Solve Recurrence Relation: a_{n+2}+3a_{n+1}+2a_n=3

[k3N70n]

[SOLVED] Recurrence Relation

Homework Statement

Solve the recurrence relation
$$a_{n+2}+3a_{n+1}+2a_n=3$$

Homework Equations

add the homogeneous solution to the particular solution

The Attempt at a Solution

Characteristic: $s^2+3s+2=0 \Rightarrow x=-2,-1$

Homogeneous: $a_n=A(-2)^n+Bn(-1)^n$

(*here is where I'm not sure if I've gone wrong. Should it be
$a_n=A(-1)^n+Bn(-2)^n$ instead? How would I know?*)
anyway, not sure I've right up to here but onwards->

Particular: $a_n=C3^n$

then $$C3^{n+2}+3C3^{n+1}+2C3^n=3^n$$

letting n=0 yeilds$$a_n=\frac{3^n}{20}$$ for the particular solution

then the solution is $$a_n=A(-2)^n+Bn(-1)^n+\frac{3^n}{20}$$

solving for A and B I got A=-1/20 and B=-3/4 and putting it all together seemed to give me nonsense.

Anyway, as you can probably see I'm not really understanding what's happening here. I would really appreciate nudge in the right direction.

Thank you kindly for you time
-kentt

 

[lippyka]

SOLUTION:Characteristic: s^2+3s+2=0 \Rightarrow x=-2,-1Homogeneous: a_n=A(-2)^n+Bn(-1)^nParticular: a_n=C3^nthen C3^{n+2}+3C3^{n+1}+2C3^n=3^nletting n=0 yeildsa_n=\frac{3^n}{20} for the particular solutionthen the solution is a_n=A(-2)^n+Bn(-1)^n+\frac{3^n}{20}solving for A and B I got A=-1/20 and B=-3/4 so the final solution is a_n=(-1/20)(-2)^n+(-3/4)n(-1)^n+\frac{3^n}{20}