I was wondering if you would point me to a book where the theory of second order homogeneous linear difference equation with variable coefficients is discussed. I am having difficulties in getting rigorous methods to solve some equations, see an example below.

In particular, given the recurrence relation

$X_{n+2} = \frac{3n-2}{n-1}X_{n+1} - \frac{2n}{n-1}X_n$,

two solutions are

$X(n)= n$ and $X(n) = 2^n$.

Is there an "elementary" way of arriving at these solutions? (i.e. without using transforms, etc.)

Thanks in advance.


HINT $\ $ Factor the difference operator. With the shift operator $\rm\ S\ X_n = X_{n+1}\ $ we have

$$\rm\ ((n-1)\ S^2 - (3\ n-2)\ S + 2\ n)\ \ X_n\ =\ ((n-1)\ S - n)\ (S - 2)\ \ X_n$$ Now put $\rm\ Y_n = (S - 2)\ X_n = X_{n+1} - 2\ X_n\:.\ $ Then the above second-order equation reduces to $\rm\ (n-1)\ Y_{n+1} - n\ Y_n = 0\:.\ $ Solve that for $\rm\:Y_n\:$ and then plug it into the prior equation to obtain a first-order nonhomogeneous equation for $\rm\: X_n\:.$

  • $\begingroup$ Thank you very much for your help. I can now see how it is solved. $\endgroup$ – Chulumba Jan 18 '11 at 17:18
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    $\begingroup$ Nice solution. Does anyone know whether there is a general solution technique for linear difference eqs with variable coefficients in the same way that there is a for linear difference equations with constant coefficients? The above technique, I imagine, will only work in particular instances? $\endgroup$ – Assad Ebrahim Jan 13 '14 at 10:37
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    $\begingroup$ @AKE If the coefficients are nonconstant then one is working in a noncommutative operator ring $\rm\,\Bbb C[n,S]\,$ since $\rm\,nS \ne Sn = (n\!+\!1)S.\,$ Since this is no longer isomorphic to a univariate polynomial ring, factorization becomes more complex. Nonetheless, factorizations do frequently exist in applications. $\endgroup$ – Bill Dubuque Jan 13 '14 at 14:39
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    $\begingroup$ @BillDubuque, could you please provide a reference book for this kind of solutions to this kind of problems? $\endgroup$ – richard May 21 '15 at 5:01
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    $\begingroup$ @Gone I know this is an old post, but yes, please! $\endgroup$ – Bcpicao Jun 23 '20 at 12:11

I'm not sure this is what you're looking for but do you know of H. Wilf's book "generatingfunctionology"?

He made it available online for free:


  • $\begingroup$ Thanks. I was aware of that book. As you know, it does not discuss what I am really after. $\endgroup$ – Chulumba Jan 18 '11 at 17:17

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