For the recurrence relation: an = 10an-1 – 25an-2; The general solution is: an C1 +c2 Note: The only constants in front of these functions should be 1. Given the initial conditions ao = 3 and a1 -3, solve for c and c2. %D The final solution is: an =

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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For the recurrence relation:

\[ a_n = 10a_{n-1} - 25a_{n-2}, \]

The general solution is:

\[ a_n = c_1 (\text{box})^{n} + c_2 (\text{box})^{n} \]

**Note:** The only constants in front of these functions should be 1.

Given the initial conditions \( a_0 = 3 \) and \( a_1 = -3 \), solve for \( c_1 \) and \( c_2 \).

The final solution is:

\[ a_n = (\text{box}) \]
Transcribed Image Text:For the recurrence relation: \[ a_n = 10a_{n-1} - 25a_{n-2}, \] The general solution is: \[ a_n = c_1 (\text{box})^{n} + c_2 (\text{box})^{n} \] **Note:** The only constants in front of these functions should be 1. Given the initial conditions \( a_0 = 3 \) and \( a_1 = -3 \), solve for \( c_1 \) and \( c_2 \). The final solution is: \[ a_n = (\text{box}) \]
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