If `x^(2)+xy-y^(2)=1," then "(x-2y)^(3)(d^(2)y)/(dx^(2))=`

Options

A. 1
B. 3
C. 5
D. 10

Answer

To solve the problem, we need to find \((x - 2y)^3 \frac{d^2y}{dx^2}\) given the equation \(x^2 + xy - y^2 = 1\).

### Step 1: Differentiate the given equation
We start with the equation:
\[ x^2 + xy - y^2 = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) = 0 \]
Using the product rule for \(xy\) and the chain rule for \(y^2\):
\[ 2x + \left(x \frac{dy}{dx} + y\right) - 2y \frac{dy}{dx} = 0 \]
This simplifies to:
\[ 2x + x \frac{dy}{dx} + y - 2y \frac{dy}{dx} = 0 \]

### Step 2: Rearranging to find \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ x \frac{dy}{dx} - 2y \frac{dy}{dx} = -2x - y \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(x - 2y) = -2x - y \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{-2x - y}{x - 2y} \]

### Step 3: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Now we differentiate \(\frac{dy}{dx}\) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-2x - y}{x - 2y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(x - 2y)(-2 - \frac{dy}{dx}) - (-2x - y)(1 - 2\frac{dy}{dx})}{(x - 2y)^2} \]

### Step 4: Substitute \(\frac{dy}{dx}\) into the equation
Substituting \(\frac{dy}{dx}\) back into the equation:
\[ \frac{d^2y}{dx^2} = \frac{(x - 2y)(-2 - \frac{-2x - y}{x - 2y}) - (-2x - y)(1 - 2\frac{-2x - y}{x - 2y})}{(x - 2y)^2} \]

### Step 5: Simplifying the expression
After simplifying the above expression, we can find \(\frac{d^2y}{dx^2}\).

### Step 6: Find \((x - 2y)^3 \frac{d^2y}{dx^2}\)
Finally, we multiply \((x - 2y)^3\) by the expression we found for \(\frac{d^2y}{dx^2}\).

### Final Result
After performing all the calculations, we find:
\[ (x - 2y)^3 \frac{d^2y}{dx^2} = 10 \]