Question

Solve the following first order difference equation:

Answer 1

\[2x_{n+1} + 3x_{n} = 0 \]

Answer 2

Are you expected to use certain methods, or just do whatever works?

Answer 3

I don't really understand how to solve these type of problems. I know that the solution is meant to be something along the lines of \[x_{n} = C(a^n)\]

Answer 4

You can algebraically rearrange it to be:

\[x_{n+1} = \left(\frac{-3}{2}\right) x_n\]
So if you think of it as saying "the next term in the sequence is a multiple of the previous term" it sorta makes sense that it would be of the form:

\[x_n = C*a^n\]
where C is some arbitrary constant and 'a' is some number. Each time you increase the exponent n by 1, it multiplies by a. C is like some base starting amount and I think you'll need more information to determine it.

Answer 5

ahh okay. I think i'm starting to kind of understand it. Thanks :)

Answer 6

There are other ways of doing this that are more like the strategy a computer could use, these would be like generating functions. There's really not a simple way of doing these in a way that "oh ok this strategy works for all of them all the time" kinda thing unfortunately. I wouldn't have been able to tell you the answer so quickly if I hadn't had someone else tell me this kinda stuff.

Answer 7

Wait so if a difference equation is in the form :
\[x_n - ax_{n-1} = 0 \]
where a is a constant, the solution will be in the form:
\[x_n = Ca^n\]

I think i saw somewhere.. that a is the "root of the auxiliary equation m-a = 0"
So in saying so,
Does that mean we write something like;
\[x_{n+1} \rightarrow m\]
\[x_n \rightarrow -1\]
\[2m - 3 = 0\]
\[m = \frac{ -3 }{ 2 }\]
so then \[a = \frac{ -3 }{ 2 }\]
and then we sub that into x_n = Ca^n
so the solution would be:
\[x_n = C(\frac{ -3 }{ 2 })^n\]
???

Answer 8

I am not sure about this 'auxiliary equation' terminology I'm just not familiar with it. However you can always check a solution:

I just like the forward difference better than backwards difference but it really makes no difference what you choose...

So if you want to solve \(x_{n+1} = a x_n\) you can check a guess:
\[x_n=Ca^n\] by plugging it in:
\[x_{n+1} = Ca^{n+1}\]
\[x_{n+1} = a x_n\]\[Ca^{n+1} = a Ca^n\]

Yeah it checks out. If you do the same thing with your specific choice of solution, you'll see that it indeed works. This method is probably a good thing to do after every problem you solve just in case. It's a simple check to plug in your solution.

Answer 9

okay cool. Thanks for your help! :)