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I am trying to solve the following:

$$a_n=5a_{n-1}-6a_{n-2}+2^n+3n$$

The general solution to the homogeneous equation is simple:

$$a_n=5a_{n-1}-6a_{n-2} \rightarrow \\ r^2-5r+6=0 \rightarrow \\r=3,2$$ giving $$a_n^{(h)}=C_13^n+C_22^n$$

Now for the particular solution it has been hinted that I find something in the form $$a_n^{(p)}=qn2^n+p_1n+p_2$$

But this has me pretty solidly stumped. Where can I move from here or how might I go about finding the particular solution for this?

Nick Chapman
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1 Answers1

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To see any choice for the variables $q, p_1, p_2$ makes the desired relation hold, substitute the given form into the value of $a_n, a_{n-1}, a_{n-2}$ in your recurrence equation.

Here, you must solve for $$\begin{align*}qn2^n+p_1n + p_2 & = 5(q(n-1)2^{n-1}+p_1(n-1) + p_2) \\ & - 6(q(n-2)2^{n-2}+p_1(n-2) + p_2) \\ & + 2^n + 3n\end{align*}$$

We note that every term is a multiple of $n2^n, 2^n, n, or 1$. First, the $n2^n$ terms have the same coefficient of $q$ on both sides. Next, we group the coefficients of $2^n$, and get $$0 = 5q( -1)\left(\frac{1}{2}\right) - 6q(-2)\left(\frac{1}{4}\right)+ 1$$ Then, we group the coefficients of $n$ to get $$ p_1 = 5p_1 - 6p_1 + 3 $$ Finally, looking at the constant terms we get $$ p_2 = 5(-p_1 + p_2) -6 (-2p_2 + p_2) $$ You will have a particular solution if you solve these three equations simultaneously.

Rolf Hoyer
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  • This will be a single equation with three unknowns. Where are the other two equations? – Nick Chapman Apr 22 '15 at 05:50
  • I don't want to be a pain, but if you aren't super busy could you demonstrate one of the groupings. I understand what you are saying in theory but when it comes down to the practice I seem to be missing a step. Specifically the $2^n$ grouping trips me up. – Nick Chapman Apr 22 '15 at 06:04
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    I have edited my answer to elaborate, hopefully this is enough – Rolf Hoyer Apr 22 '15 at 06:19