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 can 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 separate the variables by dividing both sides by \(y(1 - y^2)\) and \(x(1 - x^2)\):
\[ \frac{x \, dx}{1 - x^2} = \frac{y \, dy}{1 - y^2} \]

### Step 3: Multiplying by 2
To simplify the integration, we multiply both sides by 2:
\[ \frac{2x \, dx}{1 - x^2} = \frac{2y \, dy}{1 - y^2} \]

### Step 4: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{2x \, dx}{1 - x^2} = \int \frac{2y \, dy}{1 - y^2} \]

### Step 5: Substitution for Integration
For the left side, let \(t = 1 - x^2\), then \(dt = -2x \, dx\) or \(dx = -\frac{dt}{2x}\). Thus, we have:
\[ \int \frac{2x \, dx}{1 - x^2} = -\int \frac{dt}{t} = -\log |t| + C_1 = -\log |1 - x^2| + C_1 \]
Similarly, for the right side, let \(p = 1 - y^2\), then \(dp = -2y \, dy\) or \(dy = -\frac{dp}{2y}\):
\[ \int \frac{2y \, dy}{1 - y^2} = -\int \frac{dp}{p} = -\log |p| + C_2 = -\log |1 - y^2| + C_2 \]

### Step 6: Equating the Integrals
Now we equate the two integrals:
\[ -\log |1 - x^2| + C_1 = -\log |1 - y^2| + C_2 \]
Rearranging gives:
\[ \log |1 - x^2| = \log |1 - y^2| + C \]
where \(C = C_2 - C_1\).

### Step 7: Exponentiating Both Sides
Exponentiating both sides, we have:
\[ |1 - x^2| = k |1 - y^2| \]
where \(k = e^C\).

### Step 8: Final Solution
Thus, the general solution of the given differential equation is:
\[ 1 - x^2 = k(1 - y^2) \]
where \(k\) is a constant.

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