The general solution of the differential equation `(2x-y+1)dx+(2y-x+1)dy=0` is -

Options

A. `x^(2) +y^(2)+xy-x + y = c `
B. `x^(2)+y^(2) - xy + x + y = c `
C. ` x^(2) - y^(2) + 2xy - x + y =c `
D. `x^(2) - y^(2)-2xy+x-y = c`

Answer

To solve the differential equation \((2x - y + 1)dx + (2y - x + 1)dy = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (2x - y + 1)dx + (2y - x + 1)dy = 0 \]
This can be rearranged as:
\[ 2x \, dx - y \, dx + dx + 2y \, dy - x \, dy + dy = 0 \]

### Step 2: Group the terms
Now, we can group the terms:
\[ (2x \, dx + 2y \, dy) + (-y \, dx - x \, dy) + (dx + dy) = 0 \]
This simplifies to:
\[ 2x \, dx + 2y \, dy - (y \, dx + x \, dy) + (dx + dy) = 0 \]

### Step 3: Identify the total differential
Notice that the term \(- (y \, dx + x \, dy)\) is the total differential of the product \(xy\):
\[ d(xy) = y \, dx + x \, dy \]
Thus, we can rewrite the equation as:
\[ 2x \, dx + 2y \, dy - d(xy) + (dx + dy) = 0 \]

### Step 4: Integrate both sides
Now, we can integrate each term:
1. The integral of \(2x \, dx\) is \(x^2\).
2. The integral of \(2y \, dy\) is \(y^2\).
3. The integral of \(d(xy)\) is \(xy\).
4. The integral of \(dx + dy\) is \(x + y\).

Putting it all together, we have:
\[ x^2 + y^2 - xy + x + y = C \]
where \(C\) is a constant of integration.

### Step 5: Write the general solution
Thus, the general solution of the differential equation is:
\[ x^2 + y^2 - xy + x + y = C \]