### Solving a Recurrence Relation**Problem Statement:**Solve the recurrence relation $a_{n+1} = 7a_n - 10a_{n-1}$, for $ n \geq 2 $, given the initial conditions $a_1 = 10$ and $a_2 = 29$.**Solution Approach:**To solve this recurrence relation, we can use characteristic equations:1. **Form the Characteristic Equation:** \[ r^2 = 7r - 10 \] Rearrange to get: \[ r^2 - 7r + 10 = 0 \]2. **Solve the Characteristic Equation:** Factorize the quadratic equation: \[ (r - 5)(r - 2) = 0 \] Thus, the roots are: \[ r_1 = 5, \quad r_2 = 2 \]3. **General Solution:** Using the roots, the general solution for the recurrence relation is: \[ a_n = A \cdot 5^n + B \cdot 2^n \]4. **Find Constants $A$ and $B$:** Use the initial conditions to form a system of equations: \[ \begin{cases} 10 = A \cdot 5^1 + B \cdot 2^1 \\ 29 = A \cdot 5^2 + B \cdot 2^2 \\ \end{cases} \] Simplify them: \[ \begin{cases} 10 = 5A + 2B \\ 29 = 25A + 4B \\ \end{cases} \]5. **Solve the System of Equations:** Multiply the first equation by 2: \[ 20 = 10A + 4B \] Subtract this from the second equation to eliminate $B$: \[ 29 - 20 = 25A - 10A \] This simplifies to: \[ 9 = 15A \quad \Rightarrow \quad A = \

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Math

Advanced Math

Solve the recurrence relation an+1 = 7an – 1Oan-1,n > 2, given a1 = 10, a2 = 29.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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### Solving a Recurrence Relation

**Problem Statement:**
Solve the recurrence relation $a_{n+1} = 7a_n - 10a_{n-1}$, for $ n \geq 2 $, given the initial conditions $a_1 = 10$ and $a_2 = 29$.

**Solution Approach:**
To solve this recurrence relation, we can use characteristic equations:

1. **Form the Characteristic Equation:**
\[
r^2 = 7r - 10
\]
Rearrange to get:
\[
r^2 - 7r + 10 = 0
\]

2. **Solve the Characteristic Equation:**
Factorize the quadratic equation:
\[
(r - 5)(r - 2) = 0
\]
Thus, the roots are:
\[
r_1 = 5, \quad r_2 = 2
\]

3. **General Solution:**
Using the roots, the general solution for the recurrence relation is:
\[
a_n = A \cdot 5^n + B \cdot 2^n
\]

4. **Find Constants $A$ and $B$:**
Use the initial conditions to form a system of equations:
\[
\begin{cases}
10 = A \cdot 5^1 + B \cdot 2^1 \\
29 = A \cdot 5^2 + B \cdot 2^2 \\
\end{cases}
\]
Simplify them:
\[
\begin{cases}
10 = 5A + 2B \\
29 = 25A + 4B \\
\end{cases}
\]

5. **Solve the System of Equations:**
Multiply the first equation by 2:
\[
20 = 10A + 4B
\]
Subtract this from the second equation to eliminate $B$:
\[
29 - 20 = 25A - 10A
\]
This simplifies to:
\[
9 = 15A \quad \Rightarrow \quad A = \

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