2nd order homogeneous difference equation

Question

$$a_{n+2} = 9a_{n+1} - 18a_n,\quad n\geq 0,\,\,a_0=1,\,\, a_1=3$$

I got to the point where i moved all to LHS which gives me $a_{n+2} - 9 a_{n+1} + 18 a_n$ (correct me if I'm wrong). I then calculated to the point of getting the roots 3 & 6. From there i don't really know how to get the ''$a_n$''. Could anyone point if i made any mistakes so far and help me get the an?

Thank you

Could someone edit the an+2 etc, i'm new and have no idea how to do those :( — Unknownstarz, Dec 05 '13 at 20:24
Put all math equations between two $, then write a_{n+2}, etc. — Carlos Eugenio Thompson Pinzón, Dec 05 '13 at 20:30
Well if the roots are 3 and 6, then an= C1(3)^n + c2(6)^n. So now what you have to do is solve this equation for n=0 and n=1 to see what are the values of C1 and C2.I think you can handle it. — randomname, Dec 05 '13 at 20:33
@user109218 I would get a simultaneous equation then calculate for A & B? — Unknownstarz, Dec 05 '13 at 20:54
Yes.That is right.Just write down some of the sequence and you will see ti. — randomname, Dec 05 '13 at 21:17

score 1 · Answer 1 · answered Dec 05 '13 at 20:34

1

You write the characteristic equation: $r^2=9r-18$ which give you $r_1=3,r_2=6$. The general term has the form $a\cdot 3^n+b\cdot 6^n$. To find $a,b$ you use the relations $a_1=3,a_0=1$ which will give you a $2 \times 2$ system of equations.

answered Dec 05 '13 at 20:34

Beni Bogosel

23,381

score 0 · Answer 2 · answered Apr 15 '14 at 16:23

A general method uses generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$. You recognize several sums: $$ \frac{A(z) - a_0 - a_1 z}{z^2} = 9 \frac{A(z) - a_0}{z} - 18 A(z) $$ Substituting initial values, solving for $A(z)$ and writing as partial fractions: $$ A(z) = \frac{1}{1 - 3 z} $$ so that $a_n = 3^n$. This checks out substituted in the recurrence.

2nd order homogeneous difference equation

2 Answers2