Getting rid of $2^n$ when solving $a_n=8a_{n-1}-20a_{n-2}+16a_{n-3}+2^n$ by characteristic roots

Question

$a_n=8a_{n-1}-20a_{n-2}+16a_{n-3}+2^n$

For $n\ge3$, With initial conditions $a_2=1$, $a_1=1$, and $a_0=1$

I'd like the find the particular solution with characteristic roots. However when generating the characteristic equation, I get up to $x^3=8x^2-20^x+16+....$, and then I'm not sure what to with the $2^n$.

I suppose I could assume a homogeneous solution eg. $a_n=c_0n^3+c_1n^2+c_2n+c_3$ and then plug and pray, but this would be a very long equation and it seems like a cack-handed approach.

Answer 1

Setting $b_n=\frac{a_n}{2^n}$ gives you

$$\frac{a_n}{2^n}=\frac{8a_{n-1}-20a_{n-2}+16a_{n-3}+2^n}{2^n}$$ $$\iff\frac{a_n}{2^n}=\frac{2\cdot 4a_{n-1}}{2^n}-\frac{2^2\cdot 5a_{n-2}}{2^n}+\frac{2^3\cdot 2a_{n-3}}{2^n}+\frac{2^n}{2^n}$$ $$\iff \frac{a_n}{2^n}=4\frac{a_{n-1}}{2^{n-1}}-5\frac{a_{n-2}}{2^{n-2}}+2\frac{a_{n-3}}{2^{n-3}}+1$$ $$\iff b_n=4b_{n-1}-5b_{n-2}+2b_{n-3}+1$$

Answer 2

Another way: divide everythign through $2^n$ and rewrite terms on RHS, e.g. $\frac{8 a_{n-1}}{2^n} = 4 \frac{a_{n-1}}{2^{n-1}}$ and so on, then set $b_n = \frac{a_n}{2^n}$ and subtract the expression for $b_{n-1}$ from it. Then use difference equations.