Question

Hi everyone! Can someone help me solve a Laplace transform? f(t)=e^(t)sinh(t)... Pretty confused and I can't seem to find an example anywhere. Thanks! :o)

Answer 1

:D

Answer 2

sorry, I wrote the wrong question initially but it is now correct and still need help.

Answer 3

Remember how to rewrite sinh(t) in terms of exponentials? :)

Answer 4

is that the only way to solve this?

Answer 5

I don't see why you want to do any other method :D hmm

Answer 6

well what I mean is this...gimme a sec to explain and maybe you can correct my thinking :o)...

Answer 7

by definition of Laplace transform, we can write:
\[\Large F\left( s \right) = \int_0^\infty {{e^t}\sinh \left( t \right){e^{ - st}}} dt\]

Answer 8

Ahh it got cut off >.<
But something like that, ya? :)
Just taking Laplace of two exponentials.

Answer 9

\[\large\rm \mathcal{L}\{ e^t \sinh(t)\}=\mathcal{L}\{ e^t \cdot\frac{1}{2}(e^t-e^{-t})\}=\frac{1}{2}\left(\mathcal{L}\{e^{2t}\}-\mathcal L\{e^{0}\}\right)\]

Answer 10

now, using the suggestion of @zepdrix we can make this sbstitution:
\[\Large \sinh t = \frac{{{e^t} - {e^{ - t}}}}{2}\]

Answer 11

when I first tried solving the problem, I first looked at e^t and recognized it of the form L{e^at}=1/s-a
Then I looked at L{sinht} and recognized it as k=1 so therefore K/s^2-K^2 which would equal 1/s^2+1
so then I thought I could multiply them together>>>
(1/s-a) times (1/s^2+1) to get 1/(s-a)(s^2-1) but that isn't the answer...
first of all, can you see why I would want to solve it like this?
second, can you tell me exactly why this is the wrong method?
:o)

Answer 12

Hmm are you confusing sin(t) with sinh(t) maybe? :)

Answer 13

no...i'm not confusing them L{sinkt}=k/s^2+k^2 whereas L{sinhkt}=k/s^2-k^2
all I was trying to do was use simple definitions of some common Laplace transforms

Answer 14

Oh yes your formula looks good hehe :)

Answer 15

can you explain why I can't do the problem this way?

Answer 16

the only thing I can think of is that you can't take L{ab} and change it to L{a}timesL{b}
is that correct?

Answer 17

Well I mean.. you can't do that in general with integration.
Example:\[\Large\rm \int\limits \frac{1}{2}x~dx=x^2,\qquad\qquad \int\limits \cos(x)dx=\sin(x)\]But,\[\Large\rm \int\limits \frac{1}{2}x \cos(x)~dx\ne x^2 \sin(x)\]

Answer 18

Yah, the last thing you said sounds correct :)

Answer 19

Laplace Transform is a `linear operator` so it satisfies these two properties:
L{a+b} = L{a} + L{b}
L{ca} = cL{a}
(Where c is a constant).

Wish that multiplication worked that way though! hehe would be nice

Answer 20

haha...okay...bleh :o/ then I just have some simple rules mixed up...so then if I can't solve it remembering that sinht=(that e^2 thingy) then I can do the long integral form that Michele posted above correct?

Answer 21

Yah you might be able to do that integral by parts, remembering that your derivative of sinh(t) is cosh(t).

And you'd have to do parts like 2 times.. and then that weird algebra step thing... if you remember back to this integral: \(\Large\rm \int e^x\sin(x)dx\). Kinda how you dealt with that guy.

Answer 22

But converting to exponentials seems way better XD hehe

Answer 23

hint:
you vcan not multiply the single laplace transforms, since what you get is the laplace transform of the convolution product of the functions, which the single laplace transforms are referring to

Answer 24

can*

Answer 25

sorry...the website is messing up

Answer 26

okay yes...I would eventually have to add or subtract the laplace transforms and then factor it out and all that ghastly algebra hoping I don't flip a sign ?

Answer 27

okay...so the key here is remembering that the Laplace transform itself is a linear operation, so no multiplication! that's a no-no correct?

Answer 28

<8-/ ?

Answer 29

ya linear operation gives us those `two properties` but unfortunately not that other one you were hoping for :D

Answer 30

are you guys still here?

Answer 31

These are really handy to remember though, you should tryyyyy!\[\Large\rm \sin(x)=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right),\qquad \sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)\]\[\Large\rm \cos(x)=\frac{1}{2}\left(e^{ix}+e^{-ix}\right),\qquad \cosh(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)\]Just tryyyy >.< They match up so nicely!

Answer 32

that's right! @SinginDaCalc2Blues

Answer 33

sorry...I think this website is really messing up...I couldn't see anything you posted until now...gimme just a sec to review :o)

Answer 34

ya it tweaked out for a second there >.<

Answer 35

hey zepdrix, are you missing an"i" next to the 2 in cost above?

Answer 36

you mean the cos(x) identity? :o
No. no i in the denominator if that's what you're asking.

Answer 37

so the sinx has an "i" in the denominator but not the cosx?

Answer 38

ya :D
Do you remember ummm... Euler's Formula thingy?
e^(ix) = cos x + i sin (x)

Sine has the imaginary unit on it.
So when you do a little bit of fancy math... . . .
the sine identity is the only one that ends up with the i.

Answer 39

I never commited that euler thingy to memory, I always used the alternate e^rx(cos bla bla stuff)

Answer 40

ya it's not super important until you take complex analysis.
up until that point, it's really just a "neato" formula :D

Answer 41

now I can't even remember what the bla bla stuff is now that I think about it...uhg!
e^rx[(C1cos(betax) + C2sin(betax)] is that right?

Answer 42

that was for imaginary roots

Answer 43

Mmm I'm not sure what that is.


Oh oh yes, when you end up with complex roots to a ODE :)
Ya ya looks familiar.\[\Large\rm y=c_1 e^{ax}\sin(bx)+c_2 e^{ax}\cos(bx)\]Where your roots to your characteristic were something like \(\Large\rm r=a+bi\)

Answer 44

Something like that, ya >.<

Answer 45

yes that's it...i misspoke...it is complex root, not imaginary roots

Answer 46

so unless I want to do the long integral, I need to remember the sinht and cosht then

Answer 47

omg...just noticed...3000 medal for your label! wow i'm happy with my little 50 medals for human calculator :o)

Answer 48

oh haha XD ya too much free time i guess lolol

Answer 49

ya just remmeber them :UUU

Answer 50

i dont see any way around it :3

Answer 51

are you still a student also? how do you have that much time to help people?

Answer 52

very impressive!

Answer 53

that's a good question :) i dunno..
i just love teaching i guess.. always find time for it.

i do tutoring for a part time job, and i come home and tutor for free on the internet, do what you love i guess ;)

Answer 54

that's really neat! thank you for all you help and don't say thanks! :o)

Answer 55

XD

Answer 56

ok so back to my problem...
change e^tsinht into e^t(e^x-e^-x/2)...then combine the base "e's"

Answer 57

ya e^t((e^t-e^-t)/2)

Answer 58

Ya exponent rule :U

Answer 59

okay.... still working on it...

Answer 60

so I could break it down like this I suppose...
e^t(e^x - e^-x)(1/2)
then
(1/2)(e^xt) - (1/2)(e^-xt)
then
(1/2)L{e^xt} - (1/2)L{e^-xt}
a=x....................a=-x
(1/2)(1/s-x) - (1/2)(1/s+x)
how does this look so far?

Answer 61

Mmm no no no.
Now you're just being silly :o)

You were dealing with e^t sinh(t)
NOT WITH e^t sinh(x)

Answer 62

okay well replace the x with t then

Answer 63

wait no

Answer 64

And make sure you remember your exponent rules correctly! :D\[\Large\rm e^t\cdot e^x\ne e^{xt}\]Even though that's not what we were looking for anyway :d

Answer 65

ugh...just a sec...lemme redo

Answer 66

haha k XD

Answer 67

okay...hopefully I just found my misplaced remaining brain cell...
e^t(e^x - e^-x)(1/2)
then
(1/2)(e^(t+x)) - (1/2)(e^(t-x))
then
(1/2)L{e^(t+x)} - (1/2)L{e^(t-x)}
a=(t+x)....................a=(t-x)
(1/2)(1/s-(t+x)) - (1/2)(1/s+(t-x))
how does this look so far?

Answer 68

wait no

Answer 69

AHH I wish you would get those stupid x's out :U

Answer 70

they're all t's!
x does exist! >.<

Answer 71

i needed to factor out a t right?

Answer 72

I dunno, you're being a silly billy.
We need to slow this down maybe..

Answer 73

what do you mean the x's don't exist?
I thought sinht was e^x-e^-x/2 ?
should it be instead e^t-e^-t/2 ?

Answer 74

Yes silly! :U\[\Large\rm \sinh(x)=\frac{1}{2}(e^x-e^{-x}),\qquad\sinh(t)=\frac{1}{2}(e^t-e^{-t})\]

Answer 75

omg! of course

Answer 76

okay...and just to make sure, to get it in the form of L{e^at} i need to factor out the t's right?

Answer 77

well maybe since I got rid of the x's, if I back the problem up, it may become more simple...one sec

Answer 78

i'm not gonna look at your work...I need to figure it out...can you erase that please?

Answer 79

okay...one more try here...gimme a sec

Answer 80

SinginDaODE Blueeeees

Answer 81

shaaa

Answer 82

Figure it out lady blues? :D

Answer 83

e^t(e^t - e^-t)(1/2)
then
(1/2)(e^(t+t)) - (1/2)(e^(t-t))
then
(1/2)L{e^(2t)} - (1/2)L{e^(0)}
then
(1/2)L{e^(2t)} - (1/2)L{1}
a=(2)
(1/2)(1/s-2) - (1/2)(1/s)
then I suppose you could write the answer in many different ways but...
(1/2) [ (1/s-2) -(1/s) ]

Answer 84

yayyy \c:/ good job!

Answer 85

OMG...whew :o)

Answer 86

ok im goin to bed -_- its too late

Answer 87

thanks @zepdrix ! :o)

Answer 88

3001 medals for u! :o)