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the equation $a_n = 4a_{n-1}-4a_{n-2}+2^n$ have the homogeneous part of $a_n=A_1 n2^n + A_2 2^n$

but i dont know how to solve the particular part.

my method is

As $\beta(n)=2^n$

by the guessing method,

guessing the particular solution of the relation is $a_n=A2^n$

so, $A2^n-4A2^{n-1}+4A2^{n-2}=2^n$

and clearly, the the equation wouldn't hold.

how can I solve this problem?

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    Also see https://math.stackexchange.com/questions/535946/solve-the-following-non-homogeneous-recurrence-relation2017-09-28

2 Answers 2

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HINT: You already have a $2^n$ term in the homogeneous solution, so adding another $2^n$ term won’t give you anything new. Similarly, there’s not point adding another $n2^n$ term. What about $a_n=An^22^n$?

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    A=1/8, I did it. thanks2012-12-19
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What you call the guessing method is not to try solutions of the homogenous equation as solutions of the inhomogenous equation (they never are, by definition) but to keep the same geometric exponent and try a higher exponent in the polynomial. Here $a_n=2^n$ and $a_n=n2^n$ solve $a_n=4a_{n-1}-4a_{n-2}$ hence you ought to identify $(*)$ in $a_n=4a_{n-1}-4a_{n-2}+(*)$ when $a_n=n^22^n$ and to see if this helps you solve the inhomogenous case you are interested in.