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anonymous
 one year ago
the laplace tranform of tcosx=cosx//s^2 am i right???
anonymous
 one year ago
the laplace tranform of tcosx=cosx//s^2 am i right???

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.0doesnt look right to me, might be your choice of notation tho

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0do we agree that: L(y') = sL(y)  yo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my question is different

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0is it about bubble gum? because it looks to be about a laplace transform to me ... can you clarify it any?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0here 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)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0your questions is in x and t ... is that a typo? or on purpose?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0assuming cos(x) is not a typo, then its simply a constant with respect to t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sory for bieng late rep but i have some personal problems so thats y.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes cosx is treat constant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then the laplace is cosx/s^2 is right?

math&ing001
 one year ago
Best ResponseYou've already chosen the best response.1Wolfram says it's right.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0late is better then never :)

math&ing001
 one year ago
Best ResponseYou've already chosen the best response.1You can use the tool from here http://www.wolframalpha.com/widgets/view.jsp?id=3272010f63d3145699ca78bbe0db05a7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank u nice to meet u.
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