Math Problem Statement

Solve the recurrence relation 𝑎௥ − 7𝑎௥ିଵ + 12𝑎௥ିଶ = 𝑟. 4 ௥ .

Solution

We are tasked with solving the recurrence relation:

ar7ar1+12ar2=ra_r - 7a_{r-1} + 12a_{r-2} = r

where r4r \geq 4.

Step 1: Solve the Homogeneous Equation

The first step is to solve the homogeneous part of the recurrence relation:

ar7ar1+12ar2=0a_r - 7a_{r-1} + 12a_{r-2} = 0

Assume a solution of the form ar=xra_r = x^r. Substituting into the homogeneous equation:

xr7xr1+12xr2=0x^r - 7x^{r-1} + 12x^{r-2} = 0

Dividing through by xr2x^{r-2} gives the characteristic equation:

x27x+12=0x^2 - 7x + 12 = 0

We solve this quadratic equation:

x=(7)±(7)24(1)(12)2(1)x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)} x=7±49482x = \frac{7 \pm \sqrt{49 - 48}}{2} x=7±12x = \frac{7 \pm 1}{2}

So, the roots are:

x=4andx=3x = 4 \quad \text{and} \quad x = 3

Thus, the general solution to the homogeneous equation is:

ar(h)=C14r+C23ra_r^{(h)} = C_1 4^r + C_2 3^r

Step 2: Solve the Non-Homogeneous Equation

Now, we solve the non-homogeneous equation:

ar7ar1+12ar2=ra_r - 7a_{r-1} + 12a_{r-2} = r

We need to find a particular solution, ar(p)a_r^{(p)}, to this equation. Since the right-hand side is linear in rr, we try a particular solution of the form:

ar(p)=Ar+Ba_r^{(p)} = Ar + B

Substitute this into the recurrence relation:

(Ar+B)7(A(r1)+B)+12(A(r2)+B)=r(Ar + B) - 7(A(r-1) + B) + 12(A(r-2) + B) = r

Simplifying both sides:

Ar+B7A(r1)7B+12A(r2)+12B=rAr + B - 7A(r - 1) - 7B + 12A(r - 2) + 12B = r Ar+B7Ar+7A7B+12Ar24A+12B=rAr + B - 7Ar + 7A - 7B + 12Ar - 24A + 12B = r (Ar7Ar+12Ar)+(B7B+12B)+(7A24A)=r(Ar - 7Ar + 12Ar) + (B - 7B + 12B) + (7A - 24A) = r 6Ar+(6B17A)=r6Ar + (6B - 17A) = r

Equating the coefficients of rr and the constant terms:

  1. 6A=16A = 1, which gives A=16A = \frac{1}{6}
  2. 6B17A=06B - 17A = 0, which gives 6B17×16=06B - 17 \times \frac{1}{6} = 0, or 6B=1766B = \frac{17}{6}, so B=1736B = \frac{17}{36}

Thus, the particular solution is:

ar(p)=16r+1736a_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:

ar=C14r+C23r+16r+1736a_r = C_1 4^r + C_2 3^r + \frac{1}{6}r + \frac{17}{36}

Conclusion

The solution to the recurrence relation is:

ar=C14r+C23r+16r+1736a_r = C_1 4^r + C_2 3^r + \frac{1}{6}r + \frac{17}{36}

To find the constants C1C_1 and C2C_2, you would need initial conditions (values for a0a_0, a1a_1, etc.).


Do you have any questions or would you like further details? Here are some questions that expand on the topic:

  1. How do we handle initial conditions in recurrence relations?
  2. Can we apply the same method for non-homogeneous terms of higher degree?
  3. What happens if the characteristic equation has repeated roots?
  4. How does this method apply to second-order linear recurrence relations in general?
  5. How would you solve a recurrence relation using matrix methods?

Tip: Always check the form of the non-homogeneous term before assuming a particular solution!

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Linear Homogeneous Equations
Characteristic Equation
Non-Homogeneous Solutions

Formulas

Characteristic equation x^2 - 7x + 12 = 0
General solution of a linear recurrence relation: a_r = C_1 4^r + C_2 3^r + particular solution

Theorems

Method of Solving Linear Recurrence Relations
Homogeneous and Non-Homogeneous Linear Recurrence Theorem

Suitable Grade Level

Grades 11-12 or early undergraduate level