The differential equation of the family of curves,` y^2 = 4a ( x + b) , a b in R ,` has order and degree respectively equal to :

Options

A. `1,2`
B. `2,1`
C. `2,2,`
D. `1,1`

Explanation

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

### Step 1: Differentiate the given equation
We start with the equation:
\[ y^2 = 4a(x + b) \]
Differentiating 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:
\[ 2y \frac{dy}{dx} = 4a \cdot 1 + 0 \]
This simplifies to:
\[ 2y \frac{dy}{dx} = 4a \]

### Step 2: Isolate \( a \)
To eliminate \( a \), we can rearrange the equation:
\[ y \frac{dy}{dx} = 2a \]
Now, we will differentiate again to eliminate \( a \).

### Step 3: Differentiate again
Differentiating \( y \frac{dy}{dx} = 2a \) with respect to \( x \):
Using the product rule:
\[ \frac{d}{dx}(y) \cdot \frac{dy}{dx} + y \cdot \frac{d^2y}{dx^2} = 2 \frac{da}{dx} \]
Since \( a \) is a constant with respect to \( x \), \( \frac{da}{dx} = 0 \):
\[ \frac{dy}{dx} \cdot \frac{dy}{dx} + y \cdot \frac{d^2y}{dx^2} = 0 \]
This can be rewritten as:
\[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} = 0 \]

### Step 4: Identify the order and degree
Now we have a differential equation:
\[ y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \]
- The highest derivative present is \( \frac{d^2y}{dx^2} \), which indicates that the **order** of the differential equation is **2**.
- The highest power of the highest order derivative \( \frac{d^2y}{dx^2} \) in this equation is **1**, indicating that the **degree** of the differential equation is **1**.

### Final Answer
Thus, the order and degree of the differential equation are:
- **Order = 2**
- **Degree = 1**

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