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 and D are constants and A is parameter
What is B equal to ?

Options

A. (A) `-1`
B. (B) 1
C. (C) 2
D. (D) None of these

Answer

To solve the differential equation \((x^2 + x + 1) dy + (y^2 + y + 1) dx = 0\) and find the value of \(B\) in the general solution \((x + y + 1) = A(1 + Bx + Cy + Dxy)\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ (x^2 + x + 1) dy + (y^2 + y + 1) dx = 0 \]
This can be rearranged to:
\[ (x^2 + x + 1) dy = - (y^2 + y + 1) dx \]

### Step 2: Separating Variables
We can separate the variables by dividing both sides:
\[ \frac{dy}{y^2 + y + 1} = -\frac{dx}{x^2 + x + 1} \]

### Step 3: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{dy}{y^2 + y + 1} = -\int \frac{dx}{x^2 + x + 1} \]

### Step 4: Completing the Square
For the left side, we complete the square for \(y^2 + y + 1\):
\[ y^2 + y + 1 = \left(y + \frac{1}{2}\right)^2 + \frac{3}{4} \]
For the right side, we do the same for \(x^2 + x + 1\):
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]

### Step 5: Using the Integral Formula
Using the integral formula for \(\int \frac{1}{a^2 + u^2} du = \frac{1}{a} \tan^{-1} \left(\frac{u}{a}\right) + C\), we find:
\[ \int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \frac{3}{4}} = \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y + 1}{\sqrt{3}}\right) + C_1 \]
\[ -\int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}} = -\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) + C_2 \]

### Step 6: Combining Constants
Combining the constants \(C_1\) and \(C_2\) into a single constant \(C\):
\[ \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y + 1}{\sqrt{3}}\right) + \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) = C \]

### Step 7: Rearranging to General Form
Rearranging the equation gives us:
\[ x + y + 1 = A(1 + Bx + Cy + Dxy) \]
From the structure of the equation, we can identify the coefficients \(B\), \(C\), and \(D\).

### Step 8: Identifying B
By comparing coefficients, we find that \(B\) corresponds to the coefficient of \(x\) in the rearranged equation. From the integration and the form of the solution, we can deduce that:
\[ B = -1 \]

Thus, the value of \(B\) is:
\[ \boxed{-1} \]