The solution of ` (dy)/(dx) = 1+ y + y^(2) + x+ xy + xy^(2)` is

Options

A. `tan^(-1)((2y+1)/(sqrt(3)))=x+x^(2)+C`
B. ` 4 tan^(-1) ((4y+1)/(sqrt(3)))= sqrt(3)(2x+x^(2))+C`
C. ` sqrt(3) tan^(-1) ((3y+1)/3)=4 (1+x+x^(2))+C`
D. ` 4 tan^(-1) ((2y+1)/(sqrt(3)))= sqrt(3)(2x+x^(2))+C`

Explanation

To solve the differential equation \( \frac{dy}{dx} = 1 + y + y^2 + x + xy + xy^2 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \frac{dy}{dx} = 1 + y + y^2 + x + xy + xy^2 \]
We can rearrange the equation to isolate \( dy \) on one side:
\[ dy = (1 + y + y^2 + x + xy + xy^2) dx \]

### Step 2: Grouping Terms
Notice that we can group the terms involving \( y \):
\[ dy = (1 + x + y + xy + y^2 + xy^2) dx \]
This can be factored as:
\[ dy = (1 + y + y^2)(1 + x) dx \]

### Step 3: Separating Variables
Now, we can separate the variables:
\[ \frac{dy}{1 + y + y^2} = (1 + x) dx \]

### Step 4: Integrating Both Sides
Next, we integrate both sides. The left side requires integrating \( \frac{1}{1 + y + y^2} \) with respect to \( y \) and the right side \( (1 + x) \) with respect to \( x \):
\[ \int \frac{dy}{1 + y + y^2} = \int (1 + x) dx \]

### Step 5: Solving the Integrals
1. For the left side, we can use partial fraction decomposition or complete the square. The integral becomes:
\[ \int \frac{dy}{1 + y + y^2} = \text{(use a suitable method to integrate)} \]
Let's denote this integral as \( I(y) \).

2. For the right side:
\[ \int (1 + x) dx = x + \frac{x^2}{2} + C \]

### Step 6: Equating the Integrals
After integrating, we have:
\[ I(y) = x + \frac{x^2}{2} + C \]

### Step 7: Solving for y
Now we need to express \( y \) in terms of \( x \). This might involve some algebraic manipulation depending on the form of \( I(y) \).

### Step 8: Final Solution
The final solution will be in the form:
\[ I(y) = x + \frac{x^2}{2} + C \]
Where \( I(y) \) is the integral we computed earlier.