Question

Please check, Find annihilator of
\[F(x)= e^{-3x}(2sinx +7cosx)\]

Answer 1

since sinx, and cosx are linearly independent, I break them out and got the annihilator for each of them is A(D) = (D^2+6D +10)
so, the final answer is \(A_total (D) = (D^2 +6D +10)^2\) but not sure about ^2
Please help

Answer 2

what subject is this for?

Answer 3

differential equation

Answer 4

@jdoe0001

Answer 5

no one help me???? why??

Answer 6

@hartnn @phi @terenzreignz

Answer 7

@hba @austinL

Answer 8

I don't even know what a bloody annihilator is... googling... brb :>

Answer 9

^ I have the same issue.... what the deuce is an annihilator?

Answer 10

do you want to see my book? it's differential equation, I sssswwweaar

Answer 11

Have I ever been known to doubt you L66? :>
...of course I don't :D

Answer 12

@austinL do you know linear operator in dff?

Answer 13

It sounds vaguely familiar....

Answer 14

ok, let me ask my prof, then. thanks anyway.

Answer 15

I will ask my professor in just a moment... I will see what he has to say.

Answer 16

wait, I think I may have gotten something...

Answer 17

this what my book says

Answer 18

Okay, I did get the same answer as you did... pretty much, and I share your suspicion that maybe the ^2 isn't necessary... so you know what, let's just differentiate the blasted thing XD

Answer 19

@terenzreignz you mean no need to take square?

Answer 20

Let's find out.
After all, both terms have the same annihilator, so why bother doing it twice?
But let's try to find out the hard way, see if it works...\[\Large (D^2+6D +10)\left[e^{-3x}(2\sin x + 7 \cos x)\right]\]

Answer 21

because sinx, and cos x are linearly independent

Answer 22

Okay, let's do the first term...
\[\Large (D^2+6D+10)2e^{-3x}\sin(x) \]

Answer 23

so, we try both way, OK? you do just one, I do ^2 , or else?

Answer 24

Let's do just 1, and it it doesn't work, then it's probably ^2.
Don't kill yourself, LOL

Answer 25

oh no!! for sure, friend, you see, just at the first line, you pick 2 as coefficient, I can conclude that 2 \(\neq\)7 of cos.

Answer 26

therefore, they cannot be the same. hehehe... is my logic reasonable?

Answer 27

Those 2 and 7 are irrelevant, they are simply constants multiplied to the function and will of course have no effect on the annihilator (since if the function is annihilated [to zero] that constant multiplied to it can do nothing)

Answer 28

how can it be irrelevant?? because at the end up , you must let them =to calculate A, B the coefficient of cos and sin from partial solution.right?

Answer 29

Just hang on :>\[\Large (D^2+6D+10)2e^{-3x}\sin(x)\]
amounts to
\[-4e^{-3x}(3\cos(x) - 4\sin(x) )+12e^{-3x}(\cos(x) - 3\sin(x))+20e^{-3x}\sin(x) \]
which is zero...which makes sense, because the differential operator was the annihilator of \(\large 2e^{-3x}\sin x\)
right?

Answer 30

yes

Answer 31

Now, two things, first, if I'm your friend, acknowledge that I have a name, and it's TJ ^_^

Secondly, let me demonstrate to you the irrelevance of constants.
Suppose we have a function f(x) and it has an annihilator a(D)

You following? :>

Answer 32

I follow both (your 2 things) hehehe..TJ

Answer 33

Well, anyway, it only follows that
\[\large a(D)f(x)=0\]
But what if f(x) is multiplied by some constant k?
Would it matter?
\[\Large a(D) kf(x) = ka(D)f(x) =k(0) = 0\]The constant factor won't matter.
Got it? :>

Answer 34

just in the case you need to find out the annihilator of f (x). When putting it into the whole problem to find \(y_c~~and~~y_p\) it is not irrelevant, am I right ?TJ

Answer 35

Get to that later. For now, we know that \[\Large (D^2+6D+10)2e^{-3x}\sin(x)=0\]
But what about \[\Large (D^2+6D+10)7e^{-3x}\cos(x)\qquad \color{red}?\]

Answer 36

=0 , too

Answer 37

ah, we have \(A_1(D)*A_2(D) \) is annihilator of \(F_1(x) +F_2(x)\) right ? TJ

Answer 38

Well yes, it should.
Now, that can only mean that \[\Large (D^2+6D+10)[2e^{-3x}\sin(x)+7e^{-3x}\cos(x)]=0\]\[\Large =\Large (D^2+6D+10)e^{-3x}(2\sin(x)++7\cos(x))=0\]

Answer 39

Hold on a sec, let's have another theoretical approach... what if we have \(f_1\) and \(f_2\) as functions of x with *the same* annihilator a(D) ?
Then...
\[\Large a(D)[f_1(x)+f_2(x)]= a(D)f_1(x)+ a(D) f_2(x)\]\[\Large = 0+0\]since a(D) was an annihilator to both of them, individually.
\[\Large = 0\]and therefore, a(D) is the annihilator to their sum as well.

Answer 40

Such is the case with \(\large 2e^{-3x}\sin(x)\) and \(\large 7e^{-3x}\cos(x)\)
They do, in fact, have the same annihilator.

Answer 41

Everything cool? :>

Answer 42

I still have another question, can I ask you here? the same topic

Answer 43

But this one question is closed, then?
Not bad for a kid that didn't know what annihilators were just half an hour ago, right? ;)

That said, that was LUCK. So I can't guarantee your second question, but hit me up anyway :P

Answer 44

ok, I will post in a new post. Thanks a lot TJ

Answer 45

You do that, we'll have a look see :>