A second order problem on recurrence relation equals 3^n

Question

I had this Recurrence Relation problem:

$a_{n+2} + a_{n+1} - 12a_n = 0$

And I solved in a form like this $a_n = A(r_1)^n + B(r_2)^n$

$r^{n+2} + r^{n+1} - 12r^n = 0$ -> Divided the hole eqation by the lowest $r$

Result of this:

$r^2+r -12 = 0$

And then i found the roots: $r_1 = 3$ and $r_2=-4$

And the solution of this problem is: $a_n = 2/7*(3)^n -7/2*(-4)^n $

The next one, is what i don´t get. It is exact like this one in stead it is equal to $3^n$, like is: $a_{n+2} + a_{n+1} - 12a_n = 3^n$

How do I solve this one ?

Answer 1

You didn't mentioned the initial conditions of the recurrence relation, but by the solution that you obtained $$a_n = \frac{2}{7}(3)^n − \frac{7}{2}(−4)^n$$ one can infer the initial conditions $$a_0 = -\frac{45}{14} \;\;\;\;\;\;\;\; a_1 = \frac{104}{7}$$ That made me think that maybe there is a typo and you intended $$a_n = \frac{2}{7}(3)^n − \frac{2}{7}(−4)^n$$ so $$a_0 = 0 \;\;\;\;\;\;\;\; a_1 = 2$$ I'll solve both cases just in case.

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What you are trying to do is to solve a non-homogeneous linear recurrence relation. $$ a_{n+2} + a_{n+1} - 12a_{n} = 3^n = f(n)$$ The first step is to solve the the associated homogeneous linear recurrence: $$ a_{n+2} + a_{n+1} - 12a_{n} = 0 $$ The associated characteristic polynomial is $$r^2 + r - 12$$ with roots $r_1 = 3$ and $r_2 = -4$.

Hence the homogeneous solution is of the form $$h_n = A 3^n + B(-4)^n$$

Instead to plug-in the initial conditions, we now need to find a particular solution to the non-homogeneous equation.

As $f(n) = 3^n$, we would guess a solution of the form $g_n = C 3^n$, where $C$ is a constant. But $3$ is a root of characteristic polynomial of the homogeneous part, hence the correct guess is $g_n = C n 3^n$. We now find $C$ by plugging it on the recurrence:

$$ C(n+2)3^{n+2} + C(n+1)3^{n+1} - 12Cn3^n = 3^n$$ $$ 9C(n+2) + 3C(n+1) - 12Cn = 1 $$

have to hold for all $n$, in particular for $n = 0$, hence:

$$ 18 C + 3 C = 1 \Rightarrow C = \frac{1}{21} $$

Hence we have the particular solution $g_n = \frac{1}{21} n 3^n$.

Therefore, the final solution is of the form: $$ a_n = h_n + g_n = \left( A + \frac{n}{21} \right) 3^n + B(-4)^n$$ where the constants $A$ and $B$ are to be determined from the initial conditions through the linear system:

$$\begin{cases} a_0 = A + B \\ a_1 = 3A - 4B + \frac{1}{7} \end{cases}$$

Using $a_0 = -\frac{45}{14}$ and $a_1 = \frac{104}{7}$ we have: $A = \frac{13}{49}$ and $B = - \frac{341}{91}$ hence the solution:

$$ a_n = \left( \frac{13}{49} + \frac{n}{21} \right) 3^n - \frac{341}{91}(-4)^n $$

Using $a_0 = 0$ and $a_1 = 2$ we have: $A = \frac{13}{49}$ and $B = - \frac{13}{49}$ hence the solution:

$$ a_n = \left( \frac{13}{49} + \frac{n}{21} \right) 3^n - \frac{13}{49}(-4)^n $$

If you had another set of initial values in mind, you can just plug in and find $A$ and $B$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{a_{n + 2} + a_{n + 1} - 12a_{n} = 3^{n}:\ {\large ?}}$.

First, write $\ds{a_{n} = \pars{b_{n} + \mu n}3^{n}}$ such that \begin{align} 3^{n}&=\pars{b_{n + 2} + \mu n + 2\mu}3^{n + 2} +\pars{b_{n + 1} + \mu n + \mu}3^{n + 1} - 12\pars{b_{n} + \mu n}3^{n} \\[5mm]&=3^{n}\pars{9b^{n + 2} + 3b_{n + 1} - 12b_{n}} + 21\mu 3^{n} \end{align}

With $\ds{\quad\mu \equiv {1 \over 21}\quad}$ we'll get $\ds{\quad9b^{n + 2} + 3b_{n + 1} - 12b_{n} = 0\quad}$ and you can use your first procedure.

Once you find $\ds{b_{n}}$, as a function of $\ds{n}$, you'll recover $\ds{a_{n}}$ as: $$ a_{n} = \pars{b_{n} + {n \over 21}}3^{n} $$