I am trying to determine the sequence of $a_n, n=0,1,2,...$ that satisfies the difference equation

$$a_{n+2}-a_{n+1}-2a_n=12 \cdot 2^n - 4n.$$

I have found the roots to the homogeneous equation, and they are $r=2$ and $r=-1$, so the general solution should be $a_n^{(h)} = A\left(2\right)^n + B\left(-1\right)^n,$ but I haven't managed to find a particular solution that works.

This is usually where I struggle when working with these problems. Often I end up finding it by basically guessing but this time I haven't been able to.

  • $\begingroup$ Have you tried the approach $a^{in}_{n}= a*n*2^{n}+ b*n +c$ with $a,b,c\in\mathbb{R}$ still to be determined? $\endgroup$ – Alex Aug 22 '16 at 19:51
  • $\begingroup$ Hm, I actually think I haven't. I have tried $a*2^n+bn+c$ but not with that extra n. Will try it now and get back to you. $\endgroup$ – a.per Aug 22 '16 at 19:55
  • $\begingroup$ Your first approach could not work, since $a*2^{n}$ is as you pointed out a homogeneous solution. So it can never catch the right hand side term $12*2^n$...In ODE theory this is called Resonance case.... $\endgroup$ – Alex Aug 22 '16 at 19:57
  • $\begingroup$ Ah, didn't think of that. Your approached worked out perfectly! $\endgroup$ – a.per Aug 22 '16 at 20:09

Since $2$ is a root of multiplicity $1$, the $12\cdot2^n$ term wants a particular solution of the form $cn^12^n$. (If $2$ were not a root, it would want $c2^n$.) $1$ is not a root, so the linear $-4n$ term just wants a particular solution of the form $d+en$. Putting the two together, we’re looking for a particular solution of the form


Substitute into the original recurrence, and you will be able to solve for $c,d$, and $e$.


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