If `y=x^(2)+x^(-2)," then "x^(2)y_(2)+4xy_(1)+2y=`

Options

A. 0
B. `12x^(2)`
C. 12x
D. 1

Answer

To solve the problem, we need to find the expression \( x^2 y'' + 4x y' + 2y \) given that \( y = x^2 + x^{-2} \).

### Step-by-Step Solution:

1. **Find \( y' \)** (First Derivative):
\[ y = x^2 + x^{-2} \]
Differentiate \( y \) with respect to \( x \):
\[ y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(x^{-2}) = 2x - 2x^{-3} \]

2. **Find \( y'' \)** (Second Derivative):
Differentiate \( y' \) with respect to \( x \):
\[ y' = 2x - 2x^{-3} \]
Now differentiate:
\[ y'' = \frac{d}{dx}(2x) + \frac{d}{dx}(-2x^{-3}) = 2 + 6x^{-4} \]

3. **Substitute \( y, y', \) and \( y'' \) into the expression \( x^2 y'' + 4x y' + 2y \)**:
\[ x^2 y'' = x^2 (2 + 6x^{-4}) = 2x^2 + 6x^{-2} \]
\[ 4x y' = 4x (2x - 2x^{-3}) = 8x^2 - 8x^{-2} \]
\[ 2y = 2(x^2 + x^{-2}) = 2x^2 + 2x^{-2} \]

4. **Combine all the terms**:
\[ x^2 y'' + 4x y' + 2y = (2x^2 + 6x^{-2}) + (8x^2 - 8x^{-2}) + (2x^2 + 2x^{-2}) \]
Combine like terms:
\[ = (2x^2 + 8x^2 + 2x^2) + (6x^{-2} - 8x^{-2} + 2x^{-2}) \]
\[ = 12x^2 + 0 \]

5. **Final Result**:
\[ x^2 y'' + 4x y' + 2y = 12x^2 \]