Question

the laplace tranform of -tcosx=-cosx//s^2 am i right???

Answer 1

doesnt look right to me, might be your choice of notation tho

Answer 2

do we agree that: L(y') = sL(y) - yo

Answer 3

my question is different

Answer 4

is it about bubble gum? because it looks to be about a laplace transform to me ... can you clarify it any?

Answer 5

here is how Im reading you question since you do not wish to clarify your responses

using the product rule of derivatives

L(tg' + g) = sL(tg) - 0go, by linearity of the laplace we get

L(tg') + L(g) = sL(tg)

let g = -cos(x)
g' = sin(x)

sL(-tcos(x)) = L(tsin(x)) + L(-cos(x))

s^2 L(-tcos(x)) = sL(tsin(x)) + sL(-cos(x))

using the same construct we can determine the sL(tsin(x))
----------------

sL(tsin(x)) = L(tcos(x)) + L(sin(x))


s^2 L(-tcos(x)) = L(tcos(x)) + L(sin(x)) + sL(-cos(x))

s^2 L(-tcos(x)) + L(-tcos(x)) = L(sin(x)) + sL(-cos(x))

(s^2+1) L(-tcos(x)) = L(sin(x)) + sL(-cos(x))

L(-tcos(x)) = [L(sin(x)) + sL(-cos(x))]/(s^2+1)

Answer 6

your questions is in x and t ... is that a typo? or on purpose?

Answer 7

assuming cos(x) is not a typo, then its simply a constant with respect to t

Answer 8

sory for bieng late rep but i have some personal problems so thats y.

Answer 9

yes cosx is treat constant.

Answer 10

then the laplace is -cosx/s^2 is right?

Answer 11

Wolfram says it's right.

Answer 12

wow its amazing.

Answer 13

late is better then never :)

Answer 14

You can use the tool from here http://www.wolframalpha.com/widgets/view.jsp?id=3272010f63d3145699ca78bbe0db05a7

Answer 15

thank u nice to meet u.