6. S(k) – S(k – 1) – 6S(k – 2) = –30, S(0) = 7, S(1) = 10

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Solve the following sets of recurrence relations and initial conditions:

**Problem 6:** Solve the recurrence relation given by:

\[ S(k) - S(k-1) - 6S(k-2) = -30 \]

with the initial conditions:

\[ S(0) = 7, \quad S(1) = 10 \]

In this problem, we are tasked with finding a sequence \( S(k) \) that satisfies the linear recurrence relation. Given initial values provide a starting point to determine subsequent terms in the sequence. 

The relation shows how each term in the sequence depends linearly on previous terms, specifically the two preceding ones, with a constant impact subtracted from each computed term. The task is to find a general expression for \( S(k) \) or to compute specific terms using these conditions.
Transcribed Image Text:**Problem 6:** Solve the recurrence relation given by: \[ S(k) - S(k-1) - 6S(k-2) = -30 \] with the initial conditions: \[ S(0) = 7, \quad S(1) = 10 \] In this problem, we are tasked with finding a sequence \( S(k) \) that satisfies the linear recurrence relation. Given initial values provide a starting point to determine subsequent terms in the sequence. The relation shows how each term in the sequence depends linearly on previous terms, specifically the two preceding ones, with a constant impact subtracted from each computed term. The task is to find a general expression for \( S(k) \) or to compute specific terms using these conditions.
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