If `xy = ae^(x) + be^(-x)` satisfies the equation `Axy'' +By' = xy`, then `|A-B|` is__________.

Answer

To solve the problem, we need to find the values of \( A \) and \( B \) such that the equation \( Axy'' + By' = xy \) is satisfied, given that \( xy = ae^x + be^{-x} \).

### Step-by-step Solution:

1. **Given Equation**:
\[ xy = ae^x + be^{-x} \]

2. **Differentiate Once**:
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(ae^x + be^{-x}) \]
Using the product rule on the left side:
\[ x \frac{dy}{dx} + y = ae^x - be^{-x} \]

3. **Differentiate Again**:
Differentiate both sides again:
\[ \frac{d}{dx}(x \frac{dy}{dx} + y) = \frac{d}{dx}(ae^x - be^{-x}) \]
Applying the product rule again:
\[ x \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = ae^x + be^{-x} \]
Simplifying gives:
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = ae^x + be^{-x} \]

4. **Set Up the Equation**:
We can express this in terms of \( xy \):
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = xy \]

5. **Compare with Given Form**:
We have:
\[ Axy'' + By' = xy \]
From our derived equation:
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = xy \]
This implies:
\[ A = 1 \quad \text{and} \quad B = 2 \]

6. **Calculate \(|A - B|\)**:
Now we find:
\[ |A - B| = |1 - 2| = |-1| = 1 \]

### Final Answer:
\[ |A - B| = 1 \]