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Agenda Section 5.6 Reminders Exam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today Office hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab office hour Sun 7-8 pm (MathLab) (Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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Page 1: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Agenda

Section 5.6

Reminders

Exam 2 on 11/168-9:30 pm (Chem 1800)

WebHW due today

Office hours Tues, Thurs1-2 pm (5852 East Hall)

MathLab office hourSun 7-8 pm (MathLab)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 2: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

§5.6 DE’s and Discontinuous Forcing Functions

Objectives

Be able to solve DE’s with discontinuous forcing terms.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 3: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + y = f (t), y(0) = 5, y ′(0) = 3

where

f (t) =

{1, 0 ≤ t < π/2,0, π/2 ≤ t.

We begin by expressing f (t) using unit step functions. That is,

f (t) = 1− uπ/2(t).

Therefore, the DE can be written as

y ′′ + y = 1− uπ/2(t).

Applying the Laplace transform to both sides gives us

L{y ′′}+ L{y} = L{1} − L{uπ/2(t)

}.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 4: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + y = f (t), y(0) = 5, y ′(0) = 3

where

f (t) =

{1, 0 ≤ t < π/2,0, π/2 ≤ t.

We begin by expressing f (t) using unit step functions. That is,

f (t) = 1− uπ/2(t).

Therefore, the DE can be written as

y ′′ + y = 1− uπ/2(t).

Applying the Laplace transform to both sides gives us

L{y ′′}+ L{y} = L{1} − L{uπ/2(t)

}.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 5: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + y = f (t), y(0) = 5, y ′(0) = 3

where

f (t) =

{1, 0 ≤ t < π/2,0, π/2 ≤ t.

We begin by expressing f (t) using unit step functions. That is,

f (t) = 1− uπ/2(t).

Therefore, the DE can be written as

y ′′ + y = 1− uπ/2(t).

Applying the Laplace transform to both sides gives us

L{y ′′}+ L{y} = L{1} − L{uπ/2(t)

}.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 6: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + y = f (t), y(0) = 5, y ′(0) = 3

where

f (t) =

{1, 0 ≤ t < π/2,0, π/2 ≤ t.

We begin by expressing f (t) using unit step functions. That is,

f (t) = 1− uπ/2(t).

Therefore, the DE can be written as

y ′′ + y = 1− uπ/2(t).

Applying the Laplace transform to both sides gives us

L{y ′′}+ L{y} = L{1} − L{uπ/2(t)

}.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 7: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 8: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 9: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 10: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 11: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 12: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 13: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

L{y ′′}

+ L{y} = L{1} − L{uπ/2(t)

}.

Looking at the table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s−

e−πs/2

s.

s2Y (s)− 5s − 3 + Y (s) =1

s−

e−πs/2

s.

Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us

s2Y (s) + Y (s) = 5s + 3 +1

s−

e−πs/2

s.

(s2 + 1)Y (s) = 5s + 3 +1

s−

e−πs/2

s.

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 14: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 15: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 16: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 17: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 18: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 19: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 20: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s(s2 + 1)−

e−πs/2

s(s2 + 1)

We need to perform completing the square on the last to terms on the right.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) give us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

As a result, A = 1, B = −1, and C = 0. Therefore,

1

s(s2 + 1)=

1

s−

s

s2 + 1

and

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 21: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 22: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 24: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 26: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 27: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) = 5s

s2 + 1+ 3

1

s2 + 1+

1

s−

s

s2 + 1− e−πs/2

(1

s−

s

s2 + 1

)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us

y(t) = 4L−1

{s

s2 + 1

}+3L−1

{1

s2 + 1

}+L−1

{1

s

}−L−1

{e−πs/2

(1

s−

s

s2 + 1

)}.

By referring to a table, we see that

L{sin at} =a

s2 + a2, L{cos at} =

s

s2 + a2, L{1} =

1

s,

and L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1

{1

s−

s

s2 + 1

}t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 28: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)

Our solution can be expressed as piecewise defined function

y(t) =

{4 cos t + 3 sin t + 1, 0 ≤ t < π/2,4 cos t + 4 sin t, π/2 ≤ t.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 29: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.

We begin by applying the Laplace transform to both sides of the DE to get

L{y (4)}

+ 5L{y ′′}

+ 4L{y} = L{1} − L{uπ(t)} .

By looking at a table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s,

andL{f (4)(t)

}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).

Therefore,

s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)

]+4Y (s) =

1

s−e−πs

s.

Plugging in the initial conditions gives us

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 30: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.

We begin by applying the Laplace transform to both sides of the DE to get

L{y (4)}

+ 5L{y ′′}

+ 4L{y} = L{1} − L{uπ(t)} .

By looking at a table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s,

andL{f (4)(t)

}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).

Therefore,

s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)

]+4Y (s) =

1

s−e−πs

s.

Plugging in the initial conditions gives us

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 31: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.

We begin by applying the Laplace transform to both sides of the DE to get

L{y (4)}

+ 5L{y ′′}

+ 4L{y} = L{1} − L{uπ(t)} .

By looking at a table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s,

andL{f (4)(t)

}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).

Therefore,

s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)

]+4Y (s) =

1

s−e−πs

s.

Plugging in the initial conditions gives us

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 32: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.

We begin by applying the Laplace transform to both sides of the DE to get

L{y (4)}

+ 5L{y ′′}

+ 4L{y} = L{1} − L{uπ(t)} .

By looking at a table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s,

andL{f (4)(t)

}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).

Therefore,

s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)

]+4Y (s) =

1

s−e−πs

s.

Plugging in the initial conditions gives us

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 33: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.

We begin by applying the Laplace transform to both sides of the DE to get

L{y (4)}

+ 5L{y ′′}

+ 4L{y} = L{1} − L{uπ(t)} .

By looking at a table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s,

andL{f (4)(t)

}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).

Therefore,

s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)

]+4Y (s) =

1

s−e−πs

s.

Plugging in the initial conditions gives us

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 34: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.

We begin by applying the Laplace transform to both sides of the DE to get

L{y (4)}

+ 5L{y ′′}

+ 4L{y} = L{1} − L{uπ(t)} .

By looking at a table, we see that

L{f ′′(t)

}= s2F (s)− sf (0)− f ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s,

andL{f (4)(t)

}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).

Therefore,

s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)

]+4Y (s) =

1

s−e−πs

s.

Plugging in the initial conditions gives us

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 35: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 36: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 38: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 39: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 40: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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s4Y (s) + 5s2Y (s) + 4Y (s) =1

s−

e−πs

s

(s4 + 5s2 + 4)Y (s) =1

s−

e−πs

s

(s2 + 1)(s2 + 4)Y (s) =1

s−

e−πs

s

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4)

We need to find the partial fraction decomposition of the two terms on the right.

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Multiplying both sides by s(s2 + 1)(s2 + 4) gives us

1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.

Expanding, moving all terms to the right-hand side, and group by power of s gives us

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 42: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)

The only way for this polynomial to be satisfied for all s is if

A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.

Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Therefore,1

s(s2 + 1)(s2 + 4)=

1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4.

Recall that

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4).

Plugging in the partial fraction decomposition for the two terms on the right gives us

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 43: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)

The only way for this polynomial to be satisfied for all s is if

A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.

Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Therefore,1

s(s2 + 1)(s2 + 4)=

1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4.

Recall that

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4).

Plugging in the partial fraction decomposition for the two terms on the right gives us

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 44: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)

The only way for this polynomial to be satisfied for all s is if

A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.

Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Therefore,1

s(s2 + 1)(s2 + 4)=

1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4.

Recall that

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4).

Plugging in the partial fraction decomposition for the two terms on the right gives us

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 45: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)

The only way for this polynomial to be satisfied for all s is if

A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.

Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Therefore,1

s(s2 + 1)(s2 + 4)=

1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4.

Recall that

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4).

Plugging in the partial fraction decomposition for the two terms on the right gives us

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 46: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)

The only way for this polynomial to be satisfied for all s is if

A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.

Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Therefore,1

s(s2 + 1)(s2 + 4)=

1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4.

Recall that

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4).

Plugging in the partial fraction decomposition for the two terms on the right gives us

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 47: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)

The only way for this polynomial to be satisfied for all s is if

A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.

Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form

1

s(s2 + 1)(s2 + 4)=

A

s+

Bs + C

s2 + 1+

Ds + 2E

s2 + 4.

Therefore,1

s(s2 + 1)(s2 + 4)=

1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4.

Recall that

Y (s) =1

s(s2 + 1)(s2 + 4)−

e−πs

s(s2 + 1)(s2 + 4).

Plugging in the partial fraction decomposition for the two terms on the right gives us

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 48: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)Applying the inverse Laplace transform to both sides gives us

y(t) =1

4L−1

{1

s

}−

1

3L−1

{s

s2 + 1

}+

1

12L−1

{s

s2 + 4

}− L−1

{e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}.

By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)L−1

{(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4−

1

3cos t +

1

12cos 2t

)t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 49: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)Applying the inverse Laplace transform to both sides gives us

y(t) =1

4L−1

{1

s

}−

1

3L−1

{s

s2 + 1

}+

1

12L−1

{s

s2 + 4

}− L−1

{e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}.

By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)L−1

{(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4−

1

3cos t +

1

12cos 2t

)t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 50: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)Applying the inverse Laplace transform to both sides gives us

y(t) =1

4L−1

{1

s

}−

1

3L−1

{s

s2 + 1

}+

1

12L−1

{s

s2 + 4

}− L−1

{e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}.

By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)L−1

{(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4−

1

3cos t +

1

12cos 2t

)t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 51: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)Applying the inverse Laplace transform to both sides gives us

y(t) =1

4L−1

{1

s

}−

1

3L−1

{s

s2 + 1

}+

1

12L−1

{s

s2 + 4

}− L−1

{e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}.

By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)L−1

{(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4−

1

3cos t +

1

12cos 2t

)t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 52: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)Applying the inverse Laplace transform to both sides gives us

y(t) =1

4L−1

{1

s

}−

1

3L−1

{s

s2 + 1

}+

1

12L−1

{s

s2 + 4

}− L−1

{e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}.

By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)L−1

{(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4−

1

3cos t +

1

12cos 2t

)t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 53: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4− e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)Applying the inverse Laplace transform to both sides gives us

y(t) =1

4L−1

{1

s

}−

1

3L−1

{s

s2 + 1

}+

1

12L−1

{s

s2 + 4

}− L−1

{e−πs

(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}.

By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)L−1

{(1

4s−

1

3

s

s2 + 1+

1

12

s

s2 + 4

)}t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4−

1

3cos t +

1

12cos 2t

)t→t−π

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 54: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

y(t) =1

4−

1

3cos t +

1

12cos 2t − uπ(t)

[1

4−

1

3cos (t − π) +

1

12cos (2(t − π))

]y(t) =

1

4−

1

3cos t +

1

12cos 2t − uπ(t)

(1

4+

1

3cos t +

1

12cos 2t

)y(t) =

{(3− 4 cos t + cos 2t) /12 0 ≤ t < π−2(cos t)/3 π ≤ t

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 55: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,

where

f (t) = 1 + 215∑k=1

(−1)kukπ(t).

We begin by applying the Laplace transform to both sides of the DE

L{y ′′}

+ L{y} = L{1}+ 215∑k=1

(−1)kL{ukπ(t)}.

By looking in a table, we’ll see that

L{f ′′(t)

}= s2F (s)− sy(0)− y ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 56: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,

where

f (t) = 1 + 215∑k=1

(−1)kukπ(t).

We begin by applying the Laplace transform to both sides of the DE

L{y ′′}

+ L{y} = L{1}+ 215∑k=1

(−1)kL{ukπ(t)}.

By looking in a table, we’ll see that

L{f ′′(t)

}= s2F (s)− sy(0)− y ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 57: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,

where

f (t) = 1 + 215∑k=1

(−1)kukπ(t).

We begin by applying the Laplace transform to both sides of the DE

L{y ′′}

+ L{y} = L{1}+ 215∑k=1

(−1)kL{ukπ(t)}.

By looking in a table, we’ll see that

L{f ′′(t)

}= s2F (s)− sy(0)− y ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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Example

Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,

where

f (t) = 1 + 215∑k=1

(−1)kukπ(t).

We begin by applying the Laplace transform to both sides of the DE

L{y ′′}

+ L{y} = L{1}+ 215∑k=1

(−1)kL{ukπ(t)}.

By looking in a table, we’ll see that

L{f ′′(t)

}= s2F (s)− sy(0)− y ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 59: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,

where

f (t) = 1 + 215∑k=1

(−1)kukπ(t).

We begin by applying the Laplace transform to both sides of the DE

L{y ′′}

+ L{y} = L{1}+ 215∑k=1

(−1)kL{ukπ(t)}.

By looking in a table, we’ll see that

L{f ′′(t)

}= s2F (s)− sy(0)− y ′(0), L{1} =

1

s, L{uc (t)} =

e−cs

s.

Therefore,

s2Y (s)− sy(0)− y ′(0) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

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s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 62: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 63: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 64: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 65: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 66: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

s2Y (s) + Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

(s2 + 1)Y (s) =1

s+ 2

15∑k=1

(−1)ke−kπs

s

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1)

We need to perform a partial fraction expansion.

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1

Multiplying both sides by s(s2 + 1) gives us

1 = A(s2 + 1) + (Bs + C)s.

Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us

0 = (A + B)s2 + Cs + (A− 1).

The only way for this equation to be satisfied for all s is if

A + B = 0, C = 0, and A− 1 = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 67: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

A + B = 0, C = 0, and A− 1 = 0.

Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1.

Plugging in our values for A, B, and C gives us

1

s(s2 + 1)=

1

s−

s

s2 + 1.

Recall that

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1).

Plugging in the result from our partial fraction decomposition gives us

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 68: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

A + B = 0, C = 0, and A− 1 = 0.

Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1.

Plugging in our values for A, B, and C gives us

1

s(s2 + 1)=

1

s−

s

s2 + 1.

Recall that

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1).

Plugging in the result from our partial fraction decomposition gives us

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 69: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

A + B = 0, C = 0, and A− 1 = 0.

Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1.

Plugging in our values for A, B, and C gives us

1

s(s2 + 1)=

1

s−

s

s2 + 1.

Recall that

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1).

Plugging in the result from our partial fraction decomposition gives us

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 70: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

A + B = 0, C = 0, and A− 1 = 0.

Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1.

Plugging in our values for A, B, and C gives us

1

s(s2 + 1)=

1

s−

s

s2 + 1.

Recall that

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1).

Plugging in the result from our partial fraction decomposition gives us

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 71: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

A + B = 0, C = 0, and A− 1 = 0.

Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was

1

s(s2 + 1)=

A

s+

Bs + C

s2 + 1.

Plugging in our values for A, B, and C gives us

1

s(s2 + 1)=

1

s−

s

s2 + 1.

Recall that

Y (s) =1

s(s2 + 1)+ 2

15∑k=1

(−1)ke−kπs

s(s2 + 1).

Plugging in the result from our partial fraction decomposition gives us

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

).

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 72: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

)Applying the inverse Laplace transform gives us

y(t) = L−1

{1

s

}− L−1

{s

s2 + 1

}+ 2

15∑k=1

(−1)kL−1

{e−kπs

(1

s−

s

s2 + 1

)}By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t)L−1

{(1

s−

s

s2 + 1

)}t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) (1− cos t)t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 73: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

)Applying the inverse Laplace transform gives us

y(t) = L−1

{1

s

}− L−1

{s

s2 + 1

}+ 2

15∑k=1

(−1)kL−1

{e−kπs

(1

s−

s

s2 + 1

)}By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t)L−1

{(1

s−

s

s2 + 1

)}t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) (1− cos t)t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 74: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

)Applying the inverse Laplace transform gives us

y(t) = L−1

{1

s

}− L−1

{s

s2 + 1

}+ 2

15∑k=1

(−1)kL−1

{e−kπs

(1

s−

s

s2 + 1

)}By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t)L−1

{(1

s−

s

s2 + 1

)}t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) (1− cos t)t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 75: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

)Applying the inverse Laplace transform gives us

y(t) = L−1

{1

s

}− L−1

{s

s2 + 1

}+ 2

15∑k=1

(−1)kL−1

{e−kπs

(1

s−

s

s2 + 1

)}By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t)L−1

{(1

s−

s

s2 + 1

)}t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) (1− cos t)t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 76: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

)Applying the inverse Laplace transform gives us

y(t) = L−1

{1

s

}− L−1

{s

s2 + 1

}+ 2

15∑k=1

(−1)kL−1

{e−kπs

(1

s−

s

s2 + 1

)}By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t)L−1

{(1

s−

s

s2 + 1

)}t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) (1− cos t)t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 77: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Y (s) =1

s−

s

s2 + 1+ 2

15∑k=1

(−1)ke−kπs

(1

s−

s

s2 + 1

)Applying the inverse Laplace transform gives us

y(t) = L−1

{1

s

}− L−1

{s

s2 + 1

}+ 2

15∑k=1

(−1)kL−1

{e−kπs

(1

s−

s

s2 + 1

)}By referring to a table, we see that

L{cos at} =s

s2 + a2, L{1} =

1

s, L{uc (t)f (t − c)} = e−csF (s).

Therefore,

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t)L−1

{(1

s−

s

s2 + 1

)}t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) (1− cos t)t→t−kπ

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 78: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

y(t) = 1− cos t + 215∑k=1

(−1)kukπ(t) [1− cos (t − kπ)].

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 79: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + 3y ′ + 2y = f (t), y(0) = 0, y ′(0) = 0,

where

f (t) =

{1, 0 ≤ t < 10,0, t ≥ 10.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 80: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + y = u3π(t), y(0) = 1, y ′(0) = 0.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 81: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + y = g(t), y(0) = 6, y ′(0) = 8,

where

g(t) =

{t/2, 0 ≤ t < 6,3, t ≥ 6.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations

Page 82: Section 5.6 Exam 2 on 11/16 WebHW due today O ce hours ...gmarple/Nov10.pdfExam 2 on 11/16 8-9:30 pm (Chem 1800) WebHW due today O ce hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab

Example

Use the Laplace transform to solve the IVP

y ′′ + 4y = uπ(t)− u3π(t), y(0) = 3, y ′(0) = 7.

(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations