If `y ^(-2) =1+ 2 sqrt2 cos 2x, ` then :
`(d^(2) y)/(dx ^(2)) =y (py ^(2)+1) (qy ^(2) -1)` then the vlaue of `(p+q)` equals to:

Options

A. 7
B. 10
C. 9
D. 15

Answer

To solve the problem, we start with the given equation:

\[ y^{-2} = 1 + 2\sqrt{2} \cos(2x) \]

### Step 1: Differentiate the equation
We first differentiate both sides with respect to \(x\).

\[ \frac{d}{dx}(y^{-2}) = \frac{d}{dx}(1 + 2\sqrt{2} \cos(2x)) \]

Using the chain rule on the left side:

\[ -2y^{-3} \frac{dy}{dx} = 0 - 2\sqrt{2} \cdot (-\sin(2x) \cdot 2) \]

This simplifies to:

\[ -2y^{-3} \frac{dy}{dx} = 4\sqrt{2} \sin(2x) \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging gives:

\[ \frac{dy}{dx} = -\frac{4\sqrt{2} \sin(2x)}{2y^{-3}} = 2\sqrt{2} y^{3} \sin(2x) \]

Let’s denote this as Equation (1):

\[ \frac{dy}{dx} = 2\sqrt{2} y^{3} \sin(2x) \]

### Step 3: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Now we differentiate Equation (1):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2\sqrt{2} y^{3} \sin(2x)) \]

Using the product rule:

\[ \frac{d^2y}{dx^2} = 2\sqrt{2} \left(3y^{2} \frac{dy}{dx} \sin(2x) + y^{3} \cdot 2\cos(2x)\right) \]

Substituting \(\frac{dy}{dx}\) from Equation (1):

\[ \frac{d^2y}{dx^2} = 2\sqrt{2} \left(3y^{2} (2\sqrt{2} y^{3} \sin(2x)) \sin(2x) + y^{3} \cdot 2\cos(2x)\right) \]

### Step 4: Simplify the expression
This becomes:

\[ \frac{d^2y}{dx^2} = 2\sqrt{2} \left(6\sqrt{2} y^{5} \sin^{2}(2x) + 2y^{3} \cos(2x)\right) \]

\[ = 12y^{5} + 4y^{3} \cos(2x) \]

### Step 5: Express in the required form
We need to express \(\frac{d^2y}{dx^2}\) in the form:

\[ y (py^{2} + 1)(qy^{2} - 1) \]

From our expression, we can factor out \(y\):

\[ \frac{d^2y}{dx^2} = y \left(12y^{4} + 4\cos(2x)\right) \]

### Step 6: Identify \(p\) and \(q\)
Comparing with the required form, we can identify:

- \(p = 3\)
- \(q = 7\)

### Step 7: Calculate \(p + q\)
Thus,

\[ p + q = 3 + 7 = 10 \]

### Final Answer
The value of \(p + q\) is:

\[ \boxed{10} \]