某个微分方程的通解为 $y = Ae^{x} + Be^{2x} + Ce^{-2x}$，其中 $A$、$B$ 和 $C$ 是任意常数。已知某特解在 $x = 0$ 时满足 $y = 2$、$\dfrac{dy}{dx} = -3$ 和 $\dfrac{d^{2}y}{dx^{2}} = 11$，求 $A$、$B$ 和 $C$ 的值。

1. Differentiate the given general solution to obtain expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

$\frac{dy}{dx} = Ae^x + 2Be^{2x} - 2Ce^{-2x}$
$\frac{d^2y}{dx^2} = Ae^x + 4Be^{2x} + 4Ce^{-2x}$

2. Substitute $x = 0$ and the given values to form a system of equations.

Since $e^0 = 1$, substituting $y = 2$, $\frac{dy}{dx} = -3$, and $\frac{d^2y}{dx^2} = 11$ gives:
$A + B + C = 2 \quad \text{--- (1)}$
$A + 2B - 2C = -3 \quad \text{--- (2)}$
$A + 4B + 4C = 11 \quad \text{--- (3)}$

3. Solve the system of linear equations to find $A$, $B$, and $C$.

Subtract (1) from (2):
$(A + 2B - 2C) - (A + B + C) = -3 - 2$
$B - 3C = -5 \quad \text{--- (4)}$

Subtract (1) from (3):
$(A + 4B + 4C) - (A + B + C) = 11 - 2$
$3B + 3C = 9 \implies B + C = 3 \quad \text{--- (5)}$

Multiply (5) by 3 and add to (4):
$(B - 3C) + 3(B + C) = -5 + 3(3)$
$4B = 4 \implies B = 1$

Substitute $B = 1$ into (5):
$1 + C = 3 \implies C = 2$

Substitute $B = 1$ and $C = 2$ into (1):
$A + 1 + 2 = 2 \implies A = -1$