A curve satisfies the differential equation `(dy)/(dx)=(x+1-xy^2)/(x^2y-y)` and passes through `(0,0)` (1) The equation of the curve is `x^2+y^2+2x=x^2y^2` (2) The equation of the curve is `x^2+y^2+2x+2y=x^2y^2` (3) `x=0` is a tangent to curve (4) `y=0` is a tangent to curve

Options

A. The equation of the curve is `x^(2)+y^(2)+2x = x^(2)y^(2)`
B. The equation of the curve is `x^(2)+y^(2)+2x+2y = x^(2)y^(2)`
C. x=0 is a tangent to curve
D. y = 0 is a tangent to curve

Answer

To solve the given differential equation and find the equation of the curve, we will follow these steps:

### Step 1: Write down the given differential equation
The differential equation is given as:

\[ \frac{dy}{dx} = \frac{x + 1 - xy^2}{x^2y - y} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ (x^2y - y) dy = (x + 1 - xy^2) dx \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:

\[ x^2y \, dy - y \, dy = (x + 1 - xy^2) \, dx \]

### Step 4: Group terms for integration
Now, we can rearrange the terms to isolate the differentials:

\[ x^2y \, dy + xy^2 \, dx = x \, dx + dy + dx \]

### Step 5: Recognize the left-hand side as a derivative
The left-hand side can be recognized as the derivative of a product:

\[ \frac{d}{dx} \left( \frac{x^2y^2}{2} \right) = x \, dx + y \, dy \]

### Step 6: Integrate both sides
Integrating both sides gives us:

\[ \frac{x^2y^2}{2} = \frac{x^2}{2} + \frac{y^2}{2} + x + C \]

### Step 7: Multiply through by 2 to eliminate the fraction
Multiplying through by 2 results in:

\[ x^2y^2 = x^2 + y^2 + 2x + 2C \]

### Step 8: Apply the initial condition
Since the curve passes through the origin (0,0), we substitute \(x = 0\) and \(y = 0\):

\[ 0 = 0 + 0 + 0 + 2C \implies C = 0 \]

### Step 9: Substitute back to find the equation of the curve
Substituting \(C = 0\) back into the equation gives us:

\[ x^2y^2 = x^2 + y^2 + 2x \]

### Step 10: Rearranging the equation
Rearranging gives us the final form of the equation:

\[ x^2y^2 - x^2 - y^2 - 2x = 0 \]

### Conclusion
Thus, the equation of the curve is:

\[ x^2y^2 = x^2 + y^2 + 2x \]