Thread: Difference Equations.. Again :D

Given that the following sequence, Q_{n} , satisfies a homogeneous second order difference equation with constant coefficients find Q_{n} :



I reckon that it would have to start with and do i then just put in the various numbers and solve them simultaneiously?

exactly, and after you get the constant coefficients you can solve the homogeneous equation to find Qn.

just a quick question, when solving them simultaneously, would i just have

18A + 11B + 7C = 0
4A + 3B + C = 0

you have to have 3 equations in order to find an unambiguous solution for the constants.

Ok, so ill add 7A +4B +3C to that, but how do i re-arrange for solving?

Originally Posted by brd_7

Ok, so ill add 7A +4B +3C to that, but how do i re-arrange for solving?

I'm sorry if I sound rude, but if you deal with a topic in "advanced" algebra such as difference equations you must at least know how to solve a system of linear equations. If you don't know or forgot, there are plenty of resources on the internet such as:

System of linear equations - Wikipedia, the free encyclopedia

Nah you wern't rude by any means, its just i get given work that hasnt been explained properly, i then look at the notes and the examples are really poor, and so out of frustration i forget how to do the simplest of things.

But, through simultaneous equations i keep getting -

-5A - 5B = 0
-5A - 5B = 0

Which gets me nowhere.. Or at least i think..

to get the first equation substitute n=3:






now n=4:




now n=5:

to get the first equation substitute n=3:






now n=4:




now n=5:





as you can see the third equation is the sum of the forst two, so the equations are linearly dependent, that's why you had a problem.

after solving the system we get that the general solution is:

A = -B = -C

let's choose A=1, B=C=-1

thus:



you can notice that we've got the fibonacci sequence.

Ok, i should be able to do the rest then. Thanks for the help