To solve the recurrence relation:

$a_r = -2a_{r-1} - a_{r-2} + 5,$

we can break it into two parts: solving the homogeneous part and finding a particular solution.

1. Homogeneous Solution

Consider the homogeneous part of the recurrence relation:

$a_r = -2a_{r-1} - a_{r-2}.$

The characteristic equation associated with this is:

$x^2 + 2x + 1 = 0.$

This factors as:

$(x + 1)^2 = 0.$

Thus, the solution to the homogeneous equation is:

$a_r^{(h)} = (C_1 + C_2r)(-1)^r,$ where $C_1$ and $C_2$ are constants to be determined by initial conditions.

2. Particular Solution

Now, let's find a particular solution to the non-homogeneous recurrence:

$a_r = -2a_{r-1} - a_{r-2} + 5.$

Since the non-homogeneous part is a constant (5), we can try a particular solution of the form $a_r^{(p)} = A$, where $A$ is a constant. Substituting this into the original recurrence gives:

$A = -2A - A + 5.$

Simplifying this:

$A = -3A + 5 \implies 4A = 5 \implies A = \frac{5}{4}.$

Thus, the particular solution is $a_r^{(p)} = \frac{5}{4}$.

3. General Solution

The general solution to the recurrence relation is the sum of the homogeneous and particular solutions:

$a_r = (C_1 + C_2r)(-1)^r + \frac{5}{4}.$

4. Initial Conditions

If initial conditions are provided, we can determine $C_1$ and $C_2$. Without them, this is the most general form of the solution.

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Do you want to proceed with applying specific initial conditions or ask for any clarifications?

Here are 5 related questions to explore further:

1. How do you solve recurrence relations with complex roots in the characteristic equation?
2. What method is used to solve recurrence relations with non-constant coefficients?
3. How would the solution change if the non-homogeneous term were a function of $r$ instead of a constant?
4. How do recurrence relations connect to difference equations in discrete mathematics?
5. Can you explain the role of generating functions in solving recurrence relations?

Tip: Recurrence relations are closely related to linear difference equations. Understanding the characteristic equation is key to solving them efficiently.