Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 26 views

    MODULE 7 - Laplace Transforms of Integrals

    Uploaded by

    Irene Hugo

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    MODULE 7 - Laplace Transforms of Integrals

    Uploaded by

    Irene Hugo
    26 views4 pages

    Copyright

    © © All Rights Reserved

    Available Formats

    PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    26 views4 pages

    MODULE 7 - Laplace Transforms of Integrals

    Uploaded by

    Irene Hugo

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    M222C | ADVANCED MATHEMATICS FOR CE

    MODULE 7
    LAPLACE TRANFORMS OF INTEGRALS

    If ℒ{𝑓(𝑡)} = 𝐹(𝑠), then

    𝑡
    𝐹(𝑠)
    ℒ{∫ 𝑓(𝑢) 𝑑𝑢} =
    0 𝑠

    Proof of Laplace Transforms of Integrals

    𝑡 ∞ 𝑡
    −𝑠𝑡
    ℒ{∫ 𝑓(𝑢) 𝑑𝑢} = ∫ 𝑒 ∫ 𝑓(𝑢) 𝑑𝑢 𝑑𝑡
    0 0 0

    Using integration by parts,


    𝑡
    𝑢 = ∫0 𝑓(𝑢) 𝑑𝑢
    𝑑 𝑡
    𝑑𝑢 = [∫0 𝑓(𝑢) 𝑑𝑢]
    𝑑𝑡
    𝑑𝑣 = 𝑒 −𝑠𝑡 𝑑𝑡
    1
    𝑣 = − 𝑒 −𝑠𝑡
    𝑠

    𝑡 𝑡 ∞
    1 1 ∞ 𝑑 𝑡
    ℒ{∫ 𝑓(𝑢) 𝑑𝑢} = [∫ 𝑓(𝑢) 𝑑𝑢 ∙ − 𝑒 −𝑠𝑡 ] + ∫ 𝑒 −𝑠𝑡 ∙ [∫ 𝑓(𝑢) 𝑑𝑢] 𝑑𝑡
    0 0 𝑠 0
    𝑠 0 𝑑𝑡 0

    𝑡 𝑑 𝑡
    Let: ∫0 𝑓(𝑢) 𝑑𝑢 = 𝑔(𝑡) [∫0 𝑓(𝑢) 𝑑𝑢] = 𝑓(𝑡)
    𝑑𝑡

    𝑡 ∞
    1 1 ∞
    ℒ{∫ 𝑓(𝑢) 𝑑𝑢} = [− 𝑒 −𝑠𝑡 ∙ 𝑔(𝑡)] + ∫ 𝑒 −𝑠𝑡 ∙ 𝑓(𝑡) 𝑑𝑡
    0 𝑠 0 𝑠 0

    𝑡
    1
    ℒ{∫ 𝑓(𝑢) 𝑑𝑢} = 0 + 0 + 𝐹(𝑠)
    0 𝑠

    𝑡
    𝐹(𝑠)
    ℒ{∫ 𝑓(𝑢) 𝑑𝑢} =
    0 𝑠

    1 of 4
    M222C | ADVANCED MATHEMATICS FOR CE

    Examples:

    2 of 4
    M222C | ADVANCED MATHEMATICS FOR CE

    𝑡
    Find the Laplace transform of ∫0 (6𝑢2 − 𝑒 2𝑢 ) 𝑑𝑢

    𝑡 ℒ(6𝑡 2 −𝑒 2𝑡 )
    ℒ [∫0 (6𝑢2 − 𝑒 2𝑢 ) 𝑑𝑢] =
    𝑠

    1
    = ℒ (6𝑡 2 − 𝑒 2𝑡 )
    𝑠

    1 2 1
    = [6 ( 3 ) − ]
    𝑠 𝑠 𝑠−2

    1 12(𝑠−2)−𝑠 3
    = [ ]
    𝑠 𝑠 3 (𝑠−2)

    1 12𝑠−24−𝑠 3
    = [ ]
    𝑠 𝑠 3 (𝑠−2)

    𝑡 −𝑠 3 12𝑠−24
    ℒ [∫0 (6𝑢2 − 𝑒 2𝑢 ) 𝑑𝑢] = answer
    𝑠 4 (𝑠−2)

    𝑡 1
    Find the Laplace transform of ∫0 ( 𝑢3 + 3𝑒 −3𝑢 ) 𝑑𝑢
    2

    1
    𝑡
    1 ℒ ( 𝑡3 + 3𝑒−3𝑡 )
    2
    ℒ [∫ ( 𝑢3 + 3𝑒−3𝑢 ) 𝑑𝑢] =
    0 2 𝑠

    1 1
    = ℒ ( 𝑡3 + 3𝑒−3𝑡 )
    𝑠 2

    1 1 6 1
    = [ ( 4) + 3 ( )]
    𝑠 2 𝑠 𝑠+3

    1 3(𝑠+3)+3𝑠4
    = [ ]
    𝑠 𝑠4 (𝑠+3)

    1 3𝑠+9+3𝑠4
    = [ ]
    𝑠 𝑠4 (𝑠+3)

    𝑡 1 3𝑠4 +3𝑠+9
    ℒ [∫0 ( 𝑢3 + 3𝑒−3𝑢 ) 𝑑𝑢] = answer
    2 𝑠5 (𝑠+3)

    3 of 4
    M222C | ADVANCED MATHEMATICS FOR CE

    Disclaimer and Fair Use Statement

    This module may contain copyrighted material, the use of which may not have
    been specifically authorized by the copyright owner. The material contained in
    this module is distributed without profit for educational purposes. This should
    constitute a ‘fair use’ of any such copyrighted material. If you wish to use any
    copyrighted material from this module for purposes of your own that go beyond
    ‘fair use’, you must obtain expressed permission from the copyright owner.

    4 of 4

    You might also like