The general solution of the differential equation ` (x^(2) +x+1) dy + (y^(2) +y+1) dx =0 " is " (x+y+1) =A (1 + Bx +Cy +Dxy)` where B,C,D are constants and A is parameter.
What is D equal to ?

Options

A. -1
B. 1
C. -2
D. None of these

Explanation

To find the value of \( D \) in the given differential equation, we will follow these steps:

### Step 1: Write the given differential equation
The differential equation is given as:
\[ (x^2 + x + 1) dy + (y^2 + y + 1) dx = 0 \]

### Step 2: Rearrange the equation
We can rearrange the equation to isolate \( dy \) and \( dx \):
\[ (x^2 + x + 1) dy = - (y^2 + y + 1) dx \]
Dividing both sides by \( (x^2 + x + 1)(y^2 + y + 1) \):
\[ \frac{dy}{y^2 + y + 1} = -\frac{dx}{x^2 + x + 1} \]

### Step 3: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{dy}{y^2 + y + 1} = -\int \frac{dx}{x^2 + x + 1} \]

### Step 4: Simplify the integrals
To integrate, we can complete the square for both \( y^2 + y + 1 \) and \( x^2 + x + 1 \).

For \( y^2 + y + 1 \):
\[ y^2 + y + 1 = \left(y + \frac{1}{2}\right)^2 + \frac{3}{4} \]
Thus,
\[ \int \frac{dy}{y^2 + y + 1} = \int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \]
This integral results in:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) + C_1 \]

For \( x^2 + x + 1 \):
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]
Thus,
\[ -\int \frac{dx}{x^2 + x + 1} = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C_2 \]

### Step 5: Combine the results
Setting the two integrals equal gives us:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) + C_1 = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C_2 \]
This can be rearranged to form:
\[ \frac{2}{\sqrt{3}} \left( \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) \right) = C \]

### Step 6: Express the solution in the required form
We can express this in the form:
\[ x + y + 1 = A(1 + Bx + Cy + Dxy) \]
By comparing coefficients, we can find the values of \( B \), \( C \), and \( D \).

### Step 7: Determine the value of \( D \)
From the comparison:
- The coefficient of \( xy \) gives us \( D = -2 \).

Thus, the value of \( D \) is:
\[ \boxed{-2} \]