how to solve a First Order inhomogenous difference/recurrence equation with a constant on the right side

Question

Suppose we have the difference equation

$aq_n +bq_{n-1} = c$,

where $a,b, c$ are constants, and we want to find the function $q_n$.

My book says, first, we find the solution to the homogeneous equation

$aq_n +bq_{n-1} = 0$,

which is $q_n = q_0(\frac{-b}{a})^n$

Then we solve the inhomogenous equation by guessing $q_n = d $ to be a constant, so we get

$ad +bd = c$, $d=\frac{c}{a+b}$

Putting together these two solutions we get the general solution:

$q_n = C(\frac{-b}{a})^n +\frac{c}{a+b}$

My question is could this have been done another way (to be specified in a moment), and if yes, how are these two solutions equivalent?

Could I solve it by writing

$q_n = \frac{c-bq_{n-1}}{a} = \frac{c}{a} - \frac{b}{a}q_{n-1}$

let $\alpha = \frac{c}{a}$ and $\beta = \frac{b}{a}$

so,

$q_n = \alpha - \beta q_{n-1}$

but, recursively, we have

$q_n = \alpha - \beta *(\alpha - \beta q_{n-2}) = \alpha - \alpha \beta + \beta^2 q_{n-2}$

again, we get

$q_n = \alpha - \alpha \beta +\alpha \beta^2 - \beta ^3 q_{n-3}$

and if we keep going,

$q_n= \alpha (1 - \beta + \beta^2 -\beta^3 +\beta^4 - ... + (-1)^{n-1}\beta^{n-1}) + (-1)^{n} \beta^nq_0$

Is this a correct solution? and more importantly if yes, how is it equivalent to the first one, and if it's not correct, where did I go wrong?

Answer 1

The alternate solution is correct and equivalent to the book solution.

$q_n = \alpha(1-\beta +\beta^2 -\beta^3 +\beta^4 - ...+(-1)^{n-1}\beta^{n-1}) + (-\beta)^{n}q_0$

The sum in the parentheses is a finite geometric sum which becomes

$S = \alpha(\frac{1-(-\beta)^{n}}{1-(-\beta)}) = \frac{c}{a}(\frac{1-(\frac{-b}{a})^n}{1+\frac{b}{a}}) = \frac{c}{a+b} - \frac{c}{a+b}(\frac{-b}{a})^n$

Thus, we have

$q_n = \frac{c}{a+b} -\frac{c}{a+b}(\frac{-b}{a})^n+(\frac{-b}{a})^nq_0=\frac{c}{a+b}+(\frac{-b}{a})^n(q_0-\frac{c}{a+b})$

This is equivalent to the book solution letting $C=(q_0-\frac{c}{a+b})$

Answer 2

I do not know if this is the kind of answer you are waiting for.

Let us make the problem a bit more general considering $$aq_n +bq_{n-1} = c+dn$$

To get rid of the problem of constants $c$ and $d$, start defining $$q_n=u_n+\alpha+\beta n$$ and replace in the original equation. This hould lead to $$a u_n+b u_{n-1}=(-a \alpha -\alpha b+b \beta +c)+n (-a \beta -b \beta +d)$$ Cancelling the extra terms in the rhs leads to $$-a \alpha -\alpha b+b \beta +c =0$$ $$-a \beta -b \beta +d=0$$ leading to $$\alpha=\frac{a c+b c+b d}{(a+b)^2}\qquad , \qquad \beta=\frac{d}{a+b}$$ and you are just left with the recurrence equation $$a u_n+b u_{n-1}=0$$ Introducing, just as you did $\beta = \frac{b}{a}$ then you just need to focus on $$u_n+\beta u_{n-1}=0$$ the solution of which being $$u_n=c_1 (-\beta )^{n-1}$$ and then $$q_n=c_1 (-\beta )^{n-1}+\frac{a c+b c+b d}{(a+b)^2}+\frac{d}{a+b}n$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ A 'particular solution' $\ds{\quad p_{s}\quad }$ is given by $\ds{\quad a\,p_{s} + b\,p_{s} = c \quad\imp\quad p_{s} = {c \over a + b}.\quad}$ Then, $\ds{q_{n} - p_{s}}$ satisfies the homogeneous $\ds{\quad a\pars{q_{n} - p_{s}} + b\pars{q_{n - 1} - p_{s}} = 0}$ \begin{align} \mbox{and}\quad q_{n} - p_{s} & = -\,{b \over a}\,\pars{q_{n - 1} - p_{s}} = \pars{b \over a}^{2}\,\pars{q_{n - 2} - p_{s}} = -\pars{b \over a}^{3}\,\pars{q_{n - 3} - p_{s}} \\[5mm] & = \cdots = \pars{-1}^{n}\pars{b \over a}^{n}\,\pars{q_{0} - p_{s}} \end{align}

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If $\ds{\,\,\,\verts{b \over a} < 1}$, $\ds{\lim_{n \to \infty}q_{n} = p_{s}}$. In any case, $\ds{q_{n}}$ is given by $$ q_{n} = {c \over a + b} + \pars{-1}^{n}\,\pars{b \over a}^{n}\,\pars{q_{0} - {c \over a + b}} $$