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ranyai12

  • 3 years ago

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

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  1. ranyai12
    • 3 years ago
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    @Spacelimbus

  2. ranyai12
    • 3 years ago
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    @TuringTest

  3. sami-21
    • 3 years ago
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    @ranyai12 you there ?

  4. ranyai12
    • 3 years ago
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    yes i am

  5. sami-21
    • 3 years ago
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    ok

  6. sami-21
    • 3 years ago
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    you have given piecewise function .first the definition of Laplace is \[[\Large F(s)=\int\limits\limits_{0}^{\infty}e^{-st}f(t)dt\]\]

  7. ranyai12
    • 3 years ago
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    sorry openstudy logged me out!

  8. ranyai12
    • 3 years ago
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    what values from the graph would we incorporate?

  9. ranyai12
    • 3 years ago
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    @sami-21 are you still there?

  10. sami-21
    • 3 years ago
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    yes i am here . openstudy crashed me :( i was typing everything went off :( let me try again.

  11. Spacelimbus
    • 3 years ago
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    OpenStudy is a bit of a pain today, I was opening this question about 20 minutes ago, and by now I am here hehe.

  12. ranyai12
    • 3 years ago
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    lol yea i understand it kept on logging me out!

  13. sami-21
    • 3 years ago
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    it is piece wise function can be written as

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  14. ranyai12
    • 3 years ago
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    do you get the integration to be (-3e^(-st))/s

  15. sami-21
    • 3 years ago
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    yes apply the limits

  16. ranyai12
    • 3 years ago
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    and the limits are the s right

  17. sami-21
    • 3 years ago
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    nopes!! they are for t :P

  18. ranyai12
    • 3 years ago
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    oh lol

  19. ranyai12
    • 3 years ago
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    ok so the answer would be (3e^(-5s)(3e^s-1))/s

  20. sami-21
    • 3 years ago
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    you did a mistake . first take 3 common also the second one should be e^(-2s)

  21. ranyai12
    • 3 years ago
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    what do you mean first take 3 common

  22. ranyai12
    • 3 years ago
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    so the answer is 3(e^5s/s-e^2s/s)

  23. sami-21
    • 3 years ago
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    \[\Large \frac{-3}{s}|e^{-st}|_{2}^{5}=\frac{-3}{s}[e^{-5s}-e^{-2s}]\]

  24. sami-21
    • 3 years ago
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    @ranyai12 yes it is just you missed minus sign with the s in the exponent.

  25. ranyai12
    • 3 years ago
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    oh ok thank you

  26. sami-21
    • 3 years ago
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    you 're welcome :)

  27. ranyai12
    • 3 years ago
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    it ended up being wrong

  28. ranyai12
    • 3 years ago
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    wait nevermind i got it!

  29. ranyai12
    • 3 years ago
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    it was right! i plugged it in wrong

  30. Spacelimbus
    • 3 years ago
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    (-: !

  31. sami-21
    • 3 years ago
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    you must have missed minus sign with 3 :P

  32. ranyai12
    • 3 years ago
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    I did! lol

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