Recurrence relations discrete math problem

[charmedbeauty]

Homework Statement

Find the general solution to the following recurrence relations (defined n≥2).

c) a_n=6a_n-1-9a_n-2+8n+4

Homework Equations

The Attempt at a Solution

a_n=6a_n-1-9a_n-2+8n+4

8n+4= a_n -6a_n-1+9a_n-2

R²-6R+9=0

R=3,3

So h_n=A(3)ⁿ+B(3)ⁿ

Assume p_n=Cn+Cn² → This is where I got stuck!

What should my assumption be??

I thought this because RHS =8n+4

so for the 4 I used Cn since 4 is a constant so I put an extra n term...likewise for 8n.

but it turned out to be a mess!

HELP

 

[tiny-tim]

hi charmedbeauty!

R²-6R+9=0

R=3,3

So h_n=A(3)ⁿ+B(3)ⁿ

no, if a recurrence relation has a root R repeated k times, the k independent solutions are nⁱRⁿ (0≤i<k)

 

[charmedbeauty]

hi charmedbeauty! no, if a recurrence relation has a root R repeated k times, the k independent solutions are nⁱRⁿ (0≤i<k)

hmm I am a little confused what should i be?

so i replace hn=A(3)ⁿ+B(3)ⁿ

with hn=A(3ⁿ)+Bn(3ⁿ)??

 

[tiny-tim]

hmm I am a little confused what should i be?

so i replace hn=A(3)ⁿ+B(3)ⁿ

with hn=A(3ⁿ)+Bn(3ⁿ)??

yup!

(and you can leave out those brackets … A3ⁿ + Bn3ⁿ )

 

[charmedbeauty]

yup!

(and you can leave out those brackets … A3ⁿ + Bn3ⁿ )

Thanks.