2nd order homogeneous repeated roots

Question

$$a_{n+2}-6a_{n+1}+9a_{n}= 3^n \quad n\geq 0 \quad a_{0}=2 \quad a_{1}=3$$ Got repeated roots of 3, so $a_{n}= A\cdot3^n+b\cdot (n\cdot3^n)$ how would i calculate A+B when there is $3^n$?

Edit: So i calculated $a_{n}=6\cdot3^n - n\cdot3^n$ can anyone check if thats right?

Thanks!

score 0 · Answer 1 · answered Dec 05 '13 at 21:18

0

Your answer IS correct. To compute the constant, just try $n=0, 1$ (two equations for two unknowns).

answered Dec 05 '13 at 21:18

Igor Rivin

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so the 3^n doesnt change anything? – Unknownstarz Dec 05 '13 at 21:32
I am not sure what you mean. To find the constants, do $n=0, 1.$ – Igor Rivin Dec 05 '13 at 22:06

2nd order homogeneous repeated roots

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