If `y = A/x - Bx ^(2),` then `x ^(2) (d ^(2) y )/( dx ^(2)) =`

Options

A. `2y `
B. `y ^(2)`
C. `y ^(3)`
D. `y ^(4)`

Answer

To solve the problem, we need to find the expression \( x^2 \frac{d^2y}{dx^2} \) given that \( y = \frac{A}{x} - Bx^2 \).

### Step-by-Step Solution:

1. **Differentiate \( y \) to find \( \frac{dy}{dx} \)**:
\[ y = \frac{A}{x} - Bx^2 \]
We can rewrite \( \frac{A}{x} \) as \( Ax^{-1} \). Now, we differentiate:
\[ \frac{dy}{dx} = \frac{d}{dx}(Ax^{-1}) - \frac{d}{dx}(Bx^2) \]
Using the power rule:
\[ \frac{dy}{dx} = -A x^{-2} - 2Bx \]
Thus, we have:
\[ \frac{dy}{dx} = -\frac{A}{x^2} - 2Bx \]

2. **Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)**:
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{A}{x^2}\right) - \frac{d}{dx}(2Bx) \]
For the first term, we apply the power rule again:
\[ \frac{d}{dx}\left(-\frac{A}{x^2}\right) = -A \cdot (-2)x^{-3} = \frac{2A}{x^3} \]
For the second term:
\[ \frac{d}{dx}(2Bx) = 2B \]
Therefore, we have:
\[ \frac{d^2y}{dx^2} = \frac{2A}{x^3} - 2B \]

3. **Multiply \( \frac{d^2y}{dx^2} \) by \( x^2 \)**:
Now we need to find \( x^2 \frac{d^2y}{dx^2} \):
\[ x^2 \frac{d^2y}{dx^2} = x^2\left(\frac{2A}{x^3} - 2B\right) \]
Distributing \( x^2 \):
\[ x^2 \frac{d^2y}{dx^2} = \frac{2A x^2}{x^3} - 2B x^2 = \frac{2A}{x} - 2B x^2 \]

4. **Factor out the common term**:
We can factor out 2 from the expression:
\[ x^2 \frac{d^2y}{dx^2} = 2\left(\frac{A}{x} - Bx^2\right) \]

5. **Substitute back for \( y \)**:
Notice that \( \frac{A}{x} - Bx^2 = y \):
\[ x^2 \frac{d^2y}{dx^2} = 2y \]

### Final Answer:
\[ x^2 \frac{d^2y}{dx^2} = 2y \]