For the recurrence relation: a n = −3an−1 + 4an−2 The general solution is: an=c1( ) + c2( ) Note: The only constants in front of these functions should be 1. Given the initial conditions a0=−1 and a1= −4, solve for c1 and c2. The final solution is: an=
For the recurrence relation: a n = −3an−1 + 4an−2 The general solution is: an=c1( ) + c2( ) Note: The only constants in front of these functions should be 1. Given the initial conditions a0=−1 and a1= −4, solve for c1 and c2. 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
Problem 1RQ
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For the recurrence relation:
a n = −3an−1 + 4an−2
The general solution is:
an=c1( ) + c2( )
Note: The only constants in front of these functions should be 1.
Given the initial conditions a0=−1 and a1= −4, solve for c1 and c2. The final solution is:
an=
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