Question

Hi everyone! Can anyone explain to me why x+3xe^(6x) has a multiplicity of 2? Thanks! :o)

Answer 1

Hmm, I'm not sure I know what you mean by multiplicity 2 in this context. Is this the solution of a set of differential equations?

Answer 2

Hi @Kainui...sorry, I was working on some problems...
The actual question asked to find the differential operator that annihilates the function: x+3xe^(6x)
I know that the annihilator for the "x" is D^2
and for the "3xe^(6x)" you need to use the annihilator form of (D-alpha)^n
I just don't know how to find the "n" which is what the book is calling it's multiplicity...I'm really confused! x-(

Answer 3

here is the answer, I just don't know how they know that n=2...I have no idea at all!

Answer 4

Oh ok I'll check it out, give me a minute.

Answer 5

Thanks! :o)

Answer 6

It's hard to believe but I can't find much on youtube about this and what I can find I can't understand their penmanship or accents.

Answer 7

Ahh ok, the multiplicity is just one more than the exponent on the x multiplying the exponential.

So a similar example where I've ONLY changed the multiplicity \[\Large (D-6)^5 3x^4e^{6x}\]

Answer 8

well what if the question didn't have an "x" and instead said:
find the differential operator that annihilates the function: 3xe^(6x)...how do you find "n"?

Answer 9

It should make sense that it's only 1 more than the exponent on x since when the exponent on x is 0 you still have a function to get rid of, so you have to have at least 1 there: \[\Large (D-6)3e^{6x} = 3*6e^{6x}-6*3e^{6x}=0\]

Does that help?

Answer 10

I'm sorry...I'm still confused...I would like to back-up for a sec and ask in a different way...
what is the multiplicity of 3xe^(6x)? is it n=2, which is 1 more than x^1?

Answer 11

Yes, exactly. If it was 3x^2e^(6x) since you have x^2 you have multiplicity n=3.

Answer 12

Is this a different type of multiplicity than like saying y'''+2y'=0 has multiplicity 3?

Answer 13

It seems weird or different somehow...I'm not sure but something about this is bothering me and I don't know why.

Answer 14

Yeah, the term multiplicity depends on the context in which we're talking about, it's just a fairly general word for meaning how many multiples of something we have. For example, a matrix with 3 eigenvalues and 2 of them are the same means that it has multiplicity 2. It doesn't really mean too much haha.

I guess one simple way to determine multiplicity is to simply just expand this equation:

\[\Large (D-6)3xe^{6x}\] Then, if it's 0 you know the multiplicity is 1 cause it only took you one of these (D-6) to get rid of it. If it doesn't go away, multiply what you get by another (D-6) operator. If you get 0 then you know it's multiplicity 2. Keep going until you get 0, the number of times you use (D-6) is the multiplicity. =D

Answer 15

well what if your function is for example: e^(-x)sinx-e^(2x)cosx
how do you figure out n=?

Answer 16

my best guess would be to say that you don't look at the "x" up in the exponent because it is part of an exponential function so next I would look at sinx and say well since it isn't squared, it must be to the 1st power so then perhaps n=2?

Answer 17

is this the correct way to reason through these or am I completely bonkers? :o/

Answer 18

Well I don't think you have the proper rules there, you have to find some new differential operator to get rid of sines and cosines since they'll behave differently.

Answer 19

I got it now!
Thanks @Kainui !