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 B equal to ?

Options

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

Explanation

To solve the given differential equation and find the value of B, we will follow these steps:

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

### Step 2: Rearrange the Equation
Rearranging the equation gives:
\[ (x^2 + x + 1) dy = -(y^2 + y + 1) dx \]
Dividing both sides by \((y^2 + y + 1)(x^2 + x + 1)\) leads to:
\[ \frac{dy}{y^2 + y + 1} = -\frac{dx}{x^2 + x + 1} \]

### 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 simplify the integrals, we can complete the square for both \(y^2 + y + 1\) and \(x^2 + x + 1\).

For \(y^2 + y + 1\):
\[ y^2 + y + 1 = \left(y + \frac{1}{2}\right)^2 + \frac{3}{4} \]

For \(x^2 + x + 1\):
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]

### Step 5: Perform the Integration
Using the formula \(\int \frac{1}{a^2 + u^2} du = \frac{1}{a} \tan^{-1} \left(\frac{u}{a}\right)\), we can integrate both sides:
\[ \int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = -\int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \]

This gives:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C \]

### Step 6: Rearranging the Result
After rearranging and simplifying, we can express the solution in the form:
\[ x + y + 1 = A(1 + Bx + Cy + Dxy) \]

### Step 7: Identify Coefficients
From the rearranged form, we need to identify the coefficients of \(x\) to find \(B\). The coefficient of \(x\) in the final expression will give us the value of \(B\).

### Step 8: Find the Value of B
From the integration and rearrangement, we find that the coefficient of \(x\) is \(-1\). Thus:
\[ B = -1 \]

### Final Answer
The value of \(B\) is:
\[ \boxed{-1} \]