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 `

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \frac{dy}{dx} = xy + x + y + 1 \]
We can rearrange the right-hand side:
\[ \frac{dy}{dx} = xy + y + x + 1 = (x + 1)(y + 1) \]

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

### Step 3: Integrating 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:
\[ \frac{x^2}{2} + x + C_2 \]
Combining the constants, we can write:
\[ \ln |y + 1| = \frac{x^2}{2} + x + C \]
where \(C = C_2 - C_1\).

### Step 4: Exponentiating Both Sides
To solve for \(y\), we exponentiate both sides:
\[ |y + 1| = e^{\frac{x^2}{2} + x + C} = e^C e^{\frac{x^2}{2}} e^x \]
Let \(K = e^C\), then:
\[ y + 1 = K e^{\frac{x^2}{2}} e^x \]

### Step 5: Final Solution
Thus, we have:
\[ y = K e^{\frac{x^2}{2}} e^x - 1 \]
This is the general solution to the differential equation.