To solve this problem, we need to solve the fourth-order differential equation:

$y^{(4)} - 12y^{(3)} + 36y'' = 0$

with initial conditions:

$y(0) = 2, \quad y'(0) = 16, \quad y''(0) = 36, \quad y^{(3)}(0) = 0$

Step 1: Solve the Differential Equation

The differential equation $y^{(4)} - 12y^{(3)} + 36y'' = 0$ suggests a constant-coefficient characteristic equation: $r^4 - 12r^3 + 36r^2 = 0$

We can factor this as: $r^2 (r^2 - 12r + 36) = 0$

This further factors into: $r^2 (r - 6)^2 = 0$

Thus, the roots of the characteristic equation are $r = 0$ (with multiplicity 2) and $r = 6$ (with multiplicity 2).

Step 2: Form the General Solution

Given the roots, the general solution for $y(x)$ is: $y(x) = (C_1 + C_2 x) + (C_3 + C_4 x)e^{6x}$

Step 3: Apply Initial Conditions

Now, we apply the initial conditions to find $C_1$, $C_2$, $C_3$, and $C_4$.

1. Using $y(0) = 2$: $y(0) = C_1 + C_3 = 2$

2. Using $y'(0) = 16$:

   The derivative of $y(x)$ is: $y'(x) = C_2 + (6C_3 + C_4)e^{6x} + 6C_4 x e^{6x}$

   At $x = 0$: $y'(0) = C_2 + 6C_3 = 16$

3. Using $y''(0) = 36$:

   The second derivative of $y(x)$ is: $y''(x) = (6C_4 + 36C_3)e^{6x} + 36C_4 x e^{6x}$

   At $x = 0$: [ y''(0) =