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

Options

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

Explanation

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y^2 dx + (x^2 - xy + y^2) dy = 0 \]
We can rearrange it to express \( dx \) in terms of \( dy \):
\[ y^2 dx = - (x^2 - xy + y^2) dy \]
Dividing both sides by \( y^2 dy \):
\[ \frac{dx}{dy} = -\frac{x^2 - xy + y^2}{y^2} \]

### Step 2: Simplifying the Right Side
Now, we simplify the right-hand side:
\[ \frac{dx}{dy} = -\left(\frac{x^2}{y^2} - \frac{xy}{y^2} + \frac{y^2}{y^2}\right) = -\left(\frac{x^2}{y^2} - \frac{x}{y} + 1\right) \]

### Step 3: Substituting Variables
Let \( v = \frac{x}{y} \), then \( x = vy \). We differentiate \( x \) with respect to \( y \):
\[ \frac{dx}{dy} = v + y \frac{dv}{dy} \]
Substituting this into our equation gives:
\[ v + y \frac{dv}{dy} = -\left(v^2 - v + 1\right) \]

### Step 4: Rearranging the Equation
Rearranging the equation:
\[ y \frac{dv}{dy} = -v^2 + v + 1 - v \]
This simplifies to:
\[ y \frac{dv}{dy} = -v^2 + 1 \]

### Step 5: Separating Variables
We can separate the variables:
\[ \frac{dv}{1 - v^2} = -\frac{dy}{y} \]

### Step 6: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{dv}{1 - v^2} = \int -\frac{dy}{y} \]
The left side integrates to:
\[ \frac{1}{2} \ln | \frac{1 + v}{1 - v} | = -\ln |y| + C \]

### Step 7: Substituting Back for \( v \)
Recall \( v = \frac{x}{y} \):
\[ \frac{1}{2} \ln \left| \frac{1 + \frac{x}{y}}{1 - \frac{x}{y}} \right| = -\ln |y| + C \]
This can be rewritten as:
\[ \ln \left| \frac{1 + \frac{x}{y}}{1 - \frac{x}{y}} \right| = -2\ln |y| + C' \]
where \( C' = 2C \).

### Step 8: Final Form
Exponentiating both sides gives us:
\[ \left| \frac{1 + \frac{x}{y}}{1 - \frac{x}{y}} \right| = K |y|^{-2} \]
where \( K = e^{C'} \) is a constant.

### Conclusion
Thus, the solution to the differential equation is:
\[ \left| \frac{1 + \frac{x}{y}}{1 - \frac{x}{y}} \right| = K |y|^{-2} \]