Solve `(x-y^(2)x)dx=(y-x^(2)y)dy`.

Explanation

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (x - y^2 x)dx = (y - x^2 y)dy \]
We can factor out \(x\) from the left side and \(y\) from the right side:
\[ x(1 - y^2)dx = y(1 - x^2)dy \]

### Step 2: Separating Variables
Next, we will separate the variables by dividing both sides by \(x(1 - y^2)\) and \(y(1 - x^2)\):
\[ \frac{dx}{x(1 - x^2)} = \frac{dy}{y(1 - y^2)} \]

### Step 3: Integrating Both Sides
Now, we will integrate both sides:
\[ \int \frac{dx}{x(1 - x^2)} = \int \frac{dy}{y(1 - y^2)} \]

### Step 4: Using Partial Fraction Decomposition
For the left side, we can use partial fraction decomposition:
\[ \frac{1}{x(1 - x^2)} = \frac{A}{x} + \frac{Bx + C}{1 - x^2} \]
Solving for \(A\), \(B\), and \(C\), we find:
\[ \frac{1}{x(1 - x^2)} = \frac{1/2}{x} + \frac{1/2}{1 - x} - \frac{1/2}{1 + x} \]
Thus, the integral becomes:
\[ \int \left( \frac{1/2}{x} + \frac{1/2}{1 - x} - \frac{1/2}{1 + x} \right) dx \]
This results in:
\[ \frac{1}{2} \ln |x| - \frac{1}{2} \ln |1 - x| - \frac{1}{2} \ln |1 + x| + C_1 \]

For the right side, we can similarly decompose:
\[ \frac{1}{y(1 - y^2)} = \frac{1/2}{y} + \frac{1/2}{1 - y} - \frac{1/2}{1 + y} \]
Integrating gives:
\[ \frac{1}{2} \ln |y| - \frac{1}{2} \ln |1 - y| - \frac{1}{2} \ln |1 + y| + C_2 \]

### Step 5: Equating the Integrals
Now we equate the two integrals:
\[ \frac{1}{2} \ln |x| - \frac{1}{2} \ln |1 - x| - \frac{1}{2} \ln |1 + x| = \frac{1}{2} \ln |y| - \frac{1}{2} \ln |1 - y| - \frac{1}{2} \ln |1 + y| + C \]

### Step 6: Simplifying the Equation
We can simplify this to:
\[ \ln \left( \frac{x}{(1 - x)(1 + x)} \right) = \ln \left( \frac{y}{(1 - y)(1 + y)} \right) + C \]
Exponentiating both sides gives:
\[ \frac{x}{(1 - x)(1 + x)} = k \cdot \frac{y}{(1 - y)(1 + y)} \]
where \(k = e^C\).

### Step 7: Final Form
Rearranging gives us the final relationship:
\[ x(1 - y)(1 + y) = k \cdot y(1 - x)(1 + x) \]