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Overview An Example Double Check Further Discussion Variation of Parameters Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double

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SLIDE 1

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

Bernd Schr¨

  • der

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 2

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 3

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp +yh

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 4

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp +yh

  • 2. Variation of Parameters is a way to obtain a particular

solution of the inhomogeneous equation.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 5

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp +yh

  • 2. Variation of Parameters is a way to obtain a particular

solution of the inhomogeneous equation.

  • 3. The particular solution can be obtained as follows.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-6
SLIDE 6

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp +yh

  • 2. Variation of Parameters is a way to obtain a particular

solution of the inhomogeneous equation.

  • 3. The particular solution can be obtained as follows.

3.1 Assume that the parameters in the solution of the homogeneous equation are functions.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 7

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp +yh

  • 2. Variation of Parameters is a way to obtain a particular

solution of the inhomogeneous equation.

  • 3. The particular solution can be obtained as follows.

3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-8
SLIDE 8

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

  • 1. The general solution of an inhomogeneous linear

differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp +yh

  • 2. Variation of Parameters is a way to obtain a particular

solution of the inhomogeneous equation.

  • 3. The particular solution can be obtained as follows.

3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 9

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

For second order equations, we obtain the general formula

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 10

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

For second order equations, we obtain the general formula

yp(x) = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 11

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

For second order equations, we obtain the general formula

yp(x) = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt,

where y1, y2 are linearly independent solutions of the homogeneous equation and W(y1,y2) := y1y′

2 −y2y′ 1.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 12

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

For second order equations, we obtain the general formula

yp(x) = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt,

where y1, y2 are linearly independent solutions of the homogeneous equation and W(y1,y2) := y1y′

2 −y2y′ 1.

W is also called the Wronskian of y1 and y2.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 13

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

For second order equations, we obtain the general formula

yp(x) = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt,

where y1, y2 are linearly independent solutions of the homogeneous equation and W(y1,y2) := y1y′

2 −y2y′ 1.

W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as W(y1,y2) = det y1 y2 y′

1

y′

2

  • .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 14

logo1 Overview An Example Double Check Further Discussion

Variation of Parameters

For second order equations, we obtain the general formula

yp(x) = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt,

where y1, y2 are linearly independent solutions of the homogeneous equation and W(y1,y2) := y1y′

2 −y2y′ 1.

W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as W(y1,y2) = det y1 y2 y′

1

y′

2

  • .

That’s it.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 15

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 16

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 17

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation. y′′ +4y′ +4y =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 18

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation. y′′ +4y′ +4y = λ 2eλx +4λeλx +4eλx =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 19

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation. y′′ +4y′ +4y = λ 2eλx +4λeλx +4eλx = λ 2 +4λ +4 =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 20

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation. y′′ +4y′ +4y = λ 2eλx +4λeλx +4eλx = λ 2 +4λ +4 = λ1,2 = −4± √ 16−16 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 21

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation. y′′ +4y′ +4y = λ 2eλx +4λeλx +4eλx = λ 2 +4λ +4 = λ1,2 = −4± √ 16−16 2 = −2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 22

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Solution of the homogeneous equation. y′′ +4y′ +4y = λ 2eλx +4λeλx +4eλx = λ 2 +4λ +4 = λ1,2 = −4± √ 16−16 2 = −2 yh = c1e−2x +c2xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 23

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 24

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the Wronskian. W(y1,y2)(t) = e−2t e−2t −2te−2t −te−2t −2e−2t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 25

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the Wronskian. W(y1,y2)(t) = e−2t e−2t −2te−2t −te−2t −2e−2t = e−4t −2te−4t +2te−4t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

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SLIDE 26

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the Wronskian. W(y1,y2)(t) = e−2t e−2t −2te−2t −te−2t −2e−2t = e−4t −2te−4t +2te−4t = e−4t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-27
SLIDE 27

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-28
SLIDE 28

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y1(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-29
SLIDE 29

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y1(t) dt =

  • 1

e−4t sin(t) 1 e−2t dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-30
SLIDE 30

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y1(t) dt =

  • 1

e−4t sin(t) 1 e−2t dt =

  • e2t sin(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-31
SLIDE 31

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-32
SLIDE 32

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t sin(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-33
SLIDE 33

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t sin(t) dt

= 1 2e2t sin(t)−

1

2e2t cos(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-34
SLIDE 34

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t sin(t) dt

= 1 2e2t sin(t)−

1

2e2t cos(t) dt = 1 2e2t sin(t)− 1 2 1 2e2t cos(t)−

1

2e2t −sin(t)

  • dt
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-35
SLIDE 35

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t sin(t) dt

= 1 2e2t sin(t)−

1

2e2t cos(t) dt = 1 2e2t sin(t)− 1 2 1 2e2t cos(t)−

1

2e2t −sin(t)

  • dt
  • =

1 2e2t sin(t)− 1 4e2t cos(t)− 1 4

  • e2t sin(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-36
SLIDE 36

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t sin(t) dt

= 1 2e2t sin(t)−

1

2e2t cos(t) dt = 1 2e2t sin(t)− 1 2 1 2e2t cos(t)−

1

2e2t −sin(t)

  • dt
  • =

1 2e2t sin(t)− 1 4e2t cos(t)− 1 4

  • e2t sin(t) dt

5 4

  • e2t sin(t) dt

= 1 2e2t sin(t)− 1 4e2t cos(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-37
SLIDE 37

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t sin(t) dt

= 1 2e2t sin(t)−

1

2e2t cos(t) dt = 1 2e2t sin(t)− 1 2 1 2e2t cos(t)−

1

2e2t −sin(t)

  • dt
  • =

1 2e2t sin(t)− 1 4e2t cos(t)− 1 4

  • e2t sin(t) dt

5 4

  • e2t sin(t) dt

= 1 2e2t sin(t)− 1 4e2t cos(t)

  • e2t sin(t) dt

= 2 5e2t sin(t)− 1 5e2t cos(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-38
SLIDE 38

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-39
SLIDE 39

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-40
SLIDE 40

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt =

  • 1

e−4t sin(t) 1 te−2t dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-41
SLIDE 41

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt =

  • 1

e−4t sin(t) 1 te−2t dt =

  • te2t sin(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-42
SLIDE 42

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt =

  • 1

e−4t sin(t) 1 te−2t dt =

  • te2t sin(t) dt

= t 2 5e2t sin(t)− 1 5e2t cos(t)

  • −

2

5e2t sin(t)− 1 5e2t cos(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-43
SLIDE 43

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt =

  • 1

e−4t sin(t) 1 te−2t dt =

  • te2t sin(t) dt

= t 2 5e2t sin(t)− 1 5e2t cos(t)

  • −

2

5e2t sin(t)− 1 5e2t cos(t) dt = 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5

  • e2t sin(t) dt + 1

5

  • e2t cos(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-44
SLIDE 44

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-45
SLIDE 45

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t cos(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-46
SLIDE 46

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t cos(t) dt

= 1 2e2t cos(t)+

1

2e2t sin(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-47
SLIDE 47

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t cos(t) dt

= 1 2e2t cos(t)+

1

2e2t sin(t) dt = 1 2e2t cos(t)+ 1 2 1 2e2t sin(t)−

1

2e2t cos(t) dt

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-48
SLIDE 48

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t cos(t) dt

= 1 2e2t cos(t)+

1

2e2t sin(t) dt = 1 2e2t cos(t)+ 1 2 1 2e2t sin(t)−

1

2e2t cos(t) dt

  • =

1 2e2t cos(t)+ 1 4e2t sin(t)− 1 4

  • e2t cos(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-49
SLIDE 49

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t cos(t) dt

= 1 2e2t cos(t)+

1

2e2t sin(t) dt = 1 2e2t cos(t)+ 1 2 1 2e2t sin(t)−

1

2e2t cos(t) dt

  • =

1 2e2t cos(t)+ 1 4e2t sin(t)− 1 4

  • e2t cos(t) dt

5 4

  • e2t cos(t) dt

= 1 2e2t cos(t)+ 1 4e2t sin(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-50
SLIDE 50

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • e2t cos(t) dt

= 1 2e2t cos(t)+

1

2e2t sin(t) dt = 1 2e2t cos(t)+ 1 2 1 2e2t sin(t)−

1

2e2t cos(t) dt

  • =

1 2e2t cos(t)+ 1 4e2t sin(t)− 1 4

  • e2t cos(t) dt

5 4

  • e2t cos(t) dt

= 1 2e2t cos(t)+ 1 4e2t sin(t)

  • e2t cos(t) dt

= 2 5e2t cos(t)+ 1 5e2t sin(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-51
SLIDE 51

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-52
SLIDE 52

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-53
SLIDE 53

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt = 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5

  • e2t sin(t) dt + 1

5

  • e2t cos(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-54
SLIDE 54

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt = 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5

  • e2t sin(t) dt + 1

5

  • e2t cos(t) dt

= 2 5te2t sin(t)− 1 5te2t cos(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-55
SLIDE 55

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt = 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5

  • e2t sin(t) dt + 1

5

  • e2t cos(t) dt

= 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5 2 5e2t sin(t)− 1 5e2t cos(t)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-56
SLIDE 56

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt = 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5

  • e2t sin(t) dt + 1

5

  • e2t cos(t) dt

= 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5 2 5e2t sin(t)− 1 5e2t cos(t)

  • +1

5 2 5e2t cos(t)+ 1 5e2t sin(t)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-57
SLIDE 57

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Computing the integrals.

  • 1

W(y1,y2)(t) f(t) a2(t)y2(t) dt = 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5

  • e2t sin(t) dt + 1

5

  • e2t cos(t) dt

= 2 5te2t sin(t)− 1 5te2t cos(t)− 2 5 2 5e2t sin(t)− 1 5e2t cos(t)

  • +1

5 2 5e2t cos(t)+ 1 5e2t sin(t)

  • =

2 5te2t sin(t)− 1 5te2t cos(t)− 3 25e2t sin(t)+ 4 25e2t cos(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-58
SLIDE 58

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-59
SLIDE 59

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Stating the general solution.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-60
SLIDE 60

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Stating the general solution. yp = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-61
SLIDE 61

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Stating the general solution. yp = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt = −e−2x 2 5xe2x sin(x)− 1 5xe2x cos(x)− 3 25e2x sin(x)+ 4 25e2x cos(x)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-62
SLIDE 62

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Stating the general solution. yp = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt = −e−2x 2 5xe2x sin(x)− 1 5xe2x cos(x)− 3 25e2x sin(x)+ 4 25e2x cos(x)

  • +xe−2x

2 5e2x sin(x)− 1 5e2x cos(x)

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-63
SLIDE 63

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Stating the general solution. yp = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt = −e−2x 2 5xe2x sin(x)− 1 5xe2x cos(x)− 3 25e2x sin(x)+ 4 25e2x cos(x)

  • +xe−2x

2 5e2x sin(x)− 1 5e2x cos(x)

  • =

3 25 sin(x)− 4 25 cos(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-64
SLIDE 64

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Stating the general solution. yp = −y1(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y2(t) dt +y2(x)

x

x0

1 W(y1,y2)(t) f(t) a2(t)y1(t) dt = −e−2x 2 5xe2x sin(x)− 1 5xe2x cos(x)− 3 25e2x sin(x)+ 4 25e2x cos(x)

  • +xe−2x

2 5e2x sin(x)− 1 5e2x cos(x)

  • =

3 25 sin(x)− 4 25 cos(x) y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-65
SLIDE 65

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-66
SLIDE 66

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-67
SLIDE 67

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-68
SLIDE 68

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-69
SLIDE 69

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-70
SLIDE 70

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-71
SLIDE 71

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-72
SLIDE 72

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25 = y′(0)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-73
SLIDE 73

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25 = y′(0) = 3 25 −2c1 +c2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-74
SLIDE 74

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25 = y′(0) = 3 25 −2c1 +c2, c2 = 2c1 − 3 25

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-75
SLIDE 75

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25 = y′(0) = 3 25 −2c1 +c2, c2 = 2c1 − 3 25 = 55 25

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-76
SLIDE 76

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25 = y′(0) = 3 25 −2c1 +c2, c2 = 2c1 − 3 25 = 55 25 = 11 5

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-77
SLIDE 77

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

Finding c1, c2. y = − 4 25 cos(x)+ 3 25 sin(x)+c1e−2x +c2xe−2x y′ = 4 25 sin(x)+ 3 25 cos(x)−2c1e−2x +c2

  • e−2x −2xe−2x

1 = y(0) = − 4 25 +c1, c1 = 29 25 = y′(0) = 3 25 −2c1 +c2, c2 = 2c1 − 3 25 = 55 25 = 11 5 y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-78
SLIDE 78

logo1 Overview An Example Double Check Further Discussion

Solve the Initial Value Problem y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0

y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-79
SLIDE 79

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-80
SLIDE 80

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-81
SLIDE 81

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-82
SLIDE 82

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-83
SLIDE 83

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

  • −16

25 + 12 25 + 4 25

  • cos(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-84
SLIDE 84

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

  • −16

25 + 12 25 + 4 25

  • cos(x)+

12 25 + 16 25 − 3 25

  • sin(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-85
SLIDE 85

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

  • −16

25 + 12 25 + 4 25

  • cos(x)+

12 25 + 16 25 − 3 25

  • sin(x)

+ 116 25 −232 25 +44 5 +116 25 −44 5

  • e−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-86
SLIDE 86

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

  • −16

25 + 12 25 + 4 25

  • cos(x)+

12 25 + 16 25 − 3 25

  • sin(x)

+ 116 25 −232 25 +44 5 +116 25 −44 5

  • e−2x+

44 5 −88 5 +44 5

  • xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-87
SLIDE 87

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

  • −16

25 + 12 25 + 4 25

  • cos(x)+

12 25 + 16 25 − 3 25

  • sin(x)

+ 116 25 −232 25 +44 5 +116 25 −44 5

  • e−2x+

44 5 −88 5 +44 5

  • xe−2x

?

= sin(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-88
SLIDE 88

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0? 4

  • − 4

25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x

  • +4

4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 11 5

  • e−2x −2xe−2x

+ 4 25 cos(x)− 3 25 sin(x)+ 116 25 e−2x + 11 5

  • −4e−2x +4xe−2x

?

= sin(x)

  • −16

25 + 12 25 + 4 25

  • cos(x)+

12 25 + 16 25 − 3 25

  • sin(x)

+ 116 25 −232 25 +44 5 +116 25 −44 5

  • e−2x+

44 5 −88 5 +44 5

  • xe−2x

√

= sin(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-89
SLIDE 89

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-90
SLIDE 90

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-91
SLIDE 91

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-92
SLIDE 92

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-93
SLIDE 93

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √ y′(x) = 4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 55 25

  • e−2x −2xe−2x

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-94
SLIDE 94

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √ y′(x) = 4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 55 25

  • e−2x −2xe−2x

y′(0) = 3 25 − 58 25 + 55 25

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-95
SLIDE 95

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √ y′(x) = 4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 55 25

  • e−2x −2xe−2x

y′(0) = 3 25 − 58 25 + 55 25 = 0

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-96
SLIDE 96

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √ y′(x) = 4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 55 25

  • e−2x −2xe−2x

y′(0) = 3 25 − 58 25 + 55 25 = 0 √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-97
SLIDE 97

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √ y′(x) = 4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 55 25

  • e−2x −2xe−2x

y′(0) = 3 25 − 58 25 + 55 25 = 0 √ Alternatively, use a computer to double check the result.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-98
SLIDE 98

logo1 Overview An Example Double Check Further Discussion

Does y = − 4 25 cos(x)+ 3 25 sin(x)+ 29 25e−2x + 11 5 xe−2x Solve y′′ +4y′ +4y = sin(x), y(0) = 1, y′(0) = 0?

y(0) = − 4 25 + 29 25 = 1 √ y′(x) = 4 25 sin(x)+ 3 25 cos(x)− 58 25e−2x + 55 25

  • e−2x −2xe−2x

y′(0) = 3 25 − 58 25 + 55 25 = 0 √ Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-99
SLIDE 99

logo1 Overview An Example Double Check Further Discussion

Double Checking with a Computer Algebra System

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-100
SLIDE 100

logo1 Overview An Example Double Check Further Discussion

Double Checking with a Computer Algebra System

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-101
SLIDE 101

logo1 Overview An Example Double Check Further Discussion

Double Checking with a Computer Algebra System

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-102
SLIDE 102

logo1 Overview An Example Double Check Further Discussion

Double Checking with a Computer Algebra System

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-103
SLIDE 103

logo1 Overview An Example Double Check Further Discussion

Double Checking with a Computer Algebra System

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-104
SLIDE 104

logo1 Overview An Example Double Check Further Discussion

Double Checking with a Computer Algebra System

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-105
SLIDE 105

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-106
SLIDE 106

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

  • 1. The method can be applied to higher order equations,

systems, and equations with non-constant coefficients.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-107
SLIDE 107

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

  • 1. The method can be applied to higher order equations,

systems, and equations with non-constant coefficients.

  • 2. Integrals get challenging or even impossible.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-108
SLIDE 108

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

  • 1. The method can be applied to higher order equations,

systems, and equations with non-constant coefficients.

  • 2. Integrals get challenging or even impossible.
  • 3. So the Derivative Form of the Fundamental Theorem of

Calculus and numerical integration are frequently used here.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-109
SLIDE 109

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

  • 1. The method can be applied to higher order equations,

systems, and equations with non-constant coefficients.

  • 2. Integrals get challenging or even impossible.
  • 3. So the Derivative Form of the Fundamental Theorem of

Calculus and numerical integration are frequently used here.

  • 4. Remember, the beauty is that we can get a solution.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-110
SLIDE 110

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

  • 1. The method can be applied to higher order equations,

systems, and equations with non-constant coefficients.

  • 2. Integrals get challenging or even impossible.
  • 3. So the Derivative Form of the Fundamental Theorem of

Calculus and numerical integration are frequently used here.

  • 4. Remember, the beauty is that we can get a solution.
  • 5. The simple solutions of toy problems are best forgotten.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters

slide-111
SLIDE 111

logo1 Overview An Example Double Check Further Discussion

Further Remarks on Variation of Parameters

  • 1. The method can be applied to higher order equations,

systems, and equations with non-constant coefficients.

  • 2. Integrals get challenging or even impossible.
  • 3. So the Derivative Form of the Fundamental Theorem of

Calculus and numerical integration are frequently used here.

  • 4. Remember, the beauty is that we can get a solution.
  • 5. The simple solutions of toy problems are best forgotten.
  • 6. We now have a tool that can handle real life situations.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Variation of Parameters