Computing Laplace Transforms (Using the Integral Definition)

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  • เผยแพร่เมื่อ 23 ธ.ค. 2024

ความคิดเห็น • 9

  • @earthworm02
    @earthworm02 4 ปีที่แล้ว +6

    I feel lucky that I've found you!
    Thanks a lot for your lectures.

    • @HoustonMathPrep
      @HoustonMathPrep  4 ปีที่แล้ว +1

      Thank you for sharing your kind words and support! You are welcome :)

  • @kavyarayidi1839
    @kavyarayidi1839 4 ปีที่แล้ว +6

    You have really helped me a lot If I score well this semester, it's all thanks to you!!!!!!!!!!!!!❤💕

  • @turbothrottletrouble4217
    @turbothrottletrouble4217 4 ปีที่แล้ว +3

    I just love your videos. These will probably prepare me for my electrical and electronic engineering course

    • @HoustonMathPrep
      @HoustonMathPrep  4 ปีที่แล้ว +1

      Really glad you like them and they are helping you! Good luck with your course :)

  • @curtpiazza1688
    @curtpiazza1688 ปีที่แล้ว +1

    Great examples. Thanx for covering this topic using limits! Very informative! 😊

  • @lynou-cats
    @lynou-cats 3 ปีที่แล้ว +4

    I'm sorry for this debutant question but I must ask : at 3:42, how come v = (-1/s)*e^(-st) ? I'm trying to wrap my head around the (-1/s) : for me the antiderivative of e^(-st), following the de^(u) = e^(u)*u' rule, would be -s*e^(-st).. I don't understand. Would someone be patient enough to explain it to me?

    • @HoustonMathPrep
      @HoustonMathPrep  3 ปีที่แล้ว +1

      Hello!
      Remember that for the derivative of e^(-st), the chain rule tells us that the constant multiple "-s" would multiply the exponential. Since this is the antiderivative, the answer would have us dividing by the "-s" multiple, or you can think of it as multiplying by the reciprocal if you like. That's where we get the -1/s.