Solve the recurrence relation, given: • a, = 4, • az = 5, %3D • a, = 5a,-1- 6a„-2 Remember: • The theorem is An = c1&n-1+ czan–2. -6+ v – 4ac The quadratic formula is 2a a, = 7(2)" – 3(3)" a a, = 7(2)" + 3(3)" а, 3 3(2)" — 7 (3)" d an = 7(3)" + 4(3)"
Solve the recurrence relation, given: • a, = 4, • az = 5, %3D • a, = 5a,-1- 6a„-2 Remember: • The theorem is An = c1&n-1+ czan–2. -6+ v – 4ac The quadratic formula is 2a a, = 7(2)" – 3(3)" a a, = 7(2)" + 3(3)" а, 3 3(2)" — 7 (3)" d an = 7(3)" + 4(3)"
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Solve the recurrence relation, given:
- \( a_0 = 4 \),
- \( a_1 = 5 \),
- \( a_n = 5a_{n-1} - 6a_{n-2} \).
Remember:
- The theorem is \( a_n = c_1 a_{n-1} + c_2 a_{n-2} \).
- The quadratic formula is:
\[
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Options:
- \( \text{a) } a_n = 7(2)^n - 3(3)^n \)
- \( \text{b) } a_n = 7(2)^n + 3(3)^n \)
- \( \text{c) } a_n = 3(2)^n - 7(3)^n \)
- \( \text{d) } a_n = 7(3)^n + 4(3)^n \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2cad0f5b-530d-4248-ba5a-31b2999fa76b%2F61ba3e0f-fa9b-422a-b443-3fab23858894%2Fcq03o7b_processed.png&w=3840&q=75)
Transcribed Image Text:Solve the recurrence relation, given:
- \( a_0 = 4 \),
- \( a_1 = 5 \),
- \( a_n = 5a_{n-1} - 6a_{n-2} \).
Remember:
- The theorem is \( a_n = c_1 a_{n-1} + c_2 a_{n-2} \).
- The quadratic formula is:
\[
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Options:
- \( \text{a) } a_n = 7(2)^n - 3(3)^n \)
- \( \text{b) } a_n = 7(2)^n + 3(3)^n \)
- \( \text{c) } a_n = 3(2)^n - 7(3)^n \)
- \( \text{d) } a_n = 7(3)^n + 4(3)^n \)
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