To solve the recurrence relation using the Characteristic Root Equation (CRE), we can follow these steps:

Step 1: Write down the recurrence relation

Given the recurrence relation:

$a_n - 6a_{n-1} + 9a_{n-2} = 0 \quad \text{for } n \geq 2$

with initial conditions:

$a_0 = 5, \quad a_1 = 12$

Step 2: Form the characteristic equation

The characteristic equation is derived by assuming a solution of the form $a_n = r^n$. Substituting this into the recurrence relation gives:

$r^n - 6r^{n-1} + 9r^{n-2} = 0$

We can factor out $r^{n-2}$ (assuming $r \neq 0$):

$r^{n-2} \left( r^2 - 6r + 9 \right) = 0$

So, the characteristic equation is:

$r^2 - 6r + 9 = 0$

Step 3: Solve the characteristic equation

Solve the quadratic equation $r^2 - 6r + 9 = 0$:

$r = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)} = \frac{6 \pm \sqrt{36 - 36}}{2} = \frac{6 \pm 0}{2} = 3$

Since the discriminant is zero, we have a repeated root:

$r_1 = r_2 = 3$

Step 4: Write the general solution

For a repeated root $r = 3$, the general solution of the recurrence relation is of the form:

$a_n = (C_1 + C_2n)3^n$

Step 5: Apply the initial conditions to find constants

Use the initial conditions $a_0 = 5$ and $a_1 = 12$ to find the constants $C_1$ and $C_2$.

1. For $n = 0$:

$a_0 = C_1 \cdot 3^0 + C_2 \cdot 0 \cdot 3^0 = C_1 = 5$

So, $C_1 = 5$.

2. For $n = 1$:

$a_1 = (C_1 + C_2 \cdot 1)3^1 = (5 + C_2)3 = 12$

Solve for $C_2$:

$15 + 3C_2 = 12 \quad \Rightarrow \quad 3C_2 = -3 \quad \Rightarrow \quad C_2 = -1$

Step 6: Write the final solution

Substitute $C_1 = 5$ and $C_2 = -1$ into the general solution:

$a_n = (5 - n)3^n$

So, the solution to the recurrence relation is:

$a_n = (5 - n)3^n$

Summary

The sequence $a_n$ is defined by the formula:

$a_n = (5 - n)3^n$

Do you want details or have any questions?

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Related Questions:

1. What would the solution be if the initial conditions were different?
2. How does the characteristic root equation help in solving linear recurrence relations?
3. What happens if the characteristic equation has distinct roots?
4. Can you derive the characteristic equation from any linear recurrence relation?
5. How does the solution change if the recurrence relation is non-homogeneous?

Tip: When solving recurrence relations, always check if the characteristic equation has repeated roots, as this will affect the form of your general solution.