What is the solution of the differential equation
`(dy)/(dx)=xy+x+y+1` ?

Options

A. `y=(x^(2))/2 +x+c`
B. `log (y+1)= x^(2)/2+x+c`
C. `y=x^(2)+x+c`
D. `log (y+1) =x^(2) +x+c`

Explanation

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

### Step 1: Rearrange the equation
We start with the given equation:
\[ \frac{dy}{dx} = xy + x + y + 1 \]
We can rearrange it as:
\[ \frac{dy}{dx} = xy + y + x + 1 \]
This can be factored as:
\[ \frac{dy}{dx} = (y + 1)(x + 1) \]

### Step 2: Separate the variables
Now, we separate the variables \(y\) and \(x\):
\[ \frac{dy}{y + 1} = (x + 1)dx \]

### Step 3: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{dy}{y + 1} = \int (x + 1)dx \]
The left side integrates to:
\[ \ln|y + 1| + C_1 \]
The right side integrates to:
\[ \frac{x^2}{2} + x + C_2 \]
So we have:
\[ \ln|y + 1| = \frac{x^2}{2} + x + C \]
where \(C = C_2 - C_1\).

### Step 4: Solve for \(y\)
To solve for \(y\), we exponentiate both sides:
\[ |y + 1| = e^{\frac{x^2}{2} + x + C} \]
This can be simplified to:
\[ y + 1 = Ce^{\frac{x^2}{2} + x} \]
where \(C\) is a constant that can take both positive and negative values.

### Step 5: Final solution
Thus, the solution for \(y\) is:
\[ y = Ce^{\frac{x^2}{2} + x} - 1 \]

### Summary
The final solution to the differential equation \(\frac{dy}{dx} = xy + x + y + 1\) is:
\[ y = Ce^{\frac{x^2}{2} + x} - 1 \]

---