$$a_{n+4} = 5a_{n+2}-4a_{n}+n(3)^n$$
The characteristic roots are given as x = -2,-1,1,2. So I have the equation:
$$a_{n} = C_1(-2)^n+C_2(-1)^n+C_3+C_4(2)^n$$
Where I'm having trouble is when I try to solve for the $n(3)^n$ part. I get: $$b_n=\beta_1n3^n+\beta_23^n$$ Then I plug that into the original relation: $$\beta_1(n+4)(3)^{n+4}+\beta_2(3)^{n+4}=5[\beta_1(n+2)(3)^{n+2}+\beta_2(3)^{n+2}]-4[\beta_1n(3)^{n}+\beta_2(3)^{n}]+n(3)^n$$ Which simplifies to: $$40\beta_1n+234\beta_1+40\beta_2 = n$$ Now I don't know how to solve for the two betas, I only have this one equation, and I don't know what I can do to get rid of one of the variables to solve for the other. Is there some trick that lets me create a second equation so I can solve the system? I looked at some other questions, but they all use linear algebra, or differential equations to find the solution to these relations, and I have to use this approach. Any help would be awesome!
The equation is true for all $n$, so you can set $n$ (e.g. to $n=0, 1$, etc) and solve for the constants that way.
$40 \beta_1 \cdot 0 + 234 \beta_1 + 40 \beta_2 = 0$
$40 \beta_1 \cdot 1 + 234 \beta_1 + 40 \beta_2 = 1$
$\beta_1 = 1/40$, $\beta_2 = -117/800$