The given recurrence relation is:

$a_r - 7a_{r-1} + 12a_{r-2} = r$

We can solve this non-homogeneous recurrence relation in two parts: the homogeneous solution and the particular solution.

Step 1: Solve the homogeneous part

The homogeneous recurrence relation is:

$a_r - 7a_{r-1} + 12a_{r-2} = 0$

This is a second-order linear recurrence with constant coefficients. To solve it, we first find the characteristic equation by assuming a solution of the form $a_r = \lambda^r$:

$\lambda^r - 7\lambda^{r-1} + 12\lambda^{r-2} = 0$

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

$\lambda^2 - 7\lambda + 12 = 0$

Now, solve this quadratic equation:

$\lambda^2 - 7\lambda + 12 = 0$

Factoring the quadratic:

$(\lambda - 3)(\lambda - 4) = 0$

Thus, the characteristic roots are $\lambda_1 = 3$ and $\lambda_2 = 4$.

The general solution to the homogeneous part is:

$a_r^{(h)} = C_1 \cdot 3^r + C_2 \cdot 4^r$

Step 2: Solve the non-homogeneous part

Now, we find a particular solution for the non-homogeneous equation:

$a_r - 7a_{r-1} + 12a_{r-2} = r$

We try a particular solution of the form $a_r^{(p)} = Ar + B$, where $A$ and $B$ are constants to be determined.

Substitute $a_r^{(p)} = Ar + B$ into the recurrence relation:

$(Ar + B) - 7(A(r-1) + B) + 12(A(r-2) + B) = r$

Simplify the left-hand side:

= Ar + B - 7A(r-1) - 7B + 12A(r-2) + 12B$$ $$= Ar + B - 7Ar + 7A - 7B + 12Ar - 24A + 12B$$ $$= (A - 7A + 12A)r + (B + 7A - 7B - 24A + 12B)$$ $$= 6Ar + (B - 7B + 12B + 7A - 24A)$$ $$= 6Ar + (6B - 17A)$$ Equate this to the right-hand side of the original equation, which is $$r$$: $$6Ar + (6B - 17A) = r$$ From this, we get two equations by comparing the coefficients of $$r$$ and the constant terms: 1. $$6A = 1$$ ⟹ $$A = \frac{1}{6}$$ 2. $$6B - 17A = 0$$ ⟹ $$6B - 17\left(\frac{1}{6}\right) = 0$$ Solve for $$B$$: $$6B - \frac{17}{6} = 0$$ $$6B = \frac{17}{6}$$ $$B = \frac{17}{36}$$ Thus, the particular solution is: $$a_r^{(p)} = \frac{1}{6}r + \frac{17}{36}$$ ### Step 3: General solution The general solution to the recurrence relation is the sum of the homogeneous and particular solutions: $$a_r = C_1 \cdot 3^r + C_2 \cdot 4^r + \frac{1}{6}r + \frac{17}{36}$$ This is the complete solution to the recurrence relation. Would you like more details or have any questions? --- ### Relative Questions: 1. How do you find the characteristic equation of a recurrence relation? 2. What are the differences between homogeneous and non-homogeneous recurrence relations? 3. How can you verify a particular solution to a recurrence relation? 4. What is the significance of characteristic roots in solving recurrence relations? 5. How would the solution change if the right-hand side was a quadratic function instead of linear? ### Tip: When solving non-homogeneous recurrence relations, always try a particular solution based on the form of the non-homogeneous term (e.g., constant, linear, quadratic, etc.).