已知 $y = 3e^{2x} + 2e^{-x}$，且 $\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} + y = Ae^{2x} + Be^{-x}$，求常数 $A$ 和 $B$ 的值。

1. Find the first derivative of $y$ with respect to $x$.

Given $y = 3e^{2x} + 2e^{-x}$,
$\frac{\text{d}y}{\text{d}x} = 3(2)e^{2x} + 2(-1)e^{-x}$
$\frac{\text{d}y}{\text{d}x} = 6e^{2x} - 2e^{-x}$

2. Find the second derivative of $y$ with respect to $x$.

$\frac{\text{d}^2 y}{\text{d}x^2} = 6(2)e^{2x} - 2(-1)e^{-x}$
$\frac{\text{d}^2 y}{\text{d}x^2} = 12e^{2x} + 2e^{-x}$

3. Substitute $y$, $\frac{\text{d}y}{\text{d}x}$, and $\frac{\text{d}^2 y}{\text{d}x^2}$ into the given expression.

$\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} + y = (12e^{2x} + 2e^{-x}) + (6e^{2x} - 2e^{-x}) + (3e^{2x} + 2e^{-x})$

4. Simplify the expression by collecting like terms.

$\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} + y = (12 + 6 + 3)e^{2x} + (2 - 2 + 2)e^{-x}$
$= 21e^{2x} + 2e^{-x}$

5. Compare coefficients to find $A$ and $B$.

Comparing $21e^{2x} + 2e^{-x}$ with $Ae^{2x} + Be^{-x}$:
$A = 21$
$B = 2$