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

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

Mathematics
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doesnt look right to me, might be your choice of notation tho
do we agree that: L(y') = sL(y) - yo
my question is different

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is it about bubble gum? because it looks to be about a laplace transform to me ... can you clarify it any?
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)
your questions is in x and t ... is that a typo? or on purpose?
assuming cos(x) is not a typo, then its simply a constant with respect to t
sory for bieng late rep but i have some personal problems so thats y.
yes cosx is treat constant.
then the laplace is -cosx/s^2 is right?
Wolfram says it's right.
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wow its amazing.
late is better then never :)
You can use the tool from here http://www.wolframalpha.com/widgets/view.jsp?id=3272010f63d3145699ca78bbe0db05a7
thank u nice to meet u.

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