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 C equal to ?

Options

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

Explanation

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

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ (x^{2} + x + 1) dy + (y^{2} + y + 1) dx = 0 \]
Rearranging gives:
\[ (x^{2} + x + 1) dy = - (y^{2} + y + 1) dx \]

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

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

### Step 4: Using Partial Fraction Decomposition
To integrate, we can use the method of completing the square for both denominators:
- For \(y^{2} + y + 1\), we complete the square:
\[ y^{2} + y + 1 = \left(y + \frac{1}{2}\right)^{2} + \frac{3}{4} \]
- For \(x^{2} + x + 1\), we do the same:
\[ x^{2} + x + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4} \]

### Step 5: Performing the Integrals
The integrals can now be computed using the formula:
\[ \int \frac{1}{a^{2} + u^{2}} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \]
Applying this gives us:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C \]

### Step 6: Simplifying the Result
After simplifying, we can express the result in terms of \(x\) and \(y\):
\[ \frac{2}{\sqrt{3}} \left(\tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)\right) = C \]

### Step 7: Comparing with the General Solution
The general solution is given as:
\[ (x+y+1) = A(1 + Bx + Cy + Dxy) \]
We need to express our result in a similar form to identify constants \(B\), \(C\), and \(D\).

### Step 8: Identifying Constants
From the comparison, we can identify:
- The coefficient of \(y\) gives us \(C\).
- The coefficient of \(x\) gives us \(B\).
- The coefficient of \(xy\) gives us \(D\).

### Conclusion
After comparing, we find that:
- \(B = -1\)
- \(C = -1\)
- \(D = -2\)

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