If ` sqrt(x+y) + sqrt(y-x) = lamda ` then ` (d^2 y)/(dx^2) ` equals

Options

A. `(-2)/(lamda^2)`
B. `(2)/(lamda^2)`
C. `(2)/(lamda)`
D. None

Answer

To solve the problem step-by-step, we need to find the second derivative of \( y \) with respect to \( x \) given the equation:

\[ \sqrt{x+y} + \sqrt{y-x} = \lambda \]

### Step 1: Isolate one of the square roots
First, we can isolate one of the square roots. Let's isolate \( \sqrt{x+y} \):

\[ \sqrt{x+y} = \lambda - \sqrt{y-x} \]

### Step 2: Square both sides
Next, we square both sides to eliminate the square root:

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

Expanding the right-hand side:

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

### Step 3: Simplify the equation
Now, we can simplify the equation:

\[ x + y = \lambda^2 - 2\lambda\sqrt{y-x} + y - x \]

Rearranging gives:

\[ 2x = \lambda^2 - 2\lambda\sqrt{y-x} \]

### Step 4: Isolate the square root again
Now, isolate the square root term:

\[ 2\lambda\sqrt{y-x} = \lambda^2 - 2x \]

Dividing both sides by \( 2\lambda \):

\[ \sqrt{y-x} = \frac{\lambda^2 - 2x}{2\lambda} \]

### Step 5: Square both sides again
Square both sides again to eliminate the square root:

\[ y - x = \left(\frac{\lambda^2 - 2x}{2\lambda}\right)^2 \]

### Step 6: Expand and rearrange
Expanding the right-hand side gives:

\[ y - x = \frac{(\lambda^2 - 2x)^2}{4\lambda^2} \]

Rearranging gives:

\[ y = x + \frac{(\lambda^2 - 2x)^2}{4\lambda^2} \]

### Step 7: Differentiate with respect to \( x \)
Now we differentiate both sides with respect to \( x \):

\[ \frac{dy}{dx} = 1 + \frac{1}{4\lambda^2} \cdot 2(\lambda^2 - 2x)(-2) \]

Simplifying gives:

\[ \frac{dy}{dx} = 1 - \frac{(\lambda^2 - 2x)}{2\lambda^2} \]

### Step 8: Differentiate again to find the second derivative
Now we differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = 0 + \frac{1}{2\lambda^2} \cdot 2 = \frac{1}{\lambda^2} \]

### Final Result
Thus, the second derivative \( \frac{d^2y}{dx^2} \) is:

\[ \frac{d^2y}{dx^2} = \frac{2}{\lambda^2} \]