Laplace Transforms of Damped Trigonometric Functions Bernd Schr - - PowerPoint PPT Presentation

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Laplace Transforms of Damped Trigonometric Functions Bernd Schr - - PowerPoint PPT Presentation

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Transforms and New Formulas An Example Double Check Laplace Transforms of Damped Trigonometric Functions Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped

slide-1
SLIDE 1

logo1 Transforms and New Formulas An Example Double Check

Laplace Transforms of Damped Trigonometric Functions

Bernd Schr¨

  • der

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-2
SLIDE 2

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-3
SLIDE 3

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-4
SLIDE 4

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-5
SLIDE 5

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Original DE & IVP

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-6
SLIDE 6

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Original DE & IVP ✲ L

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-7
SLIDE 7

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Original DE & IVP Algebraic equation for the Laplace transform ✲ L

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-8
SLIDE 8

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform ✲ L

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-9
SLIDE 9

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform ✲ L Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-10
SLIDE 10

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform Laplace transform

  • f the solution

✲ L Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-11
SLIDE 11

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform Laplace transform

  • f the solution

✲ ✛ L L −1 Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-12
SLIDE 12

logo1 Transforms and New Formulas An Example Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform Laplace transform

  • f the solution

Solution ✲ ✛ L L −1 Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-13
SLIDE 13

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-14
SLIDE 14

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-15
SLIDE 15

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution. y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-16
SLIDE 16

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution. y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0 s2Y −s+2sY −2+2Y = 2 s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-17
SLIDE 17

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution. y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0 s2Y −s+2sY −2+2Y = 2 s2 +4

  • s2 +2s+2
  • Y

= s+2+ 2 s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-18
SLIDE 18

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution. y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0 s2Y −s+2sY −2+2Y = 2 s2 +4

  • s2 +2s+2
  • Y

= s+2+ 2 s2 +4

  • s2 +2s+2
  • Y

= (s+2)

  • s2 +4
  • +2

s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-19
SLIDE 19

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution. y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0 s2Y −s+2sY −2+2Y = 2 s2 +4

  • s2 +2s+2
  • Y

= s+2+ 2 s2 +4

  • s2 +2s+2
  • Y

= (s+2)

  • s2 +4
  • +2

s2 +4 = s3 +2s2 +4s+10 s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-20
SLIDE 20

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Finding the Laplace transform of the solution. y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0 s2Y −s+2sY −2+2Y = 2 s2 +4

  • s2 +2s+2
  • Y

= s+2+ 2 s2 +4

  • s2 +2s+2
  • Y

= (s+2)

  • s2 +4
  • +2

s2 +4 = s3 +2s2 +4s+10 s2 +4 Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-21
SLIDE 21

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-22
SLIDE 22

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-23
SLIDE 23

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-24
SLIDE 24

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-25
SLIDE 25

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-26
SLIDE 26

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-27
SLIDE 27

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-28
SLIDE 28

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-29
SLIDE 29

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2 +(4A+2C +2D)s

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-30
SLIDE 30

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2 +(4A+2C +2D)s+(4B+2D)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-31
SLIDE 31

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2 +(4A+2C +2D)s+(4B+2D) A+C = 1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-32
SLIDE 32

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2 +(4A+2C +2D)s+(4B+2D) A+C = 1 B+2C +D = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-33
SLIDE 33

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2 +(4A+2C +2D)s+(4B+2D) A+C = 1 B+2C +D = 2 4A+2C +2D = 4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-34
SLIDE 34

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. Y = s3 +2s2 +4s+10 (s2 +2s+2)(s2 +4) = As+B s2 +2s+2 + Cs+D s2 +4 s3 +2s2 +4s+10 = (As+B)

  • s2+4
  • +(Cs+D)
  • s2+2s+2
  • s3 +2s2 +4s+10

= (A+C)s3 +(B+2C +D)s2 +(4A+2C +2D)s+(4B+2D) A+C = 1 B+2C +D = 2 4A+2C +2D = 4 4B+2D = 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-35
SLIDE 35

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-36
SLIDE 36

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-37
SLIDE 37

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-38
SLIDE 38

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-39
SLIDE 39

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-40
SLIDE 40

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 10D = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-41
SLIDE 41

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 10D = 2 D = −1 5,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-42
SLIDE 42

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 10D = 2 D = −1 5, C = −1 5,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-43
SLIDE 43

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 10D = 2 D = −1 5, C = −1 5, B = 13 5 ,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-44
SLIDE 44

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 10D = 2 D = −1 5, C = −1 5, B = 13 5 , A = 6 5,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-45
SLIDE 45

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Partial fraction decomposition. A + C = 1 B + 2C + D = 2 4A + 2C + 2D = 4 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 4B + 2D = 10 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 8C − 2D = 2 A + C = 1 B + 2C + D = 2 − 2C + 2D = 0 − 10D = 2 D = −1 5, C = −1 5, B = 13 5 , A = 6 5, Y =

6 5s+ 13 5

s2 +2s+2 + − 1

5s− 1 5

s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-46
SLIDE 46

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-47
SLIDE 47

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-48
SLIDE 48

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-49
SLIDE 49

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-50
SLIDE 50

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6(s+1−1)+13 (s+1)2 +1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-51
SLIDE 51

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6(s+1−1)+13 (s+1)2 +1 − 1 5 s s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-52
SLIDE 52

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6(s+1−1)+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 2· 1

2

s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-53
SLIDE 53

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6((s+1)−1)+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 2· 1

2

s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-54
SLIDE 54

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6((s+1)−1)+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 2· 1

2

s2 +4 = 1 5 6(s+1)−6+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 10 2 s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-55
SLIDE 55

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6((s+1)−1)+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 2· 1

2

s2 +4 = 1 5 6(s+1)−6+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 10 2 s2 +4 = 6 5 s+1 (s+1)2 +1 + 7 5 1 (s+1)2 +1 − 1 5 s s2 +4 − 1 10 2 s2 +4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-56
SLIDE 56

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

Inverting the Laplace transform. Y =

6 5s+ 13 5

s2 +2s+2 + −1

5s− 1 5

s2 +4 = 1 5 6s+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 1 s2 +4 = 1 5 6((s+1)−1)+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 5 2· 1

2

s2 +4 = 1 5 6(s+1)−6+13 (s+1)2 +1 − 1 5 s s2 +4 − 1 10 2 s2 +4 = 6 5 s+1 (s+1)2 +1 + 7 5 1 (s+1)2 +1 − 1 5 s s2 +4 − 1 10 2 s2 +4 y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-57
SLIDE 57

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0

y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-58
SLIDE 58

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-59
SLIDE 59

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-60
SLIDE 60

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-61
SLIDE 61

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-62
SLIDE 62

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-63
SLIDE 63

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-64
SLIDE 64

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t) √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-65
SLIDE 65

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t) √ y(0) = 1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-66
SLIDE 66

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t) √ y(0) = 1 √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-67
SLIDE 67

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t) √ y(0) = 1 √ y′(0) =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-68
SLIDE 68

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t) √ y(0) = 1 √ y′(0) = √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-69
SLIDE 69

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0? 2y+2y′ +y′′ = 2 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t)

  • +2

1 5e−t cos(t)− 13 5 e−t sin(t)+ 2 5 sin(2t)− 2 10 cos(2t)

  • +
  • −14

5 e−t cos(t)+ 12 5 e−t sin(t)+ 4 5 cos(2t)+ 4 10 sin(2t)

  • =

sin(2t) √ y(0) = 1 √ y′(0) = √

  • r use a computer algebra system.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-70
SLIDE 70

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-71
SLIDE 71

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-72
SLIDE 72

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-73
SLIDE 73

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-74
SLIDE 74

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions

slide-75
SLIDE 75

logo1 Transforms and New Formulas An Example Double Check

Does y = 6 5e−t cos(t)+ 7 5e−t sin(t)− 1 5 cos(2t)− 1 10 sin(2t) Solve y′′ +2y′ +2y = sin(2t), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Damped Trigonometric Functions