Question

Hi everyone! I have posted the answer to a 4th Order DE. I understand the solution perfectly up to m=+-(root3)i/2...could someone please explain why there are 4 roots? In this case, I thought that when using A+-Bi, A would be zero and B would be (root3)/2...therefore I thought the answer would be y=e^0x(c1cos(root3)x/2 + c2sin(root3)x/2) but that answer is wrong. I just don't see where "4" roots came from to make this a repeating complex root problem. Please help me understand. Thanks! :o)

Answer 1

Something else I should mention...
When comparing this complex root problem with another complex root problem of 3y"+2y'+y=0 where you eventually get r=-(1/3)+-(root2)i/3 and an answer of y=e^-1/3x(c1cos(root2)x/3 + c2sin(root2)x/3)..."A" was -1/3 and B was (root2)/3...I just cannot seem to figure out how my original question arises at 4 roots even after comparing it to this more simple question...Hope this made sense. :o)

Answer 2

the solutions to that fourth degree polynomial is
(1/2*I)*sqrt(3), -(1/2*I)*sqrt(3), (1/2*I)*sqrt(3), -(1/2*I)*sqrt(3)

note that the solution (1/2*I)*sqrt(3) has multiplicity 2
and you have to account for multiplicity when you solve the diff. equ

Answer 3

you have 4 solutions, with 2 solutions having multiplicity of 2 (repeated roots)

Answer 4

can you explain why the more simple example I gave doesn't have 4 roots?

Answer 5

this one :
3y"+2y'+y=0 ?
this is a second degree polynomial, so it has only 2 roots. and they are distinct roots

Answer 6

our original equation was a 4th degree equation, so it has 4 complex roots

Answer 7

An nth degree polynomial has n complex roots with possible repeated roots. That's a corollary of the fundamental theorem of algebra.

Answer 8

I'm not sure what it is but I am missing something basic here...having trouble seeing 4 roots even though you explained it as well as the answer I posted...grrr :o(

Answer 9

you want to show algebraically how you get 4 roots

Answer 10

okay...hmmm...let me think a moment

Answer 11

so are you saying if I had a 7th order DE, that I would have 7 roots?

Answer 12

ok i see where the multiple root went

Answer 13

7 complex roots with possible multiplicity. also note that with an odd degree there must be at least one real root (this is not generally true with even powers)

Answer 14

so algebraically I can show you were the two roots went

Answer 15

yes...please show me algebraically thank you!!!

Answer 16

I was at r^2=-3/4

Answer 17

$$

\Large
{ 16m^4 +24m^2 + 9 = 0\\
m^2 = t \\
16t^2 + 24t + 9 = 0\\
\\ \therefore \\
t = \frac{-24 \pm \sqrt{24^2-4(16)(9)}}{2(16)} \\
t = \frac{-24 \pm 0}{32} \\
\\ \therefore\\
t = \frac{-24 + 0}{32} , t = \frac{-24 - 0}{32} \\\\



}
$$

Answer 18

omg...i see it! a repeated root!

Answer 19

:)

Answer 20

should there technically be an "i" next to each zero just for book keeping sake?

Answer 21

you can label them t_1 and t_2

Answer 22

$$

\Large
{ 16m^4 +24m^2 + 9 = 0\\
m^2 = t \\
16t^2 + 24t + 9 = 0\\
\\ \therefore \\
t = \frac{-24 \pm \sqrt{24^2-4(16)(9)}}{2(16)} \\
t = \frac{-24 \pm 0}{32} \\
\\ \therefore\\
t = \frac{-24 + 0}{32} , t=\frac{-24 - 0}{32} \\\\

\\ \therefore \\
m^2 = \frac{-24 + 0}{32}, m^2 = \frac{-24 - 0}{32}
\\ \therefore \\
m = \pm \sqrt{ \frac{-24 + 0}{32} }, m = \pm \sqrt{ \frac{-24 - 0}{32} }
\\ \text { now you have 4 roots}
}
$$

Answer 23

okay yes...this problem is pretty tricky if you haven't done one like this before...yikes!
I will have to practice this quite a bit!
One last question...I heard my professor say the word "multiplicity" like only once during lecture and he said it like it should have been common knowledge...
1) what does multiplicity mean and
2)is that something I maybe forgot from algebra?

Answer 24

Root multiplicity refers to the number of times the root appears.
If it appears 2 times then it has multiplicity 2 (which means it is a repeating root).
If the root appears only once , then it has multiplicity 1

For example
f(x) = (x - 2)(x - 3) , this has roots x =2, x = 3 both roots have multiplicity one


f(x) = (x-7)^2 , this has the root x= 7 with multiplicity 2

Answer 25

I have never heard of multiplicity zero being used, but you could say that the root x = 8 has multiplicity zero in f(x) = (x-7)^2 , since its not a root of that equation.

Answer 26

since you said f(x) = (x-7)^2 , has the root x= 7 with multiplicity 2 ...
does this mean that 7 is a repeated root here also?

Answer 27

yes

Answer 28

Whenever you have multiplicity 2 or greater, you have a repeated root
The word multiplicity characterizes or qualifies the root, how many times does the root appear as a solution

Answer 29

that's great!
thanks soooo much for your help Perl! :o)

Answer 30

a multiplicity 1 root means that the root does not repeat

Answer 31

your welcome :)

Answer 32

multiplicity 1 root is a standalone root :P

Answer 33

how was I not already a fan of yours is beyond me! :o)
Thank you again!