`(dy)/(dx) = (2xy)/(x^(2)-1-2y)`

Explanation

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

### Step 1: Cross Multiply
We start by cross multiplying the terms to eliminate the fraction:
\[ (x^2 - 1 - 2y) dy = 2xy dx \]

### Step 2: Expand the Left Side
Now, we expand the left-hand side:
\[ x^2 dy - dy - 2y dy = 2xy dx \]

### Step 3: Rearranging Terms
Next, we rearrange the equation to isolate terms involving \(dy\) on one side and \(dx\) on the other:
\[ 2xy dx - x^2 dy = dy + 2y dy \]

### Step 4: Factor Out \(dy\)
We can factor \(dy\) from the right-hand side:
\[ 2xy dx - x^2 dy = (1 + 2y) dy \]

### Step 5: Divide by \(y^2\)
Now, we will divide the entire equation by \(y^2\):
\[ \frac{2x}{y} dx - \frac{x^2}{y^2} dy = \frac{1 + 2y}{y^2} dy \]

### Step 6: Rearranging Again
Rearranging gives us:
\[ \frac{2x}{y} dx = \left(\frac{1 + 2y}{y^2} + \frac{x^2}{y^2}\right) dy \]

### Step 7: Integrate Both Sides
Now we can integrate both sides:
\[ \int \frac{2x}{y} dx = \int \left(\frac{1 + 2y + x^2}{y^2}\right) dy \]

### Step 8: Solve the Integrals
The left-hand side integrates to:
\[ x^2 \cdot \frac{1}{y} + C_1 \]
The right-hand side integrates to:
\[ -\frac{1}{y} - 2 \ln |y| + C_2 \]

### Step 9: Combine the Results
Combining the results gives us:
\[ \frac{x^2}{y} = -\frac{1}{y} - 2 \ln |y| + C \]
Where \(C = C_2 - C_1\).

### Step 10: Final Rearrangement
Rearranging the equation leads us to our final solution:
\[ \frac{x^2}{y} + \frac{1}{y} + 2 \ln |y| = C \]