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Laplace Transforms and Integral Equations Bernd Schr oder logo1 - - PowerPoint PPT Presentation

Jan 14, 2023 692 likes •1.18k views

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

slide-1
SLIDE 1

logo1 Transforms and New Formulas An Example Double Check

Laplace Transforms and Integral Equations

Bernd Schr¨

  • der

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

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 and Integral Equations

slide-13
SLIDE 13

logo1 Transforms and New Formulas An Example Double Check

The Laplace Transform of an Integral

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-14
SLIDE 14

logo1 Transforms and New Formulas An Example Double Check

The Laplace Transform of an Integral

  • 1. Definite integrals of the form

t

0 f(τ)dτ arise in circuit

theory: The charge of a capacitor is the integral of the current over time.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-15
SLIDE 15

logo1 Transforms and New Formulas An Example Double Check

The Laplace Transform of an Integral

  • 1. Definite integrals of the form

t

0 f(τ)dτ arise in circuit

theory: The charge of a capacitor is the integral of the current over time. (We assume the capacitor is initially uncharged.)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-16
SLIDE 16

logo1 Transforms and New Formulas An Example Double Check

The Laplace Transform of an Integral

  • 1. Definite integrals of the form

t

0 f(τ)dτ arise in circuit

theory: The charge of a capacitor is the integral of the current over time. (We assume the capacitor is initially uncharged.)

  • 2. L

t

0 f(τ)dτ

  • = F(s)

s

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-17
SLIDE 17

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-18
SLIDE 18

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0 f ′(t)+f(t)−2

t

0 f(z) dz

= t, f(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-19
SLIDE 19

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0 f ′(t)+f(t)−2

t

0 f(z) dz

= t, f(0) = 0 sF +F −2F s = 1 s2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-20
SLIDE 20

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0 f ′(t)+f(t)−2

t

0 f(z) dz

= t, f(0) = 0 sF +F −2F s = 1 s2 F

  • s+1− 2

s

  • =

1 s2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-21
SLIDE 21

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0 f ′(t)+f(t)−2

t

0 f(z) dz

= t, f(0) = 0 sF +F −2F s = 1 s2 F

  • s+1− 2

s

  • =

1 s2 Fs2 +s−2 s = 1 s2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-22
SLIDE 22

logo1 Transforms and New Formulas An Example Double Check

Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0 f ′(t)+f(t)−2

t

0 f(z) dz

= t, f(0) = 0 sF +F −2F s = 1 s2 F

  • s+1− 2

s

  • =

1 s2 Fs2 +s−2 s = 1 s2 F = s s2(s−1)(s+2)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-23
SLIDE 23

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-24
SLIDE 24

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-25
SLIDE 25

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-26
SLIDE 26

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-27
SLIDE 27

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-28
SLIDE 28

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-29
SLIDE 29

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-30
SLIDE 30

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3 s = −2 : −2 = D·(−2)2 ·(−3)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-31
SLIDE 31

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3 s = −2 : −2 = D·(−2)2 ·(−3), D = 1 6

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-32
SLIDE 32

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3 s = −2 : −2 = D·(−2)2 ·(−3), D = 1 6 s = −1 : −1=A·(−1)·(−2)·1+0+1 3·(−1)2·1+1 6·(−1)2·(−2)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-33
SLIDE 33

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3 s = −2 : −2 = D·(−2)2 ·(−3), D = 1 6 s = −1 : −1=A·(−1)·(−2)·1+0+1 3·(−1)2·1+1 6·(−1)2·(−2), A = −1 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-34
SLIDE 34

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3 s = −2 : −2 = D·(−2)2 ·(−3), D = 1 6 s = −1 : −1=A·(−1)·(−2)·1+0+1 3·(−1)2·1+1 6·(−1)2·(−2), A = −1 2 F = −1 2 1 s +0· 1 s2 + 1 3 1 s−1 + 1 6 1 s+2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-35
SLIDE 35

logo1 Transforms and New Formulas An Example Double Check

Solve f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0

F = s s2(s−1)(s+2) = A s + B s2 + C s−1 + D s+2 s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1) s = 0 : 0 = B·(−1)·2, B = 0 s = 1 : 1 = C ·12 ·3, C = 1 3 s = −2 : −2 = D·(−2)2 ·(−3), D = 1 6 s = −1 : −1=A·(−1)·(−2)·1+0+1 3·(−1)2·1+1 6·(−1)2·(−2), A = −1 2 F = −1 2 1 s +0· 1 s2 + 1 3 1 s−1 + 1 6 1 s+2 f(t) = −1 2 + 1 3et + 1 6e−2t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-36
SLIDE 36

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-37
SLIDE 37

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value:

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-38
SLIDE 38

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-39
SLIDE 39

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-40
SLIDE 40

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-41
SLIDE 41

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-42
SLIDE 42

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6 = et 1 3 + 1 3 − 2 3

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-43
SLIDE 43

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6 = et 1 3 + 1 3 − 2 3

  • +e−2t
  • −1

3 + 1 6 + 1 6

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-44
SLIDE 44

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6 = et 1 3 + 1 3 − 2 3

  • +e−2t
  • −1

3 + 1 6 + 1 6

  • +t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-45
SLIDE 45

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6 = et 1 3 + 1 3 − 2 3

  • +e−2t
  • −1

3 + 1 6 + 1 6

  • +t +
  • −1

2 + 2 3 − 1 6

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-46
SLIDE 46

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6 = et 1 3 + 1 3 − 2 3

  • +e−2t
  • −1

3 + 1 6 + 1 6

  • +t +
  • −1

2 + 2 3 − 1 6

  • =

t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

slide-47
SLIDE 47

logo1 Transforms and New Formulas An Example Double Check

Does f(t) = −1 2 + 1 3et + 1 6e−2t Really Solve the Initial Value Problem f ′(t)+f(t)−2

t

0 f(z) dz = t, f(0) = 0?

Initial value: Look at f!

  • −1

2 + 1 3et + 1 6e−2t ′ +

  • −1

2 + 1 3et + 1 6e−2t

  • −2

t

0 −1

2 + 1 3ez + 1 6e−2z dz = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t −2

  • −1

2z+ 1 3ez − 1 12e−2z t = 1 3et − 1 3e−2t − 1 2 + 1 3et + 1 6e−2t +t −0− 2 3et + 2 3 + 1 6e−2t − 1 6 = et 1 3 + 1 3 − 2 3

  • +e−2t
  • −1

3 + 1 6 + 1 6

  • +t +
  • −1

2 + 2 3 − 1 6

  • =

t √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations