solve the difference equation: a (down r)+5a (down r-1)+6a (down r-2)=3r^2-2r+1

The given difference equation is:

\[
a_r + 5a_{r-1} + 6a_{r-2} = 3r^2 - 2r + 1
\]

### Step 1: Solve the corresponding homogeneous equation
The homogeneous part of the equation is:

\[
a_r + 5a_{r-1} + 6a_{r-2} = 0
\]

This is a linear recurrence relation. To solve it, we assume a solution of the form \( a_r = \lambda^r \). Substituting into the homogeneous equation:

\[
\lambda^r + 5\lambda^{r-1} + 6\lambda^{r-2} = 0
\]

Dividing by \( \lambda^{r-2} \) (assuming \( \lambda \neq 0 \)) gives:

\[
\lambda^2 + 5\lambda + 6 = 0
\]

Factoring the quadratic equation:

\[
(\lambda + 2)(\lambda + 3) = 0
\]

Thus, \( \lambda = -2 \) and \( \lambda = -3 \). Therefore, the general solution to the homogeneous equation is:

\[
a_r^{(h)} = C_1(-2)^r + C_2(-3)^r
\]

### Step 2: Find a particular solution
Next, we find a particular solution to the non-homogeneous equation. The non-homogeneous term is \( 3r^2 - 2r + 1 \), which is a quadratic polynomial.

We assume a particular solution of the form:

\[
a_r^{(p)} = Ar^2 + Br + C
\]

Substitute this into the original equation:

\[
(Ar^2 + Br + C) + 5(A(r-1)^2 + B(r-1) + C) + 6(A(r-2)^2 + B(r-2) + C) = 3r^2 - 2r + 1
\]

We now expand and simplify each term:

- \( A(r-1)^2 = A(r^2 - 2r + 1) = Ar^2 - 2Ar + A \)
- \( B(r-1) = Br - B \)
- \( A(r-2)^2 = A(r^2 - 4r + 4) = Ar^2 - 4Ar + 4A \)
- \( B(r-2) = Br - 2B \)

Substituting into the equation:

\[
Ar^2 + Br + C + 5(Ar^2 - 2Ar + A + Br - B + C) + 6(Ar^2 - 4Ar + 4A + Br - 2B + C) = 3r^2 - 2r + 1
\]

Now expand and collect terms:

\[
(Ar^2 + Br + C) + 5(Ar^2 - 2Ar + Br + A - B + C) + 6(Ar^2 - 4Ar + Br + 4A - 2B + C)
\]
\[
= (A + 5A + 6A)r^2 + (B - 2A + 5B - 10A + 6B - 24A)r + (C + 5C + 6C + 5A - 5B + 24A - 12B) = 3r^2 - 2r + 1
\]

Simplifying the coefficients:

- Coefficient of \( r^2 \): \( 12A = 3 \) → \( A = \frac{1}{4} \)
- Coefficient of \( r \): \( -12A + 12B = -2 \) → \( -3 + 12B = -2 \) → \( 12B = 1 \) → \( B = \frac{1}{12} \)
- Constant term: \( C + 11C + 29A - 17B = 1 \)

We substitute \( A = \frac{1}{4} \) and \( B = \frac{1}{12} \) into the constant term equation:

\[
12C + 29 \times \frac{1}{4} - 17 \times \frac{1}{12} = 1
\]
\[
12C + 7.25 - 1.42 = 1
\]
\[
12C + 5.83 = 1 \quad \Rightarrow \quad 12C = -4.83 \quad \Rightarrow \quad C = -\frac{4.83}{12} \approx -0.4025
\]

Thus, the particular solution is:

\[
a_r^{(p)} = \frac{1}{4}r^2 + \frac{1}{12}r - 0.4025
\]

### Step 3: General solution
The general solution to the difference equation is the sum of the homogeneous and particular solutions:

\[
a_r = C_1(-2)^r + C_2(-3)^r + \frac{1}{4}r^2 + \frac{1}{12}r - 0.4025
\]

This is the general form of the solution.

Do you have any questions or would you like more details? Here are some related questions to expand on:

1. How do you solve higher-order difference equations?
2. What is the difference between homogeneous and non-homogeneous solutions?
3. Can this method be applied to first-order difference equations?
4. How are the particular solutions influenced by the non-homogeneous term?
5. Can you explain the role of characteristic equations in solving recurrence relations?

**Tip:** Always verify your particular solution by substituting it back into the original equation to ensure it satisfies the non-homogeneous part.

solve the difference equation: a_r + 5a_{r-1} + 6a_{r-2} = 3r^2 - 2r + 1

Learn how to solve the difference equation a_r + 5a_{r-1} + 6a_{r-2} = 3r^2 - 2r + 1. This step-by-step solution covers key mathematical concepts such as homogeneous and non-homogeneous solutions, recurrence relations, and polynomial particular solutions. Suitable for university-level students or advanced high school learners.

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