The general solution of the differential equation ` (dy)/(dx) +x (x+y)=x (x+y) ^(3)-1` is:

Options

A. `ln |((x +y+1)(x+y-1))/((x+y)^(4))|=x ^(2) +C`
B. `ln |((x +y+1)(x+y-1))/((x+y)^(2))|=x ^(2) +C`
C. `2ln |((x +y+1)(x+y-1))/((x+y)^(2))|=x ^(2) +C`
D. `ln |((x +y+1)(x+y-1))/((x+y)^(2))|=x +C`

Answer

To solve the differential equation \( \frac{dy}{dx} + x(x+y) = x(x+y)^3 - 1 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rewriting the differential equation in a more manageable form:
\[ \frac{dy}{dx} + x(x+y) = x(x+y)^3 - 1 \]
This can be rearranged to:
\[ \frac{dy}{dx} = x(x+y)^3 - x(x+y) - 1 \]

### Step 2: Factoring the Right Side
Notice that the right side can be factored:
\[ \frac{dy}{dx} = x(x+y)((x+y)^2 - 1) \]
This gives us:
\[ \frac{dy}{dx} = x(x+y)((x+y-1)(x+y+1)) \]

### Step 3: Substitution
Let \( t = x + y \). Then, we have:
\[ y = t - x \quad \text{and} \quad \frac{dy}{dx} = \frac{dt}{dx} - 1 \]
Substituting into the equation gives:
\[ \frac{dt}{dx} - 1 = x t \left( (t-1)(t+1) \right) \]
This simplifies to:
\[ \frac{dt}{dx} = x t (t^2 - 1) + 1 \]

### Step 4: Separating Variables
Now, we can rearrange the equation:
\[ \frac{dt}{dx} = x t (t^2 - 1) + 1 \]
This can be separated as:
\[ \frac{dt}{t(t^2 - 1)} = x \, dx \]

### Step 5: Partial Fraction Decomposition
Next, we perform partial fraction decomposition on the left side:
\[ \frac{1}{t(t-1)(t+1)} = \frac{A}{t} + \frac{B}{t-1} + \frac{C}{t+1} \]
Multiplying through by the denominator and solving for \( A, B, C \) gives us:
- Set \( t = 0 \) to find \( A \)
- Set \( t = 1 \) to find \( B \)
- Set \( t = -1 \) to find \( C \)

After solving, we find:
\[ A = 1, \quad B = \frac{1}{2}, \quad C = \frac{1}{2} \]

### Step 6: Integrating Both Sides
Now we can integrate both sides:
\[ \int \left( \frac{1}{t} + \frac{1/2}{t-1} + \frac{1/2}{t+1} \right) dt = \int x \, dx \]
This results in:
\[ \ln |t| + \frac{1}{2} \ln |t-1| + \frac{1}{2} \ln |t+1| = \frac{x^2}{2} + C \]

### Step 7: Back Substitution
Substituting back \( t = x + y \):
\[ \ln |x+y| + \frac{1}{2} \ln |x+y-1| + \frac{1}{2} \ln |x+y+1| = \frac{x^2}{2} + C \]

### Final Step: Exponential Form
Exponentiating both sides gives us the general solution:
\[ |x+y| \sqrt{|x+y-1| |x+y+1|} = e^{\frac{x^2}{2} + C} \]
This can be expressed as:
\[ (x+y) \sqrt{(x+y-1)(x+y+1)} = k e^{\frac{x^2}{2}} \quad \text{where } k = e^C \]

### Conclusion
Thus, the general solution of the differential equation is:
\[ (x+y) \sqrt{(x+y-1)(x+y+1)} = k e^{\frac{x^2}{2}} \]