The slope of the tangent at any arbitrary point of a curve is twice the product of the abscissa and square of the ordinate of the point. Then, the equation of the curve is (where c is an arbitrary constant)

Options

A. `x^(2)y+y+c=0`
B. `x^(2)y+cy+1=0`
C. `xy+y+c=0`
D. `xy^(2)+cy+y=0`

Answer

To solve the problem, we need to find the equation of the curve given that the slope of the tangent at any point \((x, y)\) is twice the product of the abscissa and the square of the ordinate.

### Step-by-step Solution:

1. **Understand the given information**:
The slope of the tangent at any point \((x, y)\) is given by \(\frac{dy}{dx}\). According to the problem, this slope is equal to \(2x y^2\).

\[ \frac{dy}{dx} = 2x y^2 \]

**Hint**: Identify the relationship between the slope of the tangent and the coordinates of the point on the curve.

2. **Rearranging the equation**:
We can rearrange the equation to separate the variables \(y\) and \(x\):

\[ \frac{dy}{y^2} = 2x \, dx \]

**Hint**: The goal is to isolate \(dy\) and \(dx\) on opposite sides of the equation.

3. **Integrate both sides**:
Now we will integrate both sides. The left side requires the integral of \(\frac{dy}{y^2}\) and the right side is the integral of \(2x \, dx\).

\[ \int \frac{dy}{y^2} = \int 2x \, dx \]

The left side integrates to \(-\frac{1}{y}\) and the right side integrates to \(x^2 + C\):

\[ -\frac{1}{y} = x^2 + C \]

**Hint**: Remember the basic integration formulas and apply them correctly.

4. **Rearranging the equation**:
To express \(y\) in terms of \(x\), we can rearrange the equation:

\[ -1 = y(x^2 + C) \]
or,
\[ y(x^2 + C) + 1 = 0 \]

**Hint**: Rearranging the equation helps in identifying the relationship between \(x\) and \(y\).

5. **Final equation of the curve**:
The final equation of the curve can be expressed as:

\[ x^2 y + y + 1 = 0 \]

This represents the equation of the curve where \(C\) is an arbitrary constant.

**Hint**: Ensure that the final equation is in a standard form that represents a relationship between \(x\) and \(y\).

### Conclusion:
The equation of the curve is given by:

\[ x^2 y + y + 1 = 0 \]

This matches with option B from the provided choices.