Question

Whats the laplace transform of this one?

Answer 1

\[f(t)=\int\limits_{0}^{t} x*\sinh(x) dx\]

Answer 2

there is a propertie which says:

\[\int\limits_{0}^{t}f(\tau)d \tau<->F(s)/\]

Answer 3

i'm missing the "s" of F(s)/S

Answer 4

i did something wrong.. hold on..

Answer 5

\[f(t)=\int\limits_{0}^{t}xsinhx dx\]

Answer 6

but it is multiplying in t-domain

Answer 7

i know that convolution in t-domain = multiplying in s-domain

Answer 8

You'll need to use two results:
* L[y(t)] = Y(s), then L[t.y(t)] = -Y'(s)
* L[y'(t)] = sY(s) - y(0)

Set y'(t) = t.sinh(t) and now solve for Y(s) from the second equation by evaluating L[y'(t)] explicitly. To do that, use the first equation.

Answer 9

but james what do i have to do with the integral sign?

Answer 10

That's what we're working around. f(t) is your integral. But rather than trying to evaluate L[f(t)] explicitly, we're calculating L[f'(t)] and using that result to find L[f(t)].

Answer 11

i really don't understand anything what you just said, sorry

Answer 12

so f(t) = xsinhx

now i have to look for f'(t)?

Answer 13

Do you recognize at least the second of these two results?
* L[y(t)] = Y(s), then L[t.y(t)] = -Y'(s)
* L[y'(t)] = sY(s) - y(0)

Answer 14

L[t.y(t)] = -Y'(s) //This is the diff property

Answer 15

No, \[ f(t) = \int_0^t x \sinh x \ dx \] hence \[ f'(t) = t \sinh t \]

Answer 16

and the second result?

Answer 17

this one: * L[y'(t)] = sY(s) - y(0) ?

Answer 18

owhh wrong..

L[t.y(t)] = -Y'(s) //this property multiplication by t

L[y'(t)] = sY(s) - y(0) //this is the diff propertie

Answer 19

ok. Now, you want to find F(s) = L[f(t)] where f(t) is as above.

Rather than find it explicitly, I'm suggesting to you that you use the second property, the differentiation property to find it.

That is, find L[f'(t)]. Then because
L[f'(t)] = sF(s) - f(0)
solve for F(s)

Answer 20

But jamesj, i dont see the point do take the derivative of it, is it because the integral will compensate it?

Answer 21

The point is you can evaluate L[f'(t)] = L[ t. sinh t ]

Answer 22

f'(t) = t.sinht

L{f'(t)} <-> s.F(s) - f(0)

L{t.sinht} <-> -dF(s)/ds{ s.(1/(s^2-1)) - 0}

Answer 23

is this right?

Answer 24

I don't know what <-> means.

What we want first is L[ t . sinh t]. To find that we want L[ sinh t] and then apply the 'multiplication identity'

Answer 25

Now L[ sinh t ] = 1/(s^2 - 1). Hence
\[ L[ t . \sinh t] = - \frac{d \ }{ds} \frac{1}{s^2-1} \]

Answer 26

<-> means the transformation of time domain to s-domain

Answer 27

Calculate that. Now, once you have it, notice that f(0) = 0 hence
\[ L[ f'(t) ] = L[t .\sinh t] = s F(s) - f(0) = s F(s) \]
and therefore
\[ L[f(t)] = F(s) = L[t .\sinh t] / s \]

Answer 28

oke.. i think i got it..

first i have to use the diff property. then the integral property :)

Answer 29

i have calculate it and i thin the answer is:

F(s) = 2/(s^2-1)^2

Answer 30

yes

Answer 31

^^ thanksss <3

Answer 32

finally i get it ;) thank you for your help