Find the general solution of each of the following differential equations:
`(1-x^(2))(1-y)dx = xy(1+y)dy`

Explanation

To solve the differential equation \((1-x^2)(1-y)dx = xy(1+y)dy\), we will separate the variables and integrate both sides. Here’s the step-by-step solution:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (1-x^2)(1-y)dx = xy(1+y)dy \]
We can rearrange this to separate the variables:
\[ \frac{(1-x^2)}{xy(1+y)}dx = \frac{dy}{(1-y)} \]

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

### Step 3: Simplifying the Left Side
We can simplify the left side:
\[ \left(1 - \frac{x^2}{x}\right)dx = \left(1 - x\right)dx \]
Thus, we have:
\[ (1-x)dx = \frac{(1+y)}{(1-y)}dy \]

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

Calculating the left side:
\[ \int (1-x)dx = x - \frac{x^2}{2} + C_1 \]

Calculating the right side:
We can rewrite the right side as:
\[ \int \left(\frac{1+y}{1-y}\right)dy = \int \left(\frac{1}{1-y} + \frac{y}{1-y}\right)dy \]
This gives us:
\[ \int \frac{1}{1-y}dy + \int \frac{y}{1-y}dy \]
The first integral is:
\[ -\ln|1-y| \]
The second integral can be solved using substitution \(u = 1 - y\):
\[ \int \frac{y}{1-y}dy = -\int \frac{1-u}{u}du = -\int \left(1 - \frac{1}{u}\right)du = -u + \ln|u| + C_2 = - (1-y) + \ln|1-y| \]

Combining these results gives:
\[ -\ln|1-y| + (1-y) + C_2 \]

### Step 5: Equating Both Integrals
Now we equate both integrals:
\[ x - \frac{x^2}{2} + C_1 = -\ln|1-y| + (1-y) + C_2 \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ x - \frac{x^2}{2} + \ln|1-y| + (1-y) = C \]
where \(C = C_2 - C_1\).

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