Find the complementary solution of an+2 – 6 an+1 – 9 an = 2
n

The given recurrence relation is:

\[
a_{n+2} - 6a_{n+1} - 9a_n = 2^n
\]

We need to find the complementary solution, which corresponds to the homogeneous recurrence relation:

\[
a_{n+2} - 6a_{n+1} - 9a_n = 0
\]

### Step 1: Set up the characteristic equation
We assume a solution of the form \( a_n = r^n \). Substituting \( a_n = r^n \) into the homogeneous recurrence relation, we get the characteristic equation:

\[
r^2 - 6r - 9 = 0
\]

### Step 2: Solve the characteristic equation
We solve the quadratic equation \( r^2 - 6r - 9 = 0 \) using the quadratic formula:

\[
r = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-9)}}{2(1)}
\]

\[
r = \frac{6 \pm \sqrt{36 + 36}}{2}
\]

\[
r = \frac{6 \pm \sqrt{72}}{2}
\]

\[
r = \frac{6 \pm 6\sqrt{2}}{2}
\]

\[
r = 3 \pm 3\sqrt{2}
\]

### Step 3: Write the complementary solution
The general solution to the homogeneous equation is:

\[
a_n^{(c)} = C_1(3 + 3\sqrt{2})^n + C_2(3 - 3\sqrt{2})^n
\]

This is the complementary solution.

Do you want further details on finding the particular solution or have any questions?

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Here are 5 related questions to deepen understanding:

1. How do you find the particular solution of a non-homogeneous recurrence relation?
2. What is the difference between the complementary and particular solutions?
3. How do you apply the initial conditions to determine the constants \(C_1\) and \(C_2\)?
4. Can this method of solving recurrence relations be applied to higher-order recurrences?
5. What are some real-world applications of recurrence relations in mathematics?

**Tip:** Always check for repeated roots when solving characteristic equations. This will affect the form of the solution.

Find the complementary solution of an+2 – 6 an+1 – 9 an = 2^n

Learn how to find the complementary solution for the recurrence relation an+2 – 6an+1 – 9an = 2^n. This solution covers key mathematical concepts such as characteristic equations, solving quadratic equations, and recurrence relations. Suitable for advanced high school or undergraduate students.

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