The differential equation of the family of curves `y^(2)=4xa(x+1)`, is

Options

A. a.`y^(2)=4(dy)/(dx)(x+(dy)/(dx))`
B. b. `y^(2)((dy)/(dx))^(2)+2xy(dy)/(dx)-y^(2)=0`
C. c.`y = (2x+2)( dy)/(dx)
D. d.`y^(2)=(dy)/(d)+4y=0`

Answer

To find the differential equation of the family of curves given by the equation \( y^2 = 4xA(x + 1) \), we will follow these steps:

### Step 1: Differentiate the given equation with respect to \( x \).
We start with the equation:
\[ y^2 = 4xA(x + 1) \]
Differentiating both sides with respect to \( x \) using implicit differentiation gives:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4xA(x + 1)) \]
Using the chain rule on the left side and the product rule on the right side, we have:
\[ 2y \frac{dy}{dx} = 4A(x + 1) + 4xA \frac{d}{dx}(x + 1) \]
Since \( \frac{d}{dx}(x + 1) = 1 \), this simplifies to:
\[ 2y \frac{dy}{dx} = 4A(x + 1) + 4xA \]
Combining the terms on the right gives:
\[ 2y \frac{dy}{dx} = 4A(2x + 1) \]

### Step 2: Solve for \( \frac{dy}{dx} \).
Now, we can isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4A(2x + 1)}{2y} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2A(2x + 1)}{y} \]

### Step 3: Eliminate the parameter \( A \).
To eliminate the parameter \( A \), we can express \( A \) in terms of \( x \) and \( y \) using the original equation. From \( y^2 = 4xA(x + 1) \), we can solve for \( A \):
\[ A = \frac{y^2}{4x(x + 1)} \]
Substituting this expression for \( A \) back into the equation for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2 \left(\frac{y^2}{4x(x + 1)}\right)(2x + 1)}{y} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y(2x + 1)}{4x(x + 1)} \]

### Step 4: Rearranging to form the differential equation.
Multiplying both sides by \( 4x(x + 1) \) gives:
\[ 4x(x + 1) \frac{dy}{dx} = y(2x + 1) \]
This is the required differential equation of the family of curves.

### Final Answer:
The differential equation of the family of curves \( y^2 = 4xA(x + 1) \) is:
\[ 4x(x + 1) \frac{dy}{dx} = y(2x + 1) \]