anonymous
  • anonymous
the laplace tranform of -tcosx=-cosx//s^2 am i right???
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amistre64
  • amistre64
doesnt look right to me, might be your choice of notation tho
amistre64
  • amistre64
do we agree that: L(y') = sL(y) - yo
anonymous
  • anonymous
my question is different

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amistre64
  • amistre64
is it about bubble gum? because it looks to be about a laplace transform to me ... can you clarify it any?
amistre64
  • amistre64
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)
amistre64
  • amistre64
your questions is in x and t ... is that a typo? or on purpose?
amistre64
  • amistre64
assuming cos(x) is not a typo, then its simply a constant with respect to t
anonymous
  • anonymous
sory for bieng late rep but i have some personal problems so thats y.
anonymous
  • anonymous
yes cosx is treat constant.
anonymous
  • anonymous
then the laplace is -cosx/s^2 is right?
math&ing001
  • math&ing001
Wolfram says it's right.
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anonymous
  • anonymous
wow its amazing.
amistre64
  • amistre64
late is better then never :)
math&ing001
  • math&ing001
You can use the tool from here http://www.wolframalpha.com/widgets/view.jsp?id=3272010f63d3145699ca78bbe0db05a7
anonymous
  • anonymous
thank u nice to meet u.

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