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

Options

A. 7
B. 8
C. 9
D. 10

Answer

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

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

We need to find the second derivative \(\frac{d^2y}{dx^2}\) and express it in the form \(y(p y^2 + 1)\).

### Step 1: Differentiate \(y^2\)
First, we differentiate both sides of the equation 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 and the product and chain rule on the right side, we have:

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

Thus, we can express \(\frac{dy}{dx}\):

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

### Step 2: Differentiate \(\frac{dy}{dx}\)
Next, we differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):

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

Using the quotient rule:

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

Calculating \(\frac{d}{dx}(-2\sqrt{2} \sin(2x))\):

\[ \frac{d}{dx}(-2\sqrt{2} \sin(2x)) = -2\sqrt{2} \cdot 2 \cos(2x) = -4\sqrt{2} \cos(2x) \]

Substituting this back into the expression for \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = \frac{y(-4\sqrt{2} \cos(2x)) + 2\sqrt{2} \sin(2x) \cdot \frac{-2\sqrt{2} \sin(2x)}{y}}{y^2} \]

Simplifying:

\[ \frac{d^2y}{dx^2} = \frac{-4\sqrt{2} y \cos(2x) - \frac{8 \sin^2(2x)}{y}}{y^2} \]

### Step 3: Substitute \(y^2\)
From the original equation, we know:

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

Thus, we can substitute \(1 + 2\sqrt{2} \cos(2x)\) into our expression for \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = \frac{-4\sqrt{2} \cos(2x) \cdot y - \frac{8 \sin^2(2x)}{y}}{y^2} \]

### Step 4: Rearranging
Now we need to express \(\frac{d^2y}{dx^2}\) in the form \(y(p y^2 + 1)\).

From the earlier steps, we can see that:

\[ \frac{d^2y}{dx^2} = y \left( -4\sqrt{2} \frac{\cos(2x)}{y^2} + \frac{-8 \sin^2(2x)}{y^3} \right) \]

### Step 5: Identify \(p\) and \(q\)
By comparing coefficients, we can identify \(p\) and \(q\) from the expression.

From the calculations, we find:

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

### Final Step: Calculate \(p + q\)
Thus, the value of \(p + q\) is:

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

### Conclusion
The final answer is:

\[ \boxed{10} \]