How to solve this reccurence with g(n)=-9n*2^(n-1)

Question

$f(n+1)=9f(n)+5f(n-1)-9n\times2^{n-1}$

How to solve this? I have problem with $g(n)=-9n\times2^{n-1}$ I can start and calculate $\Delta$ and $x_1$ and $x_2$ but when I get $s_n=c_1\times 1^n+c_2\times5^n + g(n)$. I don't know what after to do.

Answer 1

First let us get rid of the non-homogeneous member $9n2^{n-1}$. For this look for a solution of a recurrence which has a form $f_1(n)=(an+b)2^{n-1}$. It is easy to see that we can pick $a=\tfrac 9{19}$ and $b=a^2$. Then a general solution of the recurrence is a sum of $f_1$ and a general solution of a recurrence $g(n+1)=9g(n)+5g(n-1)$. The latter is a linear recurrence with constant coefficients. Its characteristic polynomial $\lambda^2=9\lambda+5$ has distinct roots $\lambda_{1,2}=\tfrac 12\left(9\pm\sqrt{101}\right)$, so it has a general solution $C_1\lambda_1^n+ C_2\lambda_2^n$ for some constants $C_1$ and $C_2$.