non homogeneous recurrence relations

Question

how to solve the particular equation of this problem Im not able to equate the polynomial formed I have taken $a_n(p) = A_0+A_1n+A_2 n^22^n$ is it right ?
$a_n-4 a_{n-1} + 4a_{n-2} = 3n + 2^n$ ?

Answer 1

So we have found the general solution $a_n=(B+Cn)2^n$ which satisfies $a_{n}-4a_{n-1}+4a_{n-2}=0$. We now need to deal with $3n+ \color{red}{0}$ and $2^n$ ... $a_n=A_0 +A_1 n $ will deal with the first term. The second term is slightly trickier ... we would try $2^n$ but we already know that this will give zero, similarly $n2^n$ will give zero, so try $ A_{2} n^2 2^n$. \begin{eqnarray*} a_{n}-4a_{n-1}+4a_{n-2} &=&A_0+A_1 n-4( A_0+A_1 (n-1)) +4(A_0+A_1 (n-2))\\&=A&_{1}n+(A_0-4A_1) =3n+ \color{red}{0} \\ A_1=3 \\ A_0-4 A_1=0 \end{eqnarray*} So $A_0=12$ ... now the $2^n$ term ... \begin{eqnarray*} a_{n}-4a_{n-1}+4a_{n-2} = A_{2} n^2 2^n-4A_{2} (n-1)^2 2^{n-1}+4A_{2} (n-2)n^2 2^{(n-2)}=2^2 A\\=A_2 2^{n} (n^2-2(n-1)^2+(n-2)^2)=A_2 2^{n} (n^2-2n^2+4n-2+n^2-4n+4) = 2 A_2 2^{n} =2^{n} \end{eqnarray*} So $A_2=\frac{1}{2}$.