Find the differential equation for the family of curves `y^2=4a (x+b)`where c is an arbitrary constant.

Answer

To find the differential equation for the family of curves given by the equation \( y^2 = 4a(x + b) \), where \( a \) and \( b \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate the equation with respect to \( x \)

Starting with the equation:
\[ y^2 = 4a(x + b) \]

We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x + b)) \]

Using the chain rule on the left side and the product rule on the right side, we get:
\[ 2y \frac{dy}{dx} = 4a \]

### Step 2: Solve for \( a \)

From the equation \( 2y \frac{dy}{dx} = 4a \), we can isolate \( a \):
\[ a = \frac{y \frac{dy}{dx}}{2} \]

### Step 3: Differentiate again to eliminate \( a \)

Now we need to differentiate our first derivative equation again to eliminate the constant \( a \). We differentiate \( 2y \frac{dy}{dx} = 4a \) with respect to \( x \):
\[ \frac{d}{dx}(2y \frac{dy}{dx}) = \frac{d}{dx}(4a) \]

Using the product rule on the left side:
\[ 2 \left( \frac{dy}{dx} \frac{dy}{dx} + y \frac{d^2y}{dx^2} \right) = 0 \]

This simplifies to:
\[ 2 \left( \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right) = 0 \]

### Step 4: Set the equation to zero

Since \( 2 \) is a constant and does not affect the equality, we can simplify this to:
\[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} = 0 \]

### Final Differential Equation

Thus, the differential equation for the family of curves \( y^2 = 4a(x + b) \) is:
\[ y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \]