What is the question?
The question was 'How do I solve this equation' basically.
I did however do some research on the web, and ask around a bit, and
I think I got at least a bit hang of it.
If any of the following is incorrect, let me know.
Equation (1):
The characteristic equation is:
Meaning that the homogenous part of the solution to
is
Now we need to find the particular solution to the equation (1).
Since the right side is a polynomial of degree q = 1, we will look for a
polynomial of the same degree.
Now we take this polynomial and put it
into equation (1). Giving us:
Now we have a equation system which can be solved:
This gives us
So we get:
And with
Hello, Twig!
Ugh! This is a messy one!
I'll show you a method that I was taught . . .
Crank out the first few terms . . .
We have been given: . . . . . . . .[1]
Write the "next" term: . .[2]
Subtract [1] from [2]: .
Let
. . We assume that the general term is an exponential.
Then we have: . . . .[3]
"Next" equaton: . .[4]
Subtract [3] from [4]: .
Divide by
This factors: .
. . and has roots: .
The general form is: .
Plug in the first three terms:
. .
Subtract [9] - [8]: .
Therefore: .
Is there a better way? . . . I hope so!
.
Hi!
Yes, this can get quite messy.
I read through your solution and then tried to compare it to mine.
I couldnīt really figure any advantages/disadvantages with a certain method though..
The method I was taught today (I made a post higher up this page) with a different solution. I think itīs correct, and it felt like less steps.
The similarities with differential equations becomes quite clear by doing this.
You are using a "characteristic equation", just as you solve equations like
Then you look for a particular solution, that satisifies just this equation.
And the general solution becomes:
Thanks for your help!