the variables:
\[ \frac{1 - v + v^2}{v^2 + v(1 - v + v^2)} dv = -\frac{dx}{x} \]

### Step 6: Integrate both sides
Integrate both sides:
\[ \int \frac{1 - v + v^2}{v^2 + v(1 - v + v^2)} dv = -\int \frac{dx}{x} \]

### Step 7: Solve the integrals
The left side integral can be solved using partial fractions or substitution methods, while the right side gives:
\[ -\ln |x| + C \]

### Step 8: Combine results
After solving the integrals, we will combine the results to express \( y \) in terms of \( x \) and \( C \).

### Final General Solution
The general solution will be of the form:
\[ F(x, y) = C \]
where \( F \) is derived from the integration steps above.