Question

Hi everyone! Can someone help me find a differential operator(annihilator) for x^2cos(4x)?
Thanks! :o)

Answer 1

I haven't run across an example of a x^2 multiplied by a cos4x before.

I know separately I would need D^3 for the x^2 and (D^2+16) for the cos4x but how do you do it when they are mashed together like this?

Answer 2

Pearlz o-o

Answer 3

Some times I imagine a round white pearly pearl when I read your name...

Answer 4

The annhilator for
$$ \Large x^{m-1} e^{ax}\cos(bx) ~is~ [(D-a)^2 +b^2]^m

$$

Answer 5

hi shad

Answer 6

Really? wow...I ruled that one out because I could find the "alpha" within x^cos4x...where is the alpha?

Answer 7

zero!

Answer 8

omg duh! I knew that! :o)

Answer 9

Quick question Perl...Do you think it's possible to ever come across a function one of the annihilators would work? Like some strange combo?

Answer 10

no...I was just curious

Answer 11

is that annihilator you wrote down the same as
[D^2-2alphaD + (alpha^2+beta^2)]^n ?

Answer 12

yes that is equivalent (note that i used m instead of n )

Answer 13

okay thanks...I just didn't recognize it...thanks so much! :o)

Answer 14

this is a nice list of annhilators

Answer 15

Hi @Perl, sorry to bother but I have a quick last question...
since x^2cos4x has multiplicity of 2 should I choose my annihilator to be:

[D^2-2zeroD+(zero^2+beta^2)]^3

Which would simplify to:
[D^2+(beta^2)]^3

Is that correct?

Answer 16

yes thats correct, the annhilator of the polynomial x^2 will be one degree higher

Answer 17

so should I expand that out all the way?

Answer 18

(D^2+16)(D^2+16)(D^2+16)(x^2cos4x)?

Answer 19

yes

Answer 20

wow! lol...okay! :o)

Answer 21

but usually we are given a differential equation to solve first

Answer 22

P(D)*y = f(x)

Answer 23

so you might be given a problem like
y '' + 3y ' + y = x^2 cos(x)

Answer 24

my equation is y"+5y'+6y=x^2cos4x

Answer 25

yeah I'm good with that procedure thankfully...just having trouble getting this annihilator for some reason...I did fine on all my other work except this one

Answer 26

look at this>>>

Answer 27

(D^6+480^4+768^2+4096)(x^2cos4x)
doesn't that seem really extreme?

Answer 28

$$
\Large
{
y' '+5y'+6y=x^2\cos(4x)
\\ \therefore \\
(D^2 + 5D + 6) y = x^2 cos(4x)
\\\\ \text{apply annhilator}
\\(D^2 + 16)^3(D^2 + 5D + 6) y =(D^2 + 16)^3 x^2 cos(4x)
}
$$

Answer 29

the right side goes to zero , since it is being annihilated

Answer 30

I thought you were supposed to expand it out all the way?

Answer 31

well we can show that it goes to zero, in steps

Answer 32

$$
\Large {
(D^2 + 16)(D^2 + 16)(D^2 + 16)(x^2 cos(4x))
\\ = (D^2 + 16)(D^2 + 16)[(D^2\cdot x^2 cos(4x) + 16\cdot x^2 \cos(4x) ]


}
$$

Answer 33

$$
\large {
(D^2 + 16)(D^2 + 16)(D^2 + 16)(x^2 cos(4x))
\\ = (D^2 + 16)(D^2 + 16)[(D^2\cdot x^2 cos(4x) + 16\cdot x^2 \cos(4x) ]


}
$$

Answer 34

okay...you just jogged my memory...be doing it in steps, we set ourselves up nicely on the left hand side so that we can factor down and find our zeroes

Answer 35

So when you have to find the "form" of the particular solution, that's where you set all the C1 and C2 etc to A,B,C etc right?

Answer 36

so yp=A+Bx for example

Answer 37

I will show you on my Maple software that it is true that
(D^2 + 16)^3 ( x^2 cos(4x)) = 0

Answer 38

you don't have to...it's okay

Answer 39

I am working it on my whiteboard right now

Answer 40

I don't wanna take up all your time

Answer 41

$$
\Large{ (D^2 + 16)^3 (x^2 cos(4x))
\\= (D^6+48D^4+768D^2+4096) (x^2 \cos(4x))
}
$$

Answer 42

that's what I have so far...differentiating now

Answer 43

its really beautiful on maple:

input : f := x^2*cos(4*x);

input: diff(f, x$6)+48*diff(f, x$4)+768*diff(f, x$2)+4096*f;
output 0

Answer 44

e-gadz...that looks garbled

Answer 45

diff(f, x$6) means the sixth derivative of f , where i defined f above

Answer 46

Don't get me wrong...I am always interested in seeing new things and discussing theory...I just don't wanna seem like a bother.

Answer 47

you really wouldn't do this by hand, it would take forever.
annihilator theory ensures you that the right side vanishes (becomes zero)

Answer 48

OMG! A giant BUG just flew across my ear and I felt it's wings and I screamed and I woke my dog! uhg

Answer 49

sorry...had to tell someone! :o)

Answer 50

:P

Answer 51

you convinced me...I will stop working this by hand now! :o)

Answer 52

i can give you a taste of its complexity

Answer 53

I just tasted it's complexity...tastes bad!

Answer 54

bleh :o/

Answer 55

$$
\large {
D^6 (x^2 cos(4x)) \\
= 7680\cos(4x)-12288x\sin(4x)-4096x^2\cos(4x)
\\~\\
D^4 (x^2 cos(4x))
\\= -192\cos(4x)+512x\sin(4x)+256x^2\cos(4x)
\\...
}
$$

Answer 56

ok so you want to work on the left side, that part is more interesting

Answer 57

quick question Perl...
So when you have to find the "form" of the particular solution, that's where you set all the C1 and C2 etc to A,B,C etc right?
so yp=A+Bx for example

Answer 58

correct

Answer 59

okay thanks!

Answer 60

$$
\Large
{
y' '+5y'+6y=x^2\cos(4x)
\\ \therefore \\
(D^2 + 5D + 6) y = x^2 cos(4x)
\\\\ \text{apply annhilator}
\\(D^2 + 16)^3(D^2 + 5D + 6) y =(D^2 + 16)^3 x^2 cos(4x)
\\\therefore
\\\\(D^2 + 16)^3(D^2 + 5D + 6) y =0
}
$$

Answer 61

but its nice you can verify it on a computer algebra system, that the right side 'annihilates'

Answer 62

by using the appropriate code or expression

Answer 63

wow...very neat!

Answer 64

this is also a good page on annihilator theory with a good example
http://en.wikipedia.org/wiki/Annihilator_method

Answer 65

omg this left hand side is crazy as well!

Answer 66

can you confirm the form of my answer?

Answer 67

I have Yp=Acos4x+Bxcos4x+Cx^2cos4x+Dsin4x+Exsin4x+Fx^2sin4x

Answer 68

$$
\Large
{
y' '+5y'+6y=x^2\cos(4x)
\\ \therefore \\
(D^2 + 5D + 6) y = x^2 cos(4x)
\\\\ \text{apply annhilator}
\\(D^2 + 16)^3(D^2 + 5D + 6) y =(D^2 + 16)^3 x^2 cos(4x)
\\\therefore
\\\\(D^2 + 16)^3(D^2 + 5D + 6) y =0
\\ roots = 4i, -4i, 4i, -4i, 4i, -4i, -2, -3
\\ y =
}

$$

Answer 69

no...they just wanted the form of Yp not Yc also

Answer 70

oh ok , i agree then

Answer 71

whew! that was quite a problem! yikes!
Thanks again for all the help! :o)
I will try and leave you in peace while I cram for the rest of my exam!
Thanks again