已知一曲线满足 $\frac{d^2y}{dx^2} = 36e^{3x} - e^{-x}$。 该曲线与 $y$ 轴交于点 $Q(0, 5)$，且曲线在点 $Q$ 处的切线斜率为 14。 求该曲线的方程。

1. Integrate the second derivative to find the expression for the first derivative, $\frac{dy}{dx}$.

$\frac{dy}{dx} = \int (36e^{3x} - e^{-x}) \, dx$
$\frac{dy}{dx} = 12e^{3x} + e^{-x} + c_1$

2. Use the given gradient at $x = 0$ to find the constant of integration $c_1$.

At $Q(0, 5)$, the gradient $\frac{dy}{dx} = 14$.
$14 = 12e^{3(0)} + e^{-(0)} + c_1$
$14 = 12(1) + 1 + c_1$
$14 = 13 + c_1$
$c_1 = 1$
So, $\frac{dy}{dx} = 12e^{3x} + e^{-x} + 1$

3. Integrate the first derivative to find the expression for $y$.

$y = \int (12e^{3x} + e^{-x} + 1) \, dx$
$y = 4e^{3x} - e^{-x} + x + c_2$

4. Use the coordinates of point $Q$ to find the constant of integration $c_2$.

At $Q(0, 5)$, $x = 0$ and $y = 5$.
$5 = 4e^{3(0)} - e^{-(0)} + 0 + c_2$
$5 = 4(1) - 1 + c_2$
$5 = 3 + c_2$
$c_2 = 2$
Therefore, the equation of the curve is $y = 4e^{3x} - e^{-x} + x + 2$.