The solution of the differential equation `xdy + (x+y) dx=0` is

Options

A. `c =(y^(2))/(2)+xy`
B. `c= xy + (x^(2))/(2)`
C. `c= x + ((xy)^(2))/(2)`
D. None of these

Answer

To solve the differential equation \( x \, dy + (x + y) \, dx = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rewriting the given differential equation:
\[ x \, dy + (x + y) \, dx = 0 \]
This can be rearranged to:
\[ x \, dy + x \, dx + y \, dx = 0 \]
or
\[ x \, dy + (x + y) \, dx = 0 \]

### Step 2: Grouping Terms
Next, we can group the terms:
\[ x \, dy + y \, dx + x \, dx = 0 \]
This can be rewritten as:
\[ x \, dy + y \, dx = -x \, dx \]

### Step 3: Recognizing the Product Rule
Notice that the left-hand side can be recognized as the derivative of the product \( xy \):
\[ d(xy) = x \, dy + y \, dx \]
Thus, we can rewrite the equation as:
\[ d(xy) = -x \, dx \]

### Step 4: Integrating Both Sides
Now, we integrate both sides:
\[ \int d(xy) = \int -x \, dx \]
The left side integrates to \( xy \), and the right side integrates to \( -\frac{x^2}{2} + C \):
\[ xy = -\frac{x^2}{2} + C \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ x^2 + 2xy = 2C \]
Letting \( 2C = c \), we can write:
\[ x^2 + 2xy = c \]

### Final Solution
Thus, the solution to the differential equation is:
\[ x^2 + 2xy = c \]