`y(xy+1)dx+x(1+xy+x^(2)y^(2))dy=0`

Explanation

To solve the differential equation \( y(xy + 1)dx + x(1 + xy + x^2y^2)dy = 0 \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ y(xy + 1)dx + x(1 + xy + x^2y^2)dy = 0 \]
This can be rewritten as:
\[ y^2x \, dx + y \, dx + x \, dy + xy \, dy + x^2y^2 \, dy = 0 \]

### Step 2: Group the Terms
Next, we group the terms involving \( dx \) and \( dy \):
\[ y^2x \, dx + (y + xy + x^2y^2) \, dy = 0 \]
This can be rearranged to:
\[ y^2x \, dx + (y(1 + x) + x^2y^2) \, dy = 0 \]

### Step 3: Factor Out Common Terms
We can factor out \( y \) from the \( dy \) terms:
\[ y^2x \, dx + y(1 + x + x^2y) \, dy = 0 \]

### Step 4: Divide by \( xy^2 \)
Assuming \( x \neq 0 \) and \( y \neq 0 \), we can divide the entire equation by \( xy^2 \):
\[ \frac{dx}{y} + \frac{(1 + x + x^2y)}{x} \, dy = 0 \]

### Step 5: Separate Variables
Now, we can separate the variables:
\[ \frac{dx}{y} = -\frac{(1 + x + x^2y)}{x} \, dy \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dx}{y} = -\int \frac{(1 + x + x^2y)}{x} \, dy \]
This leads to:
\[ \ln |x| = -\left( \ln |y| + xy + \frac{1}{2}x^2y^2 \right) + C \]

### Step 7: Solve for the General Solution
Rearranging gives us the general solution:
\[ -\frac{1}{2}(xy)^2 - \frac{1}{xy} + \ln |y| = C \]

### Final General Solution
Thus, the general solution of the differential equation is:
\[ -\frac{1}{2}(xy)^2 - \frac{1}{xy} + \ln |y| = C \]