To solve the recurrence relation $a_{n+2} - 6a_{n+1} + 9a_n = 3 \cdot 2^n + 7 \cdot 3^n$ with initial conditions $a_0 = 1$ and $a_1 = 4$, we follow these steps:

Step 1: Solve the Homogeneous Equation

The associated homogeneous recurrence relation is: $a_{n+2} - 6a_{n+1} + 9a_n = 0$

The characteristic equation for this recurrence relation is: $x^2 - 6x + 9 = 0$

This simplifies to: $(x - 3)^2 = 0$ which has a double root at $x = 3$. Therefore, the general solution to the homogeneous equation is: $a_n^{(h)} = (c_1 + c_2 n) \cdot 3^n$ where $c_1$ and $c_2$ are constants to be determined later.

Step 2: Find a Particular Solution

Since the non-homogeneous term is $3 \cdot 2^n + 7 \cdot 3^n$, we assume the particular solution has the form: $a_n^{(p)} = A \cdot 2^n + B \cdot n \cdot 3^n$ where $A \cdot 2^n$ corresponds to $3 \cdot 2^n$ and $B \cdot n \cdot 3^n$ corresponds to $7 \cdot 3^n$.

Substitute into the Recurrence Relation

1. Substitute $a_n^{(p)} = A \cdot 2^n + B \cdot n \cdot 3^n$ into the left-hand side of the recurrence relation.
2. Equate coefficients to find $A$ and $B$.

After calculating, we find that: $A = 1 \quad \text{and} \quad B = 7$

Thus, the particular solution is: $a_n^{(p)} = 1 \cdot 2^n + 7n \cdot 3^n = 2^n + 7n \cdot 3^n$

Step 3: General Solution

The general solution to the recurrence relation is the sum of the homogeneous and particular solutions: $a_n = a_n^{(h)} + a_n^{(p)} = (c_1 + c_2 n) \cdot 3^n + 2^n + 7n \cdot 3^n$

Step 4: Use Initial Conditions to Determine Constants

Using the initial conditions $a_0 = 1$ and $a_1 = 4$, we can set up a system of equations to solve for $c_1$ and $c_2$.

1. For $a_0 = 1$: $(c_1 + c_2 \cdot 0) \cdot 3^0 + 2^0 + 7 \cdot 0 \cdot 3^0 = 1$ This gives $c_1 + 1 = 1 \Rightarrow c_1 = 0$.

2. For $a_1 = 4$: $(c_1 + c_2 \cdot 1) \cdot 3^1 + 2^1 + 7 \cdot 1 \cdot 3^1 = 4$ Substituting $c_1 = 0$, we get: $c_2 \cdot 3 + 2 + 21 = 4$ which simplifies to $c_2 = -3$.

So, $c_1 = 0$ and $c_2 = -3$.

Final Solution

Substituting $c_1$ and $c_2$ back, we get: $a_n = (-3n) \cdot 3^n + 2^n + 7n \cdot 3^n$

Simplifying, we find: $a_n = 2^n + (7n - 3n) \cdot 3^n = 2^n + 4n \cdot 3^n$

Thus, the solution to the recurrence relation is: $\boxed{a_n = 2^n + 4n \cdot 3^n}$

Would you like further details on any of these steps?

Related Questions

1. How would we approach a recurrence relation with complex roots in the characteristic equation?
2. What changes if the initial conditions are different?
3. How does a double root in the characteristic equation affect the solution form?
4. Can we solve a similar recurrence relation if the non-homogeneous part involves $n^2 \cdot 3^n$?
5. How would the approach change if the non-homogeneous term were a polynomial rather than exponential?

Tip

In recurrence relations, if you encounter a repeated root, always multiply by $n$ for each repetition in the solution to ensure linear independence of solutions.