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: Rewrite the equation
We start with the given differential equation:
\[ y^2 dx + (x^2 - xy + y^2) dy = 0 \]
We can rearrange this as:
\[ y^2 dx = - (x^2 - xy + y^2) dy \]

### Step 2: Divide by \( y^2 \)
Next, we divide the entire equation by \( y^2 \):
\[ dx + \left( \frac{x^2}{y^2} - \frac{x}{y} + 1 \right) dy = 0 \]
This simplifies to:
\[ dx + \left( \left( \frac{x}{y} \right)^2 - \frac{x}{y} + 1 \right) dy = 0 \]

### Step 3: Substitute \( x = vy \)
Now, we will use the substitution \( x = vy \), where \( v = \frac{x}{y} \). Differentiating both sides with respect to \( y \), we get:
\[ dx = v dy + y dv \]
Substituting \( dx \) into the equation gives:
\[ v dy + y dv + \left( v^2 - v + 1 \right) dy = 0 \]

### Step 4: Combine terms
Combining terms, we have:
\[ (v + v^2 - v + 1) dy + y dv = 0 \]
This simplifies to:
\[ (v^2 + 1) dy + y dv = 0 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ y dv = - (v^2 + 1) dy \]
Dividing both sides by \( y(v^2 + 1) \) yields:
\[ \frac{dv}{v^2 + 1} = -\frac{dy}{y} \]

### Step 6: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dv}{v^2 + 1} = \int -\frac{dy}{y} \]
The left-hand side integrates to \( \tan^{-1}(v) \) and the right-hand side integrates to \( -\ln|y| + C \):
\[ \tan^{-1}(v) = -\ln|y| + C \]

### Step 7: Substitute back for \( v \)
Recalling that \( v = \frac{x}{y} \), we substitute back:
\[ \tan^{-1}\left(\frac{x}{y}\right) = -\ln|y| + C \]

### Final Solution
Thus, the solution to the differential equation is:
\[ \tan^{-1}\left(\frac{x}{y}\right) + \ln|y| = C \]