The general solution of `(dy)/(dx) = 1 - x^(2) -y^(2) + x^(2) y^(2)` is

Options

A. `2 sin^(-1) y = x sqrt(1-y^(2))+c`
B. `sin^(-1) y = (1)/(2) sin^(-1) x + c`
C. `cos^(-1) y = x cos^(-1) x + c`
D. `(1)/(2)"log"((1+y)/(1-y))=x-(x^(3))/(3)+c`

Explanation

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

### Step 1: Rearrange the equation
We start with the given equation:
\[ \frac{dy}{dx} = 1 - x^2 - y^2 + x^2 y^2 \]
We can rearrange it as:
\[ \frac{dy}{dx} = 1 - x^2(1 - y^2) - y^2 \]

### Step 2: Factor the right-hand side
We can factor the right-hand side:
\[ \frac{dy}{dx} = (1 - x^2)(1 - y^2) \]

### Step 3: Separate variables
Now we separate the variables \(y\) and \(x\):
\[ \frac{dy}{1 - y^2} = (1 - x^2)dx \]

### Step 4: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{dy}{1 - y^2} = \int (1 - x^2)dx \]

The left-hand side can be integrated using partial fractions:
\[ \int \frac{dy}{1 - y^2} = \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| + C_1 \]

The right-hand side integrates to:
\[ \int (1 - x^2)dx = x - \frac{x^3}{3} + C_2 \]

### Step 5: Combine the results
Setting the two integrals equal gives us:
\[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x - \frac{x^3}{3} + C \]
where \(C = C_2 - C_1\) is a constant.

### Step 6: Solve for the general solution
Thus, the general solution of the differential equation is:
\[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x - \frac{x^3}{3} + C \]

### Summary
The general solution of the differential equation \(\frac{dy}{dx} = 1 - x^2 - y^2 + x^2 y^2\) is:
\[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x - \frac{x^3}{3} + C \]