anonymous
  • anonymous
Can someone please help me!!! Use the definition of the Laplace Transform to find the transform of the function shown in the link https://instruct.math.lsa.umich.edu/webwork2_course_files/ma216-u12/tmp/gif/rilayian-103-sethomework6prob13image1.png
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Spacelimbus
anonymous
  • anonymous
@TuringTest
anonymous
  • anonymous
@ranyai12 you there ?

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anonymous
  • anonymous
yes i am
anonymous
  • anonymous
ok
anonymous
  • anonymous
you have given piecewise function .first the definition of Laplace is \[[\Large F(s)=\int\limits\limits_{0}^{\infty}e^{-st}f(t)dt\]\]
anonymous
  • anonymous
sorry openstudy logged me out!
anonymous
  • anonymous
what values from the graph would we incorporate?
anonymous
  • anonymous
@sami-21 are you still there?
anonymous
  • anonymous
yes i am here . openstudy crashed me :( i was typing everything went off :( let me try again.
anonymous
  • anonymous
OpenStudy is a bit of a pain today, I was opening this question about 20 minutes ago, and by now I am here hehe.
anonymous
  • anonymous
lol yea i understand it kept on logging me out!
anonymous
  • anonymous
it is piece wise function can be written as
1 Attachment
anonymous
  • anonymous
do you get the integration to be (-3e^(-st))/s
anonymous
  • anonymous
yes apply the limits
anonymous
  • anonymous
and the limits are the s right
anonymous
  • anonymous
nopes!! they are for t :P
anonymous
  • anonymous
oh lol
anonymous
  • anonymous
ok so the answer would be (3e^(-5s)(3e^s-1))/s
anonymous
  • anonymous
you did a mistake . first take 3 common also the second one should be e^(-2s)
anonymous
  • anonymous
what do you mean first take 3 common
anonymous
  • anonymous
so the answer is 3(e^5s/s-e^2s/s)
anonymous
  • anonymous
\[\Large \frac{-3}{s}|e^{-st}|_{2}^{5}=\frac{-3}{s}[e^{-5s}-e^{-2s}]\]
anonymous
  • anonymous
@ranyai12 yes it is just you missed minus sign with the s in the exponent.
anonymous
  • anonymous
oh ok thank you
anonymous
  • anonymous
you 're welcome :)
anonymous
  • anonymous
it ended up being wrong
anonymous
  • anonymous
wait nevermind i got it!
anonymous
  • anonymous
it was right! i plugged it in wrong
anonymous
  • anonymous
(-: !
anonymous
  • anonymous
you must have missed minus sign with 3 :P
anonymous
  • anonymous
I did! lol

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