Solve the recurrence relation
an+2-6an+1+9an=3.2^n +7.3^n for n≥0; a0=1, a1 = 4

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.

Solve the recurrence relation an+2 - 6an+1 + 9an = 3.2^n + 7.3^n for n≥0; a0=1, a1 = 4

This page covers solving the recurrence relation an+2 - 6an+1 + 9an = 3 * 2^n + 7 * 3^n with initial conditions a0 = 1 and a1 = 4. It explains steps like solving the characteristic equation with repeated roots, finding the particular solution, and applying initial conditions to determine constants. Ideal for discrete mathematics students at the undergraduate level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation