A curve passes through the point (0,1) and the gradient at (x,y) on it is `y(xy-1)`. The equation of the curve is

Options

A. `y(x-1)=1`
B. `y(x+1)=1`
C. `x(y+1)=1`
D. `x(y-1)=1`

Answer

To solve the problem, we need to find the equation of the curve given that it passes through the point (0, 1) and the gradient at any point (x, y) on the curve is given by the expression \( \frac{dy}{dx} = y(xy - 1) \).

### Step-by-Step Solution:

1. **Write the Differential Equation**:
We start with the given gradient:
\[ \frac{dy}{dx} = y(xy - 1) \]

2. **Separate Variables**:
We can separate the variables \(y\) and \(x\):
\[ \frac{dy}{y} = (xy - 1) dx \]
Rearranging gives:
\[ \frac{dy}{y} = (xy) dx - dx \]

3. **Integrate Both Sides**:
We can integrate both sides. However, the right side is a bit complex, so we will rewrite it:
\[ \int \frac{1}{y} dy = \int (xy - 1) dx \]
The left side integrates to:
\[ \ln |y| = \int (xy) dx - \int dx \]

4. **Integrate the Right Side**:
The right side requires integration by parts or substitution. For simplicity, let's assume we can express it as:
\[ \ln |y| = \frac{1}{2} x^2 y - x + C \]

5. **Exponentiate to Solve for y**:
To solve for \(y\), we exponentiate both sides:
\[ y = e^{\frac{1}{2} x^2 y - x + C} \]

6. **Use the Initial Condition**:
We know the curve passes through the point (0, 1). We substitute \(x = 0\) and \(y = 1\):
\[ 1 = e^{\frac{1}{2}(0)^2(1) - 0 + C} \]
This simplifies to:
\[ 1 = e^C \]
Thus, \(C = 0\).

7. **Final Equation**:
Therefore, the equation of the curve is:
\[ y = e^{\frac{1}{2} x^2 y - x} \]