Question

Please explain me how e^(-2t)* sin t = e^(-2 +i) t

Answer 1

\[e^{-2t}sint = e^{(-2+i)t}\]

Answer 2

hey, 0 0 know it? explain me, please

Answer 3

doesn't seem very likely to me
if \(t=0\) the left side is \(0\) and the right side is \(1\)

Answer 4

Isn't that the same as saying that sin(t) = e^(it)?

Answer 5

I don't know what the logic is in that equation.

Answer 6

if you use Laplace Transforms you might be able to work through it, but that's something that I forgot how to do after my class was over with.

Answer 7

see the part I circle? that's my prof's stuff. I don't understand, so make question here

Answer 8

I didn't reach Laplace transform yet. May be next week, not now.

Answer 9

Did you look in your book, there might be some crazy formula that says that's true, but I don't remember learning about it.

Answer 10

too many things drive me crazy. this is one of them. hehehe

Answer 11

wait, I post other, please read and tell me whether you understand or not.

Answer 12

Haha yeah crazy professors. That problem is harder than any I've done, I only did stuff that had a 2x2 wronskian. Sorry I can't help you.

Answer 13

I don't understand a word. :(

Answer 14

You don't understand how he got the homogeneous part done?

Answer 15

homogeneous part is quite easy, I got that part, everybody in class know that stuff ( I thought) just partial one

Answer 16

he applies the question I ask above to this stuff, too. you see, e^(-t) sint = e^(( -1+i)t) again, that means it 's not a mistake, it's application, but what the logic is?

Answer 17

I think he is actually saying that e^(-t) sint = e^(( -1+0i)t) which would make since but I don't understand why he would do that, and his handwriting is really bad.

Answer 18

lol, but I have to correct you, it's not his handwriting, it's mine,

Answer 19

And I was looking at the wrong part. Sorry, I really don't know. I could maybe follow it if I had a hard copy in front of me.

Answer 20

it's ok, I give up. tomorrow, i will stop by his office to make it clear.

Answer 21

I tried to wolframalpha it, but nothing showed up :(

Answer 22

um looks like u have lost the 0 somewhere in the way to me o.O u have (-1+0i)t then u have e^(-1+i)t

Answer 23

Hmmmm

Answer 24

e^(-t)sint = .5ie^(-1-i)t-.5ie^(-1+i)t

Answer 25

Let me explain,
first , he separated the partial into 2 parts, y"' -y"......= e^(-t) and
y"' -y".......= sin t
then, he combined them to get the form he gave out. That 's why first (-1+0i) t then e^((-1+i)t)

Answer 26

It's probably just some crazy formula that no one really needs to know but teachers expect that everyone knows it.

Answer 27

Thanks everybody. I will ask him, ( for sure, cannot let him free with my problem). If you are curious , can stop by and see his answer. I will post if I can understand his explanation, hihihi...

Answer 28

Alright I'll stop back by tomorrow sometime late. good luck trying to figure out that stuff.

Answer 29

thnks . it's so late here, I am sleepy, too

Answer 30

Alright good night :P

Answer 31

@Arfney tomorrow, I'll show you my worst handwriting, hehehe

Answer 32

Haha I'll look forward to that!

Answer 33

@Loser66 it has to do with the fact that we can evaluate real integrals using complex values!

Answer 34

Intuitively the idea is that from Euler's formula we have:$$e^{it}=\cos t+i\sin t$$hence we can say \(\cos t=\Re\{e^{it}\},\sin t=\Im\{e^{it}\}\) i.e. \(\cos t,\sin t\) are the real and imaginary parts, respectively, of \(e^{it}\).Observe, then, that we can treat \(e^{-2t}\sin t\) as the *imaginary* part of \(e^{-2t}e^{it}=e^{(-2+i)t}\).$$\int e^{(-2+i)t}\,dt=\int\left(e^{-2t}\cos t+ie^{-2t}\sin t\right)\,dt=\int e^{-2t}\cos t\,dt+i\int e^{-2t}\sin t\,dt$$so clearly we can write \(\int e^{-2t}\sin t\,dt=\Im\{\int e^{(-2+i)t}\,dt\}\) i.e. our integral is merely the imaginary part of that latter integral (because integrals are *linear* operators)

Answer 35

So observe we can find antiderivatives as follows:$$\int e^{(-2+i)t}\,dt=\frac1{-2+i}e^{(-2+i)t}+C$$Recall that \(\dfrac1{-2+i}=\dfrac{\overline{(-2+i)}}{|-2+i|^2}=\dfrac{-2-i}5\) and so we get:$$\begin{align*}\int e^{(-2+i)t}\,dt&=\frac{-2-i}5e^{-2t}e^{it}+C_1+iC_2\\&=\frac{e^{-2t}(-2-i)(\cos t+i\sin t)}5+C_1+iC_2\\&=\frac{e^{-2t}(-2\cos t-2i\sin t-i\cos t+\sin t)}5+C_1+iC_2\\&=\left(\frac{e^{-2t}(-2\cos t+\sin t)}5+C_1\right)+i\left(\frac{-e^{-2t}(\cos t+2\sin t)}5+C_2\right)\end{align*}$$hence we have:$$\int e^{-2t}\sin t\,dt=\Im\left\{\int e^{(-2+i)t}\,dt\right\}=-\frac{e^{-2t}(\cos t+2\sin t)}5+C_2$$

Answer 36

Wow!! Thank you very much.@oldrin.bataku
I searched and saw the same problem from Stackexchange.com but just "the same" not exactly my problem.
http://math.stackexchange.com/questions/105873/integration-with-complex-numbers
@Arfney : There is no chance to show you my handwriting, sorry for that, hehehe

Answer 37

no problem @Loser66 :-) so you're correct they're not equal, but the integral with sine is just the imaginary part of the integral with just the exponential

Answer 38

Ok, I got it. thanks again

Answer 39

@oldrin.bataku.
At your second comment, line 4, after take integral, why do we have \(iC_2\)?

Answer 40

@Loser66 because our integral is complex-valued, our constant of integration may be complex-valued, hence I split into real/imaginary parts.