Engineering Mathematics II (2M03)Tutorial 8

Marina ChugunovaDepartment of Math. & Stat., office: HH403

e-mail: chugunom@math.mcmaster.ca

office hours: Math Help Centre, Thursday 1:30 - 3:30

November 1-2, 2007

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 40)

Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.

Solution:

2

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 40)

Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

3

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 40)

Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

Translation theorem:

L{g(t)U(t− a)} = e−asL{g(t + a)}

4

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 40)

Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

Translation theorem:

L{g(t)U(t− a)} = e−asL{g(t + a)}

a = 1, g(t) = 3t + 1, g(t + 1) = 3(t + 1) + 1

5

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 40)

Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

Translation theorem:

L{g(t)U(t− a)} = e−asL{g(t + a)}

a = 1, g(t) = 3t + 1, g(t + 1) = 3(t + 1) + 1

F (s) = L{(3t + 1)U(t− 1)} = e−sL{(3(t + 1) + 1)} = e−sL{(3t + 4)}

6

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 40)

Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

Translation theorem:

L{g(t)U(t− a)} = e−asL{g(t + a)}

a = 1, g(t) = 3t + 1, g(t + 1) = 3(t + 1) + 1

F (s) = L{(3t + 1)U(t− 1)} = e−sL{(3(t + 1) + 1)} = e−sL{(3t + 4)}

F (s) = 3e−s 1

s2+ 4e−s1

s

7

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 60)

Find the Laplace transform of the function:

f (t) =

{sin t, 0 ≤ t < 2π0, t ≥ 2π

Solution:

8

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 60)

Find the Laplace transform of the function:

f (t) =

{sin t, 0 ≤ t < 2π0, t ≥ 2π

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

1− U(t− a) =

{1, 0 ≤ t < a0, t ≥ a

9

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 60)

Find the Laplace transform of the function:

f (t) =

{sin t, 0 ≤ t < 2π0, t ≥ 2π

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

1− U(t− a) =

{1, 0 ≤ t < a0, t ≥ a

f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)

10

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 60)

Find the Laplace transform of the function:

f (t) =

{sin t, 0 ≤ t < 2π0, t ≥ 2π

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

1− U(t− a) =

{1, 0 ≤ t < a0, t ≥ a

f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)

F (s) = L{f (t)} =1

s2 + 1− e−2πsL{sin(t + 2π)}

11

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 60)

Find the Laplace transform of the function:

f (t) =

{sin t, 0 ≤ t < 2π0, t ≥ 2π

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

1− U(t− a) =

{1, 0 ≤ t < a0, t ≥ a

f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)

F (s) = L{f (t)} =1

s2 + 1− e−2πsL{sin(t + 2π)}

F (s) = L{f (t)} =1

s2 + 1− e−2πsL{sin t} =

1

s2 + 1− e−2πs 1

s2 + 1

12

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 60)

Find the Laplace transform of the function:

f (t) =

{sin t, 0 ≤ t < 2π0, t ≥ 2π

Solution:

U(t− a) =

{0, 0 ≤ t < a1, t ≥ a

1− U(t− a) =

{1, 0 ≤ t < a0, t ≥ a

f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)

F (s) = L{f (t)} =1

s2 + 1− e−2πsL{sin(t + 2π)}

F (s) = L{f (t)} =1

s2 + 1− e−2πsL{sin t} =

1

s2 + 1− e−2πs 1

s2 + 1

F (s) =1

s2 + 1(1− e−2πs)

13

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =

{1, 0 ≤ t < 10, t ≥ 1

Solution:

14

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =

{1, 0 ≤ t < 10, t ≥ 1

Solution:

f (t) = 1− U(t− 1), L{f (t)} =1

s− L{U(t− 1)} =

1

s− e−s

s

15

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =

{1, 0 ≤ t < 10, t ≥ 1

Solution:

f (t) = 1− U(t− 1), L{f (t)} =1

s− L{U(t− 1)} =

1

s− e−s

s

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

s(1− e−s), Y (s)[s2 + 4] =

1

s(1− e−s)− 1

16

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =

{1, 0 ≤ t < 10, t ≥ 1

Solution:

f (t) = 1− U(t− 1), L{f (t)} =1

s− L{U(t− 1)} =

1

s− e−s

s

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

s(1− e−s), Y (s)[s2 + 4] =

1

s(1− e−s)− 1

Y (s) =1

s(s2 + 4)− e−s

s(s2 + 4)− 1

s2 + 4

17

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =

{1, 0 ≤ t < 10, t ≥ 1

Solution:

f (t) = 1− U(t− 1), L{f (t)} =1

s− L{U(t− 1)} =

1

s− e−s

s

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

s(1− e−s), Y (s)[s2 + 4] =

1

s(1− e−s)− 1

Y (s) =1

s(s2 + 4)− e−s

s(s2 + 4)− 1

s2 + 4

Y (s) = 1/41

s− 1/4

s

s2 + 4− 1/4

e−s

s+ 1/4

e−ss

s2 + 4− 1/2

2

s2 + 4

18

The Laplace Transform (4.3 Translation Theorems)

Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =

{1, 0 ≤ t < 10, t ≥ 1

Solution:

f (t) = 1− U(t− 1), L{f (t)} =1

s− L{U(t− 1)} =

1

s− e−s

s

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

s(1− e−s), Y (s)[s2 + 4] =

1

s(1− e−s)− 1

Y (s) =1

s(s2 + 4)− e−s

s(s2 + 4)− 1

s2 + 4

Y (s) = 1/41

s− 1/4

s

s2 + 4− 1/4

e−s

s+ 1/4

e−ss

s2 + 4− 1/2

2

s2 + 4

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

19

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 6)Find the Laplace transform F (s) of f (t) = t2 cos t.

Solution:

20

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 6)Find the Laplace transform F (s) of f (t) = t2 cos t.

Solution:

Derivatives of the Laplace transform

L{tny(t)} = (−1)ndn

dsnY (s)

21

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 6)Find the Laplace transform F (s) of f (t) = t2 cos t.

Solution:

Derivatives of the Laplace transform

L{tny(t)} = (−1)ndn

dsnY (s)

F (s) = L{t2 cos t} = (−1)2d2

ds2L{cos t} =

d2

ds2

(s

s2 + 1

)=

2s3 − 6s

(s2 + 1)3

22

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

23

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

24

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

25

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

26

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

27

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

28

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

Y (s) = −e−π2s 1

s(s2 + 1)2+

1

s

29

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem

y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =

{1, 0 ≤ t < π/2sin t, t ≥ π/2

Solution:

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

Y (s) = −e−π2s 1

s(s2 + 1)2+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− s

(s2 + 1)2

)+

1

s30

The Laplace Transform (4.4 Additional Operational Properties )

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

Y (s) = −e−π2s 1

s(s2 + 1)2+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− s

(s2 + 1)2

)+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− 1/2(−1)

d

ds

(1

s2 + 1

))+

1

s

31

The Laplace Transform (4.4 Additional Operational Properties )

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

Y (s) = −e−π2s 1

s(s2 + 1)2+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− s

(s2 + 1)2

)+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− 1/2(−1)

d

ds

(1

s2 + 1

))+

1

s

y(t) = U(t− π/2)[−1 + cos(t− π/2) + 1/2(t− π/2) sin(t− π/2)] + 1

32

The Laplace Transform (4.4 Additional Operational Properties )

f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)

L{g(t)U(t− a)} = e−asL{g(t + a)}

F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +

1

s− 1

se−

π2s

F (s) = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s

s2 + 1+

1

s− 1

se−

π2s

Y (s) = −e−π2s 1

s(s2 + 1)2+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− s

(s2 + 1)2

)+

1

s

Y (s) = −e−π2s

(1

s− s

s2 + 1− 1/2(−1)

d

ds

(1

s2 + 1

))+

1

s

y(t) = U(t− π/2)[−1 + cos(t− π/2) + 1/2(t− π/2) sin(t− π/2)] + 1

y(t) = U(t− π/2)[−1 + sin t− 1/2(t− π/2) cos t] + 1

33

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 26)Find

L{∫ t

0

sin τ τdτ

}Solution:

34

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 26)Find

L{∫ t

0

sin τ τdτ

}Solution:

Transform of the integral

L{∫ t

0

f (τ ) dτ

}=

F (s)

s

35

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 26)Find

L{∫ t

0

sin τ τdτ

}Solution:

Transform of the integral

L{∫ t

0

f (τ ) dτ

}=

F (s)

s

L{∫ t

0

sin τ τdτ

}=

1

sL{t sin t}

36

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 26)Find

L{∫ t

0

sin τ τdτ

}Solution:

Transform of the integral

L{∫ t

0

f (τ ) dτ

}=

F (s)

s

L{∫ t

0

sin τ τdτ

}=

1

sL{t sin t} = −1

s

d

ds

(1

s2 + 1

)

37

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 26)Find

L{∫ t

0

sin τ τdτ

}Solution:

Transform of the integral

L{∫ t

0

f (τ ) dτ

}=

F (s)

s

L{∫ t

0

sin τ τdτ

}=

1

sL{t sin t} = −1

s

d

ds

(1

s2 + 1

)=

2

(s2 + 1)2

38

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 38)Use the Laplace transform to solve the integral equation

f (t) = 2t− 4

∫ t

0

sin τ f(t− τ ) dτ

Solution:

39

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 38)Use the Laplace transform to solve the integral equation

f (t) = 2t− 4

∫ t

0

sin τ f(t− τ ) dτ

Solution:By the Convolution theorem

F (s) = 21

s2− 4

(1

s2 + 1F (s)

)

40

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 38)Use the Laplace transform to solve the integral equation

f (t) = 2t− 4

∫ t

0

sin τ f(t− τ ) dτ

Solution:By the Convolution theorem

F (s) = 21

s2− 4

(1

s2 + 1F (s)

)F (s) =

2(s2 + 1)

s2(s2 + 5)=

2

s2 + 5+

2/5

s2− 2/5

s2 + 5=

41

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 38)Use the Laplace transform to solve the integral equation

f (t) = 2t− 4

∫ t

0

sin τ f(t− τ ) dτ

Solution:By the Convolution theorem

F (s) = 21

s2− 4

(1

s2 + 1F (s)

)F (s) =

2(s2 + 1)

s2(s2 + 5)=

2

s2 + 5+

2/5

s2− 2/5

s2 + 5=

=8

5√

5

√5

s2 + 5+

2

5

1

s2

42

The Laplace Transform (4.4 Additional Operational Properties )

Problem (4.4: 38)Use the Laplace transform to solve the integral equation

f (t) = 2t− 4

∫ t

0

sin τ f(t− τ ) dτ

Solution:By the Convolution theorem

F (s) = 21

s2− 4

(1

s2 + 1F (s)

)F (s) =

2(s2 + 1)

s2(s2 + 5)=

2

s2 + 5+

2/5

s2− 2/5

s2 + 5=

=8

5√

5

√5

s2 + 5+

2

5

1

s2

f (t) =8

5√

5sin 5t +

2

5t

43

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

44

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

sX(s)− 1 = 2Y (s) +1

s− 1, sY (s)− 1 = 8X(s)− 1

s2

45

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

sX(s)− 1 = 2Y (s) +1

s− 1, sY (s)− 1 = 8X(s)− 1

s2

Y (s) = 24

s2 − 16+

s

s2 − 16+

8

(s− 1)(s2 − 16)− 1

s(s2 − 16)

46

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

sX(s)− 1 = 2Y (s) +1

s− 1, sY (s)− 1 = 8X(s)− 1

s2

Y (s) = 24

s2 − 16+

s

s2 − 16+

8

(s− 1)(s2 − 16)− 1

s(s2 − 16)

Y (s) = 24

s2 − 16+

s

s2 − 16− 8/15

1

s− 1+ 1/3

1

s− 4+ 1/5

1

s + 4− 1/16

(s

s2 − 16− 1

s

)

47

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

sX(s)− 1 = 2Y (s) +1

s− 1, sY (s)− 1 = 8X(s)− 1

s2

Y (s) = 24

s2 − 16+

s

s2 − 16+

8

(s− 1)(s2 − 16)− 1

s(s2 − 16)

Y (s) = 24

s2 − 16+

s

s2 − 16− 8/15

1

s− 1+ 1/3

1

s− 4+ 1/5

1

s + 4− 1/16

(s

s2 − 16− 1

s

)y(t) = 2 sinh 4t +

15

16cosh 4t +

1

16− 8

15et +

1

3e4t +

1

5e−4t

48

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

sX(s)− 1 = 2Y (s) +1

s− 1, sY (s)− 1 = 8X(s)− 1

s2

Y (s) = 24

s2 − 16+

s

s2 − 16+

8

(s− 1)(s2 − 16)− 1

s(s2 − 16)

Y (s) = 24

s2 − 16+

s

s2 − 16− 8/15

1

s− 1+ 1/3

1

s− 4+ 1/5

1

s + 4− 1/16

(s

s2 − 16− 1

s

)y(t) = 2 sinh 4t +

15

16cosh 4t +

1

16− 8

15et +

1

3e4t +

1

5e−4t

x(t) = 1/8(y′(t) + t), y′(t) = 8 cosh 4t +15

4sinh 4t− 8

15et +

4

3e4t − 4

5e−4t

49

The Laplace Transform (4.6 Systems of Linear Differential Equations )

Problem (4.6: 2)Use the Laplace transform to solve the system:

dx

dt= 2y + et,

dy

dt= 8x− t, x(0) = 1, y(0) = 1

Solution:

sX(s)− 1 = 2Y (s) +1

s− 1, sY (s)− 1 = 8X(s)− 1

s2

Y (s) = 24

s2 − 16+

s

s2 − 16+

8

(s− 1)(s2 − 16)− 1

s(s2 − 16)

Y (s) = 24

s2 − 16+

s

s2 − 16− 8/15

1

s− 1+ 1/3

1

s− 4+ 1/5

1

s + 4− 1/16

(s

s2 − 16− 1

s

)y(t) = 2 sinh 4t +

15

16cosh 4t +

1

16− 8

15et +

1

3e4t +

1

5e−4t

x(t) = 1/8(y′(t) + t), y′(t) = 8 cosh 4t +15

4sinh 4t− 8

15et +

4

3e4t − 4

5e−4t

x(t) = cosh 4t +15

32sinh 4t− 1

15et +

1

6e4t − 1

10e−4t +

1

8t

50

See you next week :-) !

51