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CHAPTER 8 Forced Equationsand Systems

8.1 The Laplace Transform and Its Inverse

Transforms from the Definition

1. { } ( )00 1 55 5 lim5 0 1st st bbe dt e 5s s s

= = = L =

2. { } 0 20 1 1limst st b stbt te dt te e dt0 1s s s

= = + L

= (integration by parts) 3. { } ( ) ( )2 22 2 00 0 1 1lim 2 2s t s tt t st bbe e e dt e dt es s = = = = L 4. { } ( ) ( )1 1 00 0 1 1lim 1 1s t s tt t st bbe e e dt e dt es s + + = = = =+ + L 5. { } 020 2cos2 sin 2 2sin 2 sin 2 lim 4 4st st bb

t s tt e tdt es s

+ = = = 2+ + L (integration by parts twice) 6. { } 20cos3 cos3 9st

st e tdts

= = +L (integration by parts twice) 7. { }2 2 2 2 30

0

2 2limbst st st

st

b

e te et e t dt t 32

s s s

= = L s = (integration by parts twice)

8. ( ){ } ( ) ( )4 4 400 1 1 11 1 4st st sf t e dt e e es s s = = = = L s 9. ( ){ } ( ) ( )2 2 2 220 2 0 2 0 1 11 1st st st st st sf t t e dt e dt te dt e dt e dt es s = + = + = L 10. ( ){ } ( ) ( )1 2 3 20 1 2 21 1st s sf t t e dt es s s = = +L e

767

768 CHAPTER 8 Laplace Transforms Transforms with Tools

11. { } { } { } { }2 2 2 321 a b ca bt ct a b t c t s s s+ + = + + = + +L L L L 12. { } { } { } ( )1 1 2 11 1 1 1t t se e s s s s ++ = + = + =+ +L L L 13. { } { } { }2 2 2 2 21 1 22 2t t t t 4se e e e s s s + = + = + = + L L L 14. { } { } { } { } ( )22

3 1 23 sin 2 3 1 sin 21 4

t tt e t t e ts s s

+ + = + + = + + + +L L L L

15. ( ){ } { } { } ( ) ( )2 2 2 31 63 3 1 1t t te t t te t e s s + = + = ++ +L L L 16. { } { } { } ( )

( )( )

3 3 3 34 2

4 164 cos3 4 cos33 1

t t t t st e e t t e e ts s

++ = + = +9+ + +L L L

17. { } { } { } 2 22 1 32 2at at at at s ae e e e s a s a s a + = = = + L L L 18. { } { } { } ( )3 3 2 2

1 22sin 2 sin13

t tte t te tss

+ = + = + ++L L L

Linearity of the Laplace Transform 19. Using basic properties of the integral yields

{ } ( ) ( ) ( ) ( )

{ } { }1 2 1 2 1 20 0

1 2 .

st st stc f c g e c f t c g t dt c e f t dt c e g t dt

c f c g

+ = + = + = + LL L

0

Is There a Product Rule? 20. Clearly, no product rule exists, as there is no product rule for integrals. For example,

{ }2 32t s=L but { } 21t s=L , so { } { } { }241t t ts = L L L .

SECTION 8.1 The Laplace Transform and its Inverse 769

Laplace Transform of Damped Sine and Cosine Functions

21. (a) ( ){ } ( ) ( )2 20 1a ik t s a ik t s a ike e dt s a ik s a k+ += = = +L . Breaking this into real and complex parts yields the desired result.

(b) ( ){ } ( ){ } ( ){ } { }cos sin cos sina ik t at at ate e kt i kt e kt e+ = + = + iL L L L ktkt t

.

Matching real and complex parts of the solution yields the Laplace transform formulas for and . cosate sinate k

Laplace Transform of Hyperbolic Functions

22. { } { } { } ( ) ( ) 2 21 1 1 1sinh 2 2 2 2 2bt bt

bt bte e bbt e es b s b s b

= = = = +

L L L L

{ } { } { } ( ) ( ) 2 21 1 1 1cosh 2 2 2 2 2bt bt

bt bte e sbt e es b s b s

+= = + = + = b +

L L L L

Using Hyperbolic Functions

23. (a) { }( ) ( ) { }

2 22 2

2 2 2 2

cosh sinh2 2

1 12 24 4

bt bt bt bt

bt bt bt bt

e e e ebt bt

e e e e 11s

+ = = + + + = =

L L

L L

(b) { } { } { }( ) ( )

2 2 2 2 2

3 3

1 1 1 1cosh2 2 2 2 2

1 1

bt btbt bt bt bte et bt t t e t e t e t e

s b s b

+ = = + = +

= + +

L L L L L 2

Power Rule 24. (a) Given { }

0

n stt e t = L ndt ,

we integrate by parts, letting nu t= and , to get stdv e dt=

{ } 10 0limn st

n b

b

t e nt es s

= + L st nt dt .

On the right side, the left-hand term becomes 0 in the limit (for );

the integral terms become

0s >{ }1nn ts L . The result follows immediately.

770 CHAPTER 8 Laplace Transforms (b) Performing integration by parts n times yields

{ }0

!n sn

nt e t dts

= L . Integrating gives the final answer as

{ } 1!n nnt s +=L . Multiplier Rule

25. ( ){ } ( ) ( ) ( ) ( ) ( )0 0 0

st st std dt f t te f t dt e f t dt f t e dt F sds ds ds

= = = = L d

Multiplier Applications

The multiplier rule (Problem 25) says to evaluate ( ){ }t f tL , we can first ignore the t and take the transform of ( )f t , getting ( )F s . Then to get the transform we differentiate ( )F s and change the sign. That is,

{ } { }( ) ( )dtf t f tds

= L L .

26. { } { } ( )21 1at atd dte e

ds ds s a s a = = = L L

27. { } { } ( )22 23 6sin 3 sin3

9 9

d dt t tds ds s s

= = = + +L L

s

{ } { } { } { }

( ) ( )2 2

1 1 1. cosh2 2 2 2

1 1 1 2

1 1 1 2

bt btbt bt bt bte et bt t te te te te

d dds s b ds s b

s b s b

+= = + = +

= + + = + +

28 L L L L L

29. { } { } ( )2 2

22 2 2 23 cos 3 cos 3 3d d s st at at

ds ds s a s a

= = = + +LL

a


30. { } { } { }( ) ( )

2 22 2

2 2

1 12 sinh 2 22 2

1 12 2

t tt te e d dt t t te te

ds s ds s

s s

= = = + 2

+ = +

L L L L

Exponential Shift

31. ( ){ } ( ) ( ) ( ) ( )0 0

s a tat st ate f t e e f t dt e f t dt F s a = = = L

Using the Shift

To find ( ){ }ate f tL , Problem 31 says we can ignore the exponential function , take the transform of ate( )f t , and then replace s in { }( ) ( )F s f t= L by ( ).s a That is,

{ } ( )( )ate f t F s a= L .

32. { }n att eL . We first compute ( ) { } 1!n nnF s t s += =L . Then { } ( ) ( ) 1

!n atn

nt e F s as a +

= = L .

33. { }sin 2teL t . We first compute ( ) { } 2 2sin 2 4F s t s= = +L . Then { } ( ) ( )2

2sin 2 11 4

te t F ss

= = +L .

34. { }cos3teL t . We first compute ( ) { } 2cos3 9sF s t s= = +L . Then { } ( ) ( )2

1cos3 11 9

t se t F ss

+= + = + +L .

772 CHAPTER 8 Laplace Transforms 35. { }2 cosh 3teL t . We first compute ( ) { } 2cosh3 9sF s t s= = L . Then { } ( ) ( )2 2

2cosh3 22 9

t se t F ss

= = L .

36. { }3 sinhteL t . We first compute ( ) { } 21sinh 1F s t s= = L . Then { } ( ) ( )3 2

1sinh 33 1

te t F ss

= + = + L .

37. { }2 sin3tte tL . We use both the multiplier rule (Problem 25) and the law (Problem 31) for the Laplace transform. It makes no difference which one is used first. Here the transform is computed first:

ate

{ } 2 3sin 3 9t s= +L . Using the exponential law, we then find

{ } ( )2 23sin 32 9

te ts

= +L .

Finally, from the multiplier rule,

{ } ( )( )

( )2

2 2 2

6 23sin 32 9 2 9

t sdte tds s s

= = + + L .

Linearity of the Inverse 38. Using the linearity of L, we write

( ){ } ( ){ }{ } ( ){ }{ } ( ){ }{ } ( ) ( )1 1 1 1a F s b G s a F s b G s aF s bG s + = + = +L L L L L L L . Taking the inverse transform of each side yields

( ) ( ){ } ( ) ( )1 1aF s bG s a F s b G s + = +L L 1L , which proves the linearity of . 1L

Out of Order 39. For any constant , we can pick t can be large enough so that t > , which implies that 2t t> ,

which, in turn, implies that 2te e t> . Hence eventually will be greater than 2te te .


Inverse Transforms

The key for Problems 40-54 is to rewrite each function in terms of functions listed in the short Laplace transform table, on page 472 of textbook. These transforms are also included in the longer table inside the back cover of the text.

40. 1 31s

L = 1 2

3

1 2 12 2

ts

= L .

41. 1 32 3 7

1s s s + + L

1 23

1 1 12 3 7 2 31 2

te ts s s

+ + = + + = L7 .

42. 1 25

3s + L = ( )1 2

5 3 5 sin 33 33

ts

2

= + L .

43. 1 3 4 3 43 3

te es s

+ = + + L3 3t , by the linearity of the inverse transform.

44. The partial fraction decomposition ( )1

3 3A B

s s s s= ++ + is equivalent to ( )1 3s A sB= + + .

Equating coefficients of like terms and solving for A and B yields 13

A = and 13

B = .

Hence ( )1 12 1 1 1 1 13 3 3( 3) 3 tes s s s = + + =L L 3 . 45. 1 2

12 10

ss s

+ + + L

Completing the square in the denominator yields ( )221 1

2 10 1 9s s

s s s+ +=+ + + + .

Hence, by the exponential law, ( )1 1

22

1 ( 1) cos32 10 1 9

ts s e ts s s

+ + = + + + + =L L .

46. 1 214 4s s

+ + L = ( )1 2

2

12

ttes

= + L

47. 1 23 5

6 25s

s s + + L =

( )( )

12

3 3 14

3 16

s

s + + L

= ( ) ( )1 3

2 2

3 7 4 73 32 23 16 3 16

ts e ts s

+ = + + +

L cos4 sin 4t .

774 CHAPTER 8 Laplace Transforms

48. 1 21

2s

s s + + L

Factoring the denominator and writing as a partial fraction gives

( )( )21 1

2 2 1 2 1s s A B

s s s s s s+ += =+ + + + .

Solving for A and B yields 13

A = and 23

B = . Hence

1 1 121 1 1 2 1 1 2

2 3 2 3 1 3 3t ts e e

s s s s + = + = + + L L L

2 + .

49. 1 25

6s s + L


( )( )25 5

6 3 2 3A B

s s s s s s= = ++ + + 2 .

Solving for A and B yields and 1A = 1B = , so the inverse transform is

1 1 125 1 1

6 3 2t te e

s s s s = + = + + + L L L

3 2 .

50. 1 22 4

1s

s + L


( )( )22 4 2 4

1 1 1 1 1s s A B

s s s s s+ += = + + + .

Solving for A and B yields and 3A = 1B = , so the inverse transform is

1 1 122 4 1 13 3

1 1 1t ts e e

s s s + = = + L L L

.

51. 1 274 7s s

+ + L = ( ) ( )1 2227 3 7 sin 33 32 3

te ts

= + +

L .

52. 1 22 16

4 13s

s s + + + L =

( )( )

12

2 2 12

2 9

s

s + + + + L

( ) ( ) ( )1 2

2 2

2 32 4 2 cos32 9 2 9

ts e ts s

+ = + = + + + +

L 2sin 3t+ .


53. ( )1 2 24

4s s + L

Here we use partial fractions to write

( ) 2 22 24

44A B Cs Ds s ss s

+= + + ++ .

Multiplying the equation by the denominator on the left-hand side yields

( ) ( ) ( )3 24 4A C s B D s A s B= + + + + + 4 . Comparing coefficients and solving the resulting equations for A, B, C, and D yields 0A = ,

1B = , , and 0C = 1D = . Hence

( )1 1 1 12 2 2 22 24 1 1 2 1 1 2 1 sin 2

2 4 2 4 24t t

s s s ss s = = + ++

=L L L L .

54. ( )( )1 2 23

1 4s s + + L

We use partial fractions to write

( )( ) 2 22 23

1 41 4As B Cs Ds ss s

+ += ++ ++ + .

Multiplying the equation by the denominator on the left-hand side, we get

( ) ( ) ( ) ( )3 23 4A C s B D s A C s B D= + + + + + + +4 . Comparing coefficients and solving the resulting equations for A, B, C, and D yields 0A = ,

1B = , , and 0C = 1D = . Hence

( )( )1 1 1 12 2 2 22 23 1 1 2 1 1 2 sin sin 2

1 2 4 1 2 4 21 4t t

s s s ss s = = = + + + ++ + L L L L

1 .

Computer Exploration 55. Student Project

Suggested Journal Entry 56. Student Project


8.2 Solving DEs and IVPs with Laplace Transforms

First-Order Problems 1. , 1y = ( )0 1y = The Laplace transform yields the equation { } 0 1s y y s =L . Substituting in the initial condition and solving for the Laplace transform yields { } 21 1y s s= +L .

Taking the inverse transform yields ( ) 1y t t= + . 2. , 0y y = ( )0 1y = The Laplace transform yields the equation { } { }0 0s y y y =L L . Substituting in the initial condition gives { } 1

1y

s= L .

Taking the inverse transform yields ( ) ty t e= . 3. , ty y e = ( )0 1y = The Laplace transform yields the equation { } { } 11

1s y y

s = L L .

Substituting the initial condition and solving for the Laplace transform gives

{ } ( )21 1

11y

ss= + L .

Taking the inverse transform yields ( ) t ty t te e= + . 4. , ty y e + = ( )0 1y = The Laplace transform yields the equation { } { } 11

1s y y

s+ + = +L L .

Substituting the initial condition and solving for the Laplace transform gives

{ } ( )21 1

11y

ss= L .

Taking the inverse transform yields ( ) t ty t te e = .

SECTION 8.2 Solving DEs and IVPs with Laplace Transforms 777

Transformations at Work 5. , 3 2 0y y y + = ( )0 1y = , ( )0 0y = Taking the Laplace transform yields the equation

{ } { }( ) { }2 3 1 2s y s s y y 0 +L L L = . Solving for the Laplace transform yields

{ } ( )( )23 3 1 2

3 2 2 1 2 1s sy

s s s s s s = = = + 1 + L .

Hence, the solution is ( ) 2 2t ty t e e= + . 6. , 2 4y y + = ( )0 1y = , ( )0 4y = Taking the transform yields the equation

{ } { }2 44 2 2s y s s ys

+ + =L L .

Solving for the Laplace transform yields

{ } ( ) ( )24 1 2

2 2y

s s s s s= + +

2+ + +L .

Using partial fractions we write the first term of { }yL as ( )2 2 2

4 12 2

A B Cs s s s s s s s

= + + = + +2 12+ + + .

Similarly, we write the last fraction of { }yL as ( )

2 12 2

D Es s s s s s

= + = + 12+ + + .

Putting these expressions together yields the transform

{ } 22 2 3 2y s s s= + + +L .

Taking the inverse transform yields the solution of the initial-value problem

( ) 22 2 3 ty t t e= + + . 7. , 9 20 ty y e + = ( )0 0y = , ( )0 1y = Taking the transform yields the equation

{ } ( ) { }2 200 1 91

s y s ys

+ = +L L .

778 CHAPTER 8 Laplace Transforms Solving for the Laplace transform yields

{ } ( )( ) 2 2 2220 1 2 2 1 12

9 1 9 91 9sy

s s s s ss s = + = + + 2 9+ + + + ++ + L .

Combining terms and taking the inverse yields ( ) 2 2cos3 sinty t e t t= + 3

+ = (.

)8. , 9 cos3y y t 0 1y = , ( )0 1y = Taking the transform yields the equation

{ } ( ) { }2 21 1 9 9ss y s y

s + + = +L L .


{ } ( )2 22199

s syss

= + ++L .

Notice that the first term is half of the negative derivative of 21

9s + . Thus, the inverse transform

of the first term is simply 1 sin 36

t t . Breaking the second term into two parts yields

1cos3 sin33

t t . Hence, the answer is

( ) 1 1sin 3 cos3 sin 36 3

y t t t t t= + . 9. , 3 2 6y y y + + = ( )0 0y = , ( )0 2y = Taking the transform yields the equation

{ } ( ) { } { }2 60 2 3 2s y s s y ys

+ + =L L L .


{ } ( )( ) ( )( )6 2

1 2 1 2

3 6 3 2 21 2 1 2

3 4 11 2

ys s s s s

s s s s s

s s s

= ++ + + + = + + + + + + = + + +

L

.

Hence, the final result is ( ) { }{ }1 23 4 .t ty t y e e = = +L L


10. 1, (0) 0, (0) 0y y y y y + + = = = Taking the Laplace transform of the equation

2 1( ) ( ) ( )s Y s sY s Y ss

+ + =

2 22

2

2 2

1 1 1 1( ) (by partial fractions)( 1) 11 3

2 4

1 1= (by completing the square)s 1 3

2 4

3 11 1 2 2

3 1 3 1 32 4 2 4

sY ss s s s s s

s s

s

s

s

ss s

+= = = + + + + + + + + +

+= + + + +

We obtain (1/ 2) (1/ 2)1 3( ) 1 sin cos2 23

t ty t e t e t =

3

11. sin , (0) 0, (0) 0y y y t y y + + = = = Taking the Laplace transform of the equation

2 21( ) ( ) ( )

1s Y s sY s Y s

s+ + = +

2 2 2 2

22

2 22

(1/ 2) (1

1 1( ) (by partial fractions)( 1)( 1) 1 1

1= (by completing the square)s 1 1 3

2 4

3 11 2 21

1 3 1 3 1 32 4 2 4

1 3( ) cos sin23

t

s sY ss s s s s ss s

s

sss

s s

y t t e t e

+= = ++ + + + + ++++ + +

+= + ++ + + + +

= + + / 2) 3cos

2t t

780 CHAPTER 8 Laplace Transforms 12. , (0) 0, (0) 1ty y y e y y + + = = = Taking the Laplace transform of the equation

2

2 2

2

2 2

(1/ 2)

1( ) 1 ( ) ( )1

2 1 1( ) (by partial fractions)( 1)( 1) 1 11 1= (by completing the square)

s+1 1 32 4

3 11 2 21 3

1 1 3 1 32 4 2 4

3( ) 3 sint t

s Y s sY s Y ss

s sY ss s s s s s

s

s

s

ss s

y t e e

+ + = ++ = = + + + + + +

+ + +

= + + + + + + = + (1/ 2) 3cos

2 2tt e t

General Solutions 13. y y = t

Taking the Laplace transform of each side and calling ( )0y A= and ( )0y B= , we get { } { }2 21s y As B y s =L L .

Solving for { }yL yields { }

( )

21

2

2 2 2 2

2 2 2 2

11 1

1 1 1

1 1 1 (using partial fractions)1 1 1

sAs B

yssA B

s s s s

sA Bs s s s

+ += = + + = + +

L

Hence,

( ) cosh sinh sinhcosh ( 1)sinh .

y t A t B t t tA t B t t

= + += + +


14. 3 2y y y + + = 0Taking the Laplace transform of each side and calling ( )0y A= and ( )0y = B , we get

{ } { }( ) { }2 3 2s y As B s y A y 0 + + =L L L . Solving for { }yL yields

{ }

( )( ) ( )( )2 2

3 13 2 3 2

3 12 1 2 1

1 2 1 1 . (using partial fractions)2 1 2 1

sy A Bs s s s

sA Bs s s s

A Bs s s s

+= ++ + + ++= ++ + + +

= + + + + + + +

L

Hence, ( ) 2 2 1 22t t t t ty t Ae Ae Be Be C e C e 2t = + + = + B B

where and 1 2C A= + 2C A= . Raising the Stakes 15. , 6 ty y y y e + = ( ) ( ) ( )0 0 0 0y y y = = . = The transform yields { } { } { } { }3 2 6

1s y s y s y y

s + = L L L L .

Solving for the { }yL yields { } ( )( )3 2

61 1

ys s s s

= +L .

The denominator factors further, and we can then use partial fractions, so

( ) ( ) ( ) ( )

( ) ( )

3 2 3

2 3

61 11 1 1 1

3 1 3 1 3 3 1 .4 1 2 4 11 1

A B C Ds ss s s s

s ss s

= + + + + + = + +

Thus the solution is ( ) 23 3 3 34 2 2 4

t t ty t e te t e e t= + . 16. , (4) 0y y = ( )0 1y = , ( )0 0y = , ( )0 1y = , ( )0 0y = . The transform yields { } { }4 3 0s y s s y + =L L . Solving for the Laplace transform yields

{ } ( )( )( )( )( ) 221 1

11 1 1s s s sy

ss s s+ = = + + +L .

Thus the solution to the IVP is ( ) cosy t t= .

782 CHAPTER 8 Laplace Transforms Which Grows Faster? 17. For k > 0 and y(0) = 1, consider the two DEs

0

,

( ) , 0t

y ky

y k y t dt k

= = >

For the first equation it should be familiar that

( ) kty t e= but we solve the equation using Laplace transforms

( ) 1 ( )1( )

( ) .kt

sY s kY s

Y ss k

y t e

== =

The second equation can also be solved using Laplace transforms

( )2

( )( ) 1

( )

( ) cosh .2

kt kt

Y ssY s ks

sY ss k

e ey t kt

=

= += =

For values of k such that 0 < k < 1, the solution to the second equation will eventually outpace the first, while if k 1, the solution to the first equation will outpace the second.

To see this, compute the limit of the ratio of the two solutions.

( )lim lim lim 22 2

kt ktk k t

kt kt ktt t t

e e ee e e

= =+

When k k < 0, this limit is 0 indicating that the denominator is much larger than the numerator. When the reverse is true, the limit is infinite. At the value of k = 1, the limit is exactly 2, indicating that the solution to the first equation is essentially twice the solution to the second.

Laplace Transform Using Power Series 18. The power series for et is

2

0

1 1 2 30 0

1 1( ) 1 ...! 2

1 ! 1 1 1 1( ) ...!

t n

n

n nn n

f t e x x xnnF s

n s s s s s

=

+ += =

= = = + + +

= = = + +

+


This is a geometric series with the first term 1s

and a common ratio of 1s

. The closed form for

this series is 1

1( ) .1 11

sF ss

s

= =

Operator Methods 19. 3 2y y y + + =1

1

The differential equation can be rewritten as

( 1)( 2)D D y+ + = , or as the system of equations

( 1) 1( 2)

D v.D y v

+ =+ =

(Simply substituting the second equation into the first will yield the original equation.) Solving the first equation for v using Laplace transforms:

1( ) (0) ( )

(0) 1 (0) 1 1( )1 ( 1) 1

( ) ( (0) 1) 1t

sV s v V ssv vV s

1s s s s s sv t v e

+ =

= + = + + + += +

+

Substituting this into the second equation yields

( 2) ( (0) 1)(0) 1( ) (0) 2 ( )

1 ( 1)

tD y v evsY s y Y ss s s

1+ = + = +

+

+ +

Note here that there was really no need to find the expression for v(t) unless you are interested in v(t).

2 2 2

(0) (0) 1( )2 ( 1)( 2) ( 1)( 2)

1 1 1 1 1 1 1 1(0) (0) (by partial fractions)2 2 1 2 1 2 ( 2)

1 1( ) (0) (0)( ) .2 2

t t t t t

y vY ss s s s s s

y vs s s s s s

y t y e v e e e e

= + ++ + + + + = + + + + + + + +

= + + +

784 CHAPTER 8 Laplace Transforms 20. 6y y t + =

2 6

62

2 4 6 6 2

6 4 2 2 4 2

( 1)1( )

1(1 ...) (by dividing 1 + into 1)

30 360 720 0 0 0 ... 30 360 720

D y t

y t tD

D D D t Dt t t t t t

+ = = +

= + += + + + + + = +

Bessel Functions with IDE 21. Student Lab Project

Computer Exploration 22. Student Lab Project

Laplace Vibration 23. Student Lab Project


SECTION 8.3 The Step Function and the Delta Function 785

8.3 The Step Function and the Delta Function

Stepping Out 1. The function f(t) first has the value of 0 for 0t < , then a from 0 to 1. Hence we write this as ( ) ( )stepf t a t= . However, at the function shifts from a to b. We, therefore, subtract a and add b, yielding 1t = ( ) ( ) ( ) ( )step step 1f t a t b a t= + .

Finally, when , the function shifts from b to c, so we subtract b and add c. Thus for all we have

2t = 0t

( ) ( ) ( ) ( ) ( ) ( )step step 1 step 2f t a t b a t c b t= + + . 2. Following the procedure in Problem 1, we write the function as

( ) ( ) ( ) ( ) ( )1 1 step 2 2 step 3t tf t e t e t= + + . 3. Following the procedure in Problem 1, we write the function as

( ) ( ) ( ) ( ) ( )2 21 4 1 step 1 1 4 step 4f t t t t t t t= + + + . 4. Following the procedure in Problem 1, we write the function as

( ) ( ) ( )sin step 2 sin step 4f t t t t t = . Geometric Series 5. ( ) ( ) ( ) ( )step 1 step 2 step 3f t t t t= + + +" , ( ){ } ( ) ( ) ( ){ }2 3 2 3 11 1s s s s ss s s se e e e ef t e e es s s s s e = + + + = + + + + = " "L . 6. ( ) ( ) ( ) ( )1 step 1 step 2 step 3f t t t t= + + + +" , ( ){ } ( ) ( ) ( ){ }2 3 2 31 1 1 1s s s s s s 1 1 se e ef t e e es s s s s s e = + + + + = + + + + = " "L . 7. ( ) ( ) ( ) ( )1 1 11 step 1 step 2 step 3

2 4 8f t t t t= " ,

( ){ } 2 31 2 11 .2 4 8 2 (2 )

s s s s

s s

e e e ef ts s s s s e s e

= = = "L2 (1 )

8. ( ) ( ) ( )1 step 1 step 2f t t t= + " , ( ){ } 2 31 1

1

s s s 1s

e e ef ts s s s s e

= + = + "L .

786 CHAPTER 8 Laplace Transforms 9. ( ) ( ) ( )1 2step 1 2step 2f t t t= + " ,

( ){ } ( )( )2 3 21 2 2 2 1 2 1 2 11 1s s s s ss s se e e e ef t e es s s s s s s s e = + = + = + " "L . Piecewise-Continuous Functions In the following problems we use the alternate delay rule

( ) ( ){ } ( ){ }step csf t t c e f t c = +L L . 10. The function can be represented by

( ) ( ) ( ) ( ) ( ) ( )step step 1 2 step 1 step 2f t t t t t t t = + . Using the Alternate form of the Delay Theorem we get

( ){ } { } { } { }( )

2 22 2 2

2 22 2 2 2

1 1 1 1 11 1

1 2 1 1 1 2 .

2 2

1 1s s s s s s s

s s s s

sf t e t e t e t e e e es s s s s

e e e es s s s

= + + = + +

= + = +

L L L L es s

11. ( ) ( )( ) ( ) ( )( )31 step 1 step 1 step 32 2

tf t t t t = + ,

( ){ } 3 32 21 2 1 12 2 2 2s 1s s se t t sf t e e e

s s s s s

= + = + L L L e

.

12. ( ) ( ) ( ) ( ) ( )1 step 1 step 3 2step 3f t t t t t = + , ( ){ } { } { } 3 3 3 33 2 2 2 22 22 2 3s s s s s s ss s e e e e e e ef t e t e t s s s s s s s

= + + = + = L L L .

13. ( ) ( )( )sin 1 steptf t b t aa = ,

( ){ } ( )( )( )

( )( ) ( )2 22 2

sin sin

sin 1 .

as

a aas as

a a

t atf t b bea a

b btbe eas s

+ = = = + + +

L L L

L

14. ( ) ( )( ) ( ) (sin 1 step 1 step 1 step 22tf t t t t = + ) ,

( ){ } ( )2

222

12

s ss e ef t se

s ss = + +

L .


15. The two parts of the sine function can be written as

( ) ( ) ( )( ) ( ) ( ) ( )( )sin 1 step 1 2sin step 1 step 2f t t t t t t = ,

( ){ } { } ( )( ){ } ( ){ } ( )( ){ }( )2

2 22 2 2 2 2 2 2 2 2 2

sin sin 1 2 sin 1 2 sin 2

2 2 1 3 2 .

s s s

s s s s

f t t e t e t e t

e e e e es s s s s

= + + + += + + + = + ++ + + + +

L L L L L

s

Transforming Delta 16. ( ) ( ) ( ){ } 2 31 2 2 3 3 2 3s s st t t e e e + + = + +L 17. ( ) ( ) ( ){ } 22 2 1 2 s st t t e e + = +L 18. ( ) ( ) ( ){ } 21 2 1 s st t t e e + = + " "L 19. ( ) ( ) ( ){ } 22 1 s st t t e e + + + = + + +" "L Laplace Step by Step 20. ( ) ( )1 step 1f t t=

The Laplace transform of 1 is 1s

, whereas the transform of ( )step 1t is ses

. Hence, the

transform is

{ } 1 sefs s

= L .

21. ( ) ( ) ( )1 2step 1 step 2f t t t= + { } 21 2 s se ef

s s s

= +L .

22. ( ) ( ) ( )1 step 1f t t t= Note that this is the function ( )f t t= shifted to the right one unit. Thus, we take the Laplace transform, of t (which is 2

1s

) and multiply it by the shifting factor (which is se ). Hence the final

answer is

{ } 2sef

s

=L .

23. ( ) ( ) ( )sin stepf t t t = This is the sine function shifted to the right . Hence, the transform is

{ } 2 1sef

s

= +L .

788 CHAPTER 8 Laplace Transforms 24. ( ) ( )step 3tf t e t=

The function here has not been shifted, so we must perform the shift and write the function as

( )3 step 3 te t e 3 . Hence, the transform is { } 3 33

1 1

3s se ef es s

= = L .

25. ( ) ( )step 1 tf t e= Sometimes it is useful to put functions as arguments of the step function, so it switches off and on at special points depending on the function. In this case, the function in the argument 1 te is always positive (for ), so the function step 0t > ( )1 1te = for . Hence, 0t >

{ } 1fs

=L .

26. ( ) ( )2 step 2f t t t= We write the expression in the form ( ) ( )stepf t c t c . We do this by writing . ( ) ( ) ( ) ( ) ( ) (22 step 2 2 step 2 4 2 step 2 4step 2t t t t t t t = + + ) Hence, the transform is

{ } 2 3 22 4 4sf e s s s = + + L .

Inverse Transforms

27. 1se

s

L

Factoring out the se , the function is simply 1s

; the inverse transform is 1. The factor se means

there is a ( )step 1t multiplied in this term. Hence, the inverse transform is ( )step 1t . Once again, the graphs of these functions are all graphs of familiar functions delayed, and examples can be seen in the text.

28. 1 2se

s

L

The inverse transform of 21s

is t. Hence, the inverse transform is

( ) ( )1 2 1 step 1se t t

s

= L .


29. 2

1

3

ses

L

The inverse transform of 13s is . Hence, the inverse transform is

3te

( ) ( )2 3 21 step 23

ste e t

s

= L .

30. 4

1

4

ses

+ L

The inverse transform of 14s + is

4te . Hence, the inverse transform is

( ) ( )4 4 41 step 44

ste e t

s

= + L .

31.

( )1 1se

s s

+ L

Writing the partial fraction decomposition, yields

( )1 1 1

1 1s s s s= + + .

Hence, this expression has the inverse transform 1 te . The given function has the inverse transform (see figure):

( ) ( ) ( )( )1 1 step 1 11s

te t es s

= + L .

3 420.0

10

0.5

1.0

t

DelayedExponential

f t( )

32. 2 3 4

1 2 2s s se e e es

+ L

s

We break this into four parts, yielding an inverse Laplace transform

( ) ( ) ( ) (2 3 41 2 2 step 1 2step 2 2step 3 step 4s s s se e e e t t ts

+ )t= + L .

790 CHAPTER 8 Laplace Transforms Transforming Solutions 33. ( )1 step 1x t = , ( )0 0x =

The Laplace transform of the equation is

( ) 10 sesX ss s

= .

Solving for ( )X s yields ( ) 21 2seX s

s s

= . Taking the inverse

transform yields the solution of the initial-value problem

( ) ( ) ( )1 step 1x t t t t= . 3 42

0.010

1.0

t

x t t t t= ( ) ( ) ( )1 1step

f t 1 t= ( ) ( )1step

34. ( ) ( )1 2step 1 step 2x t t ( ) = + , 0 0x =

The Laplace transform of the equation is

( ) 21 20 s se esX ss s s

= + .

Solving for ( )X s yields ( ) 22 2 21 2

s se eX ss s s

= + .


( ) ( ) ( ) ( ) ( )2 1 step 1 2 step 2x t t t t t t= + .

3 42

1

1

0

0

1

t

x t t t t

t t

( ) = ( ) ( )+ ( ) ( )

2 1 1

2 2

step

step

f t 1 t

t

( ) = ( )+ ( )

2 1

2

step

step

Forcing term and response

35. ( )step 3x x t + = , ( )0 0x = , ( )0 1x = The Laplace transform of the equation is

( ) ( ) 32 1 ses X s X ss

+ = .

Solving for ( )X s yields ( ) ( )

3

2 2

11 1

seX ss s s

= ++ + .


( ) ( ) ( )sin step 3 1 cos 3x t t t t = + .

9 1260.0

30

3

t

x t t t t( ) = ( ) ( ) ( )( )sin cosstep 3 1 3

f t t( ) = ( )step 3

Forcing term and response


36. ( ) ( )step step 2x x t t + = , ( )0 0x = , ( )0 1x = The Laplace transform of the equation is

( ) ( ) 22 1 s se es X s X ss s

+ = .

Solving for ( )X s yields ( ) ( ) ( )

2

2 2 2

11 1 1

s se eX ss s s s s

= + + + + .

9 126

3

3

3

t

x t( )f t( )

Response to one square pulse

Using a partial fraction decomposition, we write

( ) 221 1

11s

s ss s= ++ .

Taking the inverse transform yields the solution of the initial-value problem,

( ) ( )( ) ( ) ( )( ) ( )sin 1 cos step 1 cos step 2x t t t t t t = + + . Periodic Formula 37. In Problem 10, we observed that the Laplace transform

of the single triangular wave on the interval [ ]0, 2 was { } ( ) 220 1 2single wave

s sst e ef t e dt

s

+= =L . Using equation (24) in the text, the Laplace transform of the periodic triangular wave on [ )0, is ( ){ } 22 21 1 21

s s

s

e ef te s

+= L .

3 420.0

10

1

2

t

f t( )

5 6

y

Triangular wave

38. From Problem 10,

{ } ( )

02

2

single wave

1 2 .

st

s s

f t e dt

e es

= +=L

Hence, the Laplace transform of the periodic triangular wave of period 3 is

( ){ } 23 21 1 21s s

s

e ef te s

+= L .

3 420.0

10

1

2

t5 6

y

Modified triangular wave

792 CHAPTER 8 Laplace Transforms 39. First, we find the Laplace transform of the single wave-form

( ) ( )2 1 step 1f t t t = 0 t, < , which is

{ } ( )0

single wave stf t e dt = L .

We determine the Laplace transform by applying the alternate form of the Delay Theorem instead of integrating. 3 420.0 10

1

2

t5 6

y

Sawtooth Wave

{ } ( ){ }2 2

2 2

1 1single wave 2 2 1 2 2

1 1 12 .

s s

s

e t e 21 2

s s s

es s s

= + = + = +

L Ls

The Laplace transform of the periodic wave-form of period 1 on [ )0, is

( ){ } { }

2 2

1 single wave1

2 1 1 1 .1

s

ss

f te

ee s s s

= = +

L L

40. From Problem 13 (with , 1a b= = ), we have

{ } ( )21single wave 11 ses = ++L . Hence, the Laplace transform of the full-rectified periodic wave with period on [ )0, is ( ){ } 21 11 1

s

s

ef te s

+ = + L .

Full wave rectification

41. From Problem 13 (with , 1a b= = ), we have { } ( )21single wave 11 ses = ++L . Hence, the Laplace transform of the half-rectified periodic wave with period 2 on [ )0, is ( ){ } 2 21 11 1

s

s

ef te s

+ = + L .

Half-wave rectification


42. From Problem 14, we have

{ } ( )( )

( )

2

222

2 2 2 2

2 2

1single wave2

2 4.

4

s ss

s s

e eses ss

s e e s

s s

= + ++ += +

L

Hence, the Laplace transform of the periodic function of period 4 on [ )0, is ( ){ } ( )

2

24 222

1 1 .1 2

s ss

s

e ef t see ss

= +s

+ L

2

1

0.0 t0

y

1 2 3 4 5 6 7


{ } 32 2

3

2

1 1 1single wave2 2

2 .2

s s

s s

e es s s

s e es

= + +=

L

Hence, the Laplace transform of the periodic function of period 4 on [ )0, is ( ){ } 34 21 21 2

s s

s

s e ef te s

+= L .

20.0 10

1

2

t543 6

y

7


{ } ( )

( )2

2 2

2

2 2

single wave 1 3 2

1 3 2.

s s

s s

e es

e es

= + +++ += +

L

Hence, the Laplace transform of the periodic function of period 2 on [ )0, is { } 22 2 21 3 21

s s

s

e efe s

+ += +L .

0.00

1

2

t

y

1 2 3 4 5 6

794 CHAPTER 8 Laplace Transforms 45. From Problem 8, we have

{ } ( ) ( )2 10 0

1single wave 1 .st st se f t dt e dt es

= = = L Hence, the Laplace transform of the periodic function of period 2 on [ )0, is ( ){ } 21 11

s

s

ef ts e

= L .

y

3 4 2

3

1t

5 6

Square wave

Sawblade 46. Input ( )x x f t + = , ( )0 0x = where

( ) ( ) ( )step 1 step 2f t t t t= " . Taking the Laplace transform of the DE yields ( ) { }

1f

X ss

= +L

,

where ( ){ } 221 s se ef t s s s

= "L . Using two partial fraction decompositions,

420.0

0

0.5

1.0

t6

y

x t( )

f t( )

sawtooth wave input

( ) ( )2 21 1 1 1 1 1 1 and ,

1 1 1s s s s s s s s s= + = + + + 1+

we obtain

( ) 2 22 21 1 1 1 1 1 1 1 1 .1 1 1s s

s se eX s e es s s s s s s s s s s

= = + + + +

" "1+

Therefore, the solution of the initial-value problem is

( ) ( )( ) ( ) ( )( ) ( )1 21 1 step 1 1 step 2 .t ttx t t e e t e t = + " Square Wave Input 47. ( )x x f t + = , ( ) ( )0 0x x= = 0 where

( ) ( ) ( )1 2step 1 2step 2f t t t= + " . Taking the Laplace transform of the DE yields ( ) { }2 1

fX s

s= +L

,

where

( ){ } 21 2 2s se ef t s s s

= + "L . Response to a square wave input


Hence

( ) ( ) 2 221 1 2 2

1sX s e e

s s = + + " .

We write the partial fraction decomposition as

( )( )

22

2 22

1 111

1 1 2 2 1 2 2 .1

s s s s

ss ss s

sX s e e e es s

= ++ = + + +" "

Therefore, the solution of the initial-value problem is

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

1 2step 1 2step 2 2step 3

cos 2cos 1 step 1 2cos 2 step 2 2cos 3 step 3

1 cos 2step 1 1 cos 1 2step 2 1 cos 2 2step 3 1 cos 3 .

x t t t t

t t t t t t t

t t t t t t t

= + + + +

= +

""

"

Solve on Impulse 48. ( )x t = , ( )0 0x =

Taking the Laplace transform of both sides of the DE yields ( ) 1sX s = or ( ) 1X ss

= .

Hence, we have ( ) 1x t = .

49. ( ) ( )1x t t = , ( )0 0x = Taking the Laplace transform of both sides of the DE yields

( ) 1 ssX s e= , or ( ) 1 seX ss s

= .

The inverse is

( ) ( )1 step 1x t t= .

50. ( )2x x t + = , ( ) ( )0 0x x= = 0 Taking the Laplace transform of both sides of the DE yields

( ) ( )2 21 ss X s e + = , or ( ) 22 1seX s

s

= + .

The inverse is

( ) ( ) ( )sin 2 step 2x t t t = ; its graph is the sine function starting at 2t = .

796 CHAPTER 8 Laplace Transforms 51. ( ) ( )2x x t t + = + , ( )0 0x = , ( )0 1x =

Taking the Laplace transform of both sides of the DE yields

( ) ( )2 21 1 s ss X s e e + = + . Solving for ( )X s yields ( ) 22 21 11 1s sX s e es s s

= + + + 21

1+ . Response to two impulses

The inverse is

( ) ( ) ( ) ( ) ( )sin sin step sin 2 step 2x t t t t t t = + . 52. ( )2x x t + = , ( )0 1x = , ( )0 0x =

Taking the Laplace transform of both sides of the DE yields

( ) ( )2 21 ss X s s e + = . Solving for ( )X s yields ( ) 22 211 1s

sX s es s

= + + + . Response to an impulse at 2 The inverse is

( ) ( )cos sin step 2x t t t t = + . Laplace with Forcing Functions 53. Student Lab Project


SECTION 8.4 The Convolution Integral and the Transfer Function 797

8.4 The Convolution Integral and the Transfer Function

Convolution Properties 1. Verify ( ) ( )f g h f g h = Method 1 Computation of these properties from the integral definition of convolution requires lengthy

substitutions and exchanging of the order of integration. It can be done from the definition, but we will rely on the convolution theorem in this method to show that it is true.

( ) [ ] [ ] ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )L f g h F s L g h F s G s H s F s G s H s L f g H s L f g h = = = = Method 2

Using the integral definition of convolution,

( )2

1 1 10

2 2 1 1 10 0

( ) ( ) ( )

( ) ( ) ( )

t

t w

f g h f g t w h w dw

2f t w g w w h w dw dw

= =

Letting u1 = w2 w1 in the inner integral (treating w2 as a constant) yields

2

2

2

0

2 1 2 1 10

2 1 2 1 10 0

2 1 2 1 1 20 0

( ) ( ) ( )( )

( ) ( ) ( )

( ) ( ) ( )

t

w

t w

t w

2

2

f t w g u h w u du dw



= =

=

Exchanging the order of the integration (a very nontrivial step) yields

12 1 2 1 20

( ) ( ) ( )t t

u 1f t w g u h w u dw du =

Letting w2 = u2 + u1 (treating u1 as a constant) in the inner integral 1

2 1 1 2 20 0( ( )) ( ) ( )

t t u

1f t u u g u h u du du = +

Again, with that nontrivial exchanging of the order of integration, 2

2 1 1 1 2 20 0

1 1 10

( ) ( ) ( )

( ) ( )

( ) .

t t u

t

f t u u g u du h u du

f t u g u du h

f g h

= = =


2. Prove ( )f g h f g f h + = + [ ]

0

0

0 0

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

t

t

t t

f g h f t w g w h w dw

f t w g w f t w h w dw

f t w g w dw f t w h w dw

f g f h

+ = += = = +

3. Prove 0 0f =

0 00 ( )(0) 0

t tf f t w dw dw = = = 0

Calculating Convolutions

4. 0

1 1 (1)(1)t

dt t = = 5. 2

0

11 1 1 1 1( )2

tt w dw = = = t

6. 20

11 1( )2

tt w dw = = t

7. 2 3

3

00

1( )2 3 6

tt w wt t t w wdw t t = = =

8. To find for any k, where k represents the number of times t appears in the product, ,t t t t " (Note that Problem 7 is the initial k = 2 case.)

Then for k = 3, :t t t

4 53 3

00

5

1 1 1( )6 6 6 4

1 .5!

tt tw wt t t t t t w w dw

t

.5

= = = =

Using this result and Problem 7, we make the following conjecture for any k:

2 11(2 1)!

kt t t t tk

= "

For k 3, we confirm this conjecture as follows: Proof by induction: Assume that 2 11*

(2 1)!kt t t t t

k = " for some k.


For the k + 1 case:

2 1

2 1

0

2 1 2

0

2 2 1

0

2 1 2 1

2 1 2 1

1( )(2 1)!

1( )( 1)!

1(2 1)!

1(2 1)! 2 2 1

1(2 1)! 2 2 1

1 1 1(2 1)! 2 (2 1) (2 1)!

k

t k

t k k

tk k

k k

k k

t t t t t t tk

t w w dwk

tw w dwk

tw wk k k

t tk k k

t tk k k k

+

+ +

+ +

= = +=

= + = + = = + +

"

which proves it true for k = 2, 3, 4,

9. ( ) 2 20 0

0

1 1 1 sinh2 2 2

tt tt t t w w t w t w t te e e e dw e dw e e e t = = = = + =

10. ( )0

tat at a t w awe e e e dw = 2 200

1 1 1 1 sinh2 2 2

tt at aw at aw at ate dw e e e a

a a a a = = = + = t

First-Order Convolution Equation

11. Is t = ba

a solution to the equation ?a t b = Lets find out.

[ ]0

2

0

( )t

t

b ba a dwa a

bbw bt ba

= = = =

This result means that t = ba

isnt a solution to a t = b. The main problem lies in the fact that 1

is no longer a multiplicative identity if you are considering to be multiplication, nor is 1a

an

inverse for the operation anymore.

800 CHAPTER 8 Laplace Transforms Convoluted Solutions 12. ( )x f t = , ( )0 0x = Taking the Laplace transform of both sides of the DE yields ( ) { }sX s f= L . Solving for ( )X s yields ( ) { } { } { }1 1X s f

s= =L L L f .

The inverse yields the solution ( ) ( ) ( )0

1t

y t f t f d = = . 13. ( )x f t = , ( )0 1x = Taking the Laplace transform of both sides of the DE yields ( ) { }1sX s f = L . Solving for ( )X s yields ( ) { } { } { }1 1 1 1X s f f

s s s= + = +L L L .

The inverse yields the solution

( ) ( ) ( )0

1 1 1t

y t f t f d = + = + . 14. ( )x x f t + = , ( )0 0x = Taking the Laplace transform of both sides of the DE yields

( ) ( ) { }sX s X s f+ = L . Solving for ( )X s yields ( ) { } { } { }1 1 tX s f es = =+ L L L f . The inverse yields the solution

( ) ( ) ( ) ( )0

t ttx t e f t e f d = = . 15. ( )x x f t + = , ( )0 1x = Taking the Laplace transform of both sides of the DE yields

( ) ( ) { }1sX s X s f + = L . Solving for ( )X s yields ( ) { } { } { }1 1 11 1 1 tX s fs s s = + = + + + + L L e fL . The inverse yields the solution

( ) ( ) ( ) ( )0

t tt t tx t e e f t e e f d = + = + .


16. ( )x x f t + = , ( )0 1x = , ( )0 0x = Taking the Laplace transform of both sides of the DE yields

( ) ( ) { }2s X s s X s f + = L . Solving for ( )X s yields ( ) { } { } {2 2 21 sin1 1 1

s s }X s fs s s

= + = + + + + L L t fL .

The inverse yields the solution

( ) ( ) ( ) ( )0

cos sin cos sint

x t t t f t t t f d = + = + .

17. ( )3 2x x x f t + + = , ( ) ( )0 0x x 0= = Taking the Laplace transform of both sides of the DE yields

( ) ( ) { }2 3 2s s X s f+ + = L . Solving for ( )X s yields

( ) { } { } { } { } { } { } { }22 3 2 1 2 t tf f fX s e fs s s s = = = + + + +L L L L L L Le f . The inverse yields the solution

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )22 20

t t tt t t tx t e f t e f t e e f t e e f d = = = .

Transfer and Impulse Response Functions 18. ( )x f t = , ( )0 0x = Taking the transform of the DE yields

( ) { }sX s f= L or ( ) { }1X s fs

= L .

The transfer function is the coefficient of { }fL or in this case, 1Transfer Function

s= .

The impulse response function ( )I t is the inverse transform of this function. Hence, ( ) 1I t = . The solution in terms of the transfer function is

( ) ( ) ( ) ( ) ( ) ( )0 0

t tx t I t f t I t f d f d = = = .

802 CHAPTER 8 Laplace Transforms 19. ( )x ax f t + = , ( )0 0x = Taking the transform of the differential equation yields

( ) ( ) { }sX s aX s f+ = L or ( ) { }1X s fs a

= + L .

The transfer function is the coefficient of { }fL , or 1Transfer Function

s a= + .

The impulse response function ( )I t is the inverse transform of this function. Hence, ( ) atI t e= . The solution in terms of the transfer function is

( ) ( ) ( ) ( ) ( ) ( ) ( )0 0

t t a tx t I t f t I t f d e f d = = = . 20. ( )x x f t + = , ( ) ( )0 0x x= = 0 Taking the transform of both sides of the equation, yields

( ) ( ) { }2s X s X s f+ = L or ( ) { }21 1X s fs= + L . The transfer function is the coefficient of { }fL , or

2

1Transfer Function 1s

= + .

The impulse response function ( )I t is the inverse transform of this function. Hence, ( ) sinI t t= . The solution in terms of the transfer function is

( ) ( ) ( ) ( ) ( ) ( ) ( )0 0

sint t

x t I t f t I t f d t f d = = = .


21. ( )4 5x x x f t + + = , ( ) ( )0 0x x 0= = . Taking the transform of the differential equation yields

( ) ( ) ( ) { }2 4 5s X s sX s X s f+ + = L or ( ) { } { }( )22 4 5 2 1f f

X ss s s

= =+ + + +L L

.

The transfer function is the coefficient of { }fL , or ( )2

1Transfer Function2 1s

= + + .

The impulse response function is

( ) ( )2

2

1 sin2 1

tI t e ts

= = + + L .

The solution is ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )20 0

sint t tx t I t f t I t f d e t f d = = = .

Inverse of Convolution Theorem

22. Find 1 21s

L . Then 1 1( ) so that ( ) 1 1 (Problem 4).F s f t ts s

= = =

23. Find 1 31s

L . Then 2

2

1 1 1( ) so that ( ) 1 (Problem 6).2

F s f t t ts s

= = =

24. Find 1 1( 1)s s

+ L

. Then 001 1( ) so that ( ) 1 (1) [ ] 1 .

1tt w w t tF s f t e e dw e e

s s = = = = = +

25. Find 1 24

( 2)s s

L

2

2 2 2 2

00

2

1 4( )2

1 1( ) 4 4 ( ) 4 ( ) (Integration by parts)2 4

2 1

ttt w w w

t

F ss s

f t t e t w e dw t w e e

e t

= = = = +

=


26. Find 1 2 21

( 1)s s +

L

2 2

0

0

1 1( )1

( ) sin ( )sin

( )( cos ) ( cos )( 1) (Integration by parts)

sin

t

t

F ss s

f t t t t w wdw

t w w w dw

t t

= += = =

= +

27. Find 1 2 21

( 1)s +

L

( )

2 2

0

0

0

1 1( )1 1

( ) sin sin sin( )sin

1 cos cos( 2 ) (by Trigonometric Identity)21 1cos sin( 2 )

2 41 1sin cos2 2

t

t

t

F ss s

f t t t t w wdw

t t w dw

w t t w

t t t

= + += =

= + =

=

28. Find 1 2 2 21

( )s k +

L

( )

2 2 2 2

0

0

0

1 1( )

( ) sin sin sin ( )sin

1 cos cos( 2 ) (by Trigonometric Identity)21 1cos sin( 2 )

2 41 1sin cos

2 2

t

t

t

F ss k s k

f t kt kt k t w kwdw

kt kt kw dw

w kt kt kwk k

kt t ktk

= + += =

= + =

=


Nonzero Initial State 29. Find the Laplace transform for the solution to 0( ), (0) , (0)ax bx cx f t x x x x1 + + = = = in terms of

h(t) where ( ),ah bh ch t + + = (0) (0) 0.h h= = First note that

2

1( ) .H sas bs c

= + + Now solving the differential equation,

20 1 0( ) ( ) ( ) ( )as X s ax s ax bsX s bx cX s F s + + =

Solving for X(s),

0 12

0 1 02

1( ) [ ( ) ]

( ) ( )

0X s F s ax s axas bs cax s ax bxH s F s

as bs c

= + ++ ++ += + + +

bx+

.

Note that solving for x(t) is not required.

Nonzero Practice 30. ( 2), (0) 1, (0) 0x x x t x x + + = = =

22

22 2

1 12 2

1( ) ( )11 32 2( ) 3

1 3 1 32 4 2 4

3 3( ) ( ) ( 2) cos 3 sin2 2

s

s

t t

sX s H s es s

sH s e

s s

x t h t t e t e t

= + + ++

= + + + + + = +

31. 4 4cos , (0) 1, (0)x x t x x + = = = 1

2 2

2 2 2

4 1( ) ( )1 4

4 1( )1 4 2

1( ) ( ) 4cos cos2 sin 2 .2

sX s H ss s

sH ss s s

24

x t h t t t t

= ++ += + + + += +


Fractional Calculus

32. 1/ 2 1/ 21/ 2 01 1(1) ( 1) 1( )

tI t w

= = dw

0

1/ 2 1/ 21/ 2 0

1/ 2 3/ 2 3/ 2

0

1 2[2 ]

1 1( ) ( 1) ( )( )

1 2 423 3

t

t

t

tw

I t t t w w dw

tw w t

= =

= =

= =

( ) ( )2 1/ 2 2 21/ 2 02 1/ 2 3/ 2 2 5/ 2

0

2 1/ 2 3/ 2 2 5/ 2

2

1 1( ) ( ) ( ) ( ) ( )

1 2 22( ) (2 ) ( )3 5

1 2 22( ) (2 ) ( )3 5

16 4 215 3

t

t

I at bt c t at bt c a t w b t w c w dw

at bt c w at b w a w

at bt c t at b t a t

t at bt c

+ + = + + = + +

= + + + + = + + + + = + +

1/ 2

33. 1/ 2 1/ 2 0( ( ))( ) ( )t

I I f t f w dw= 1/ 2

1/ 2 1/ 2 1/ 21( ( ))( ) ( ( ))I I f t t I f

= Applying the convolution theorem to each side yields: (L indicates the Laplace transform.)

[ ] [ ]1/ 2 1/ 2 1/ 21( ( ))( ) ( )1 ( )

( )

I I f t I fs

F ss s

F ss

= =

=

L L

Since Laplace transforms have unique continuous inverses, ( )1/ 2 1/ 2 0( ) ( ) ( )tI I f t f w dw=


34. (a) ( )1/ 2 1/ 21/ 2 (1) (1)2 (see problem 32)

1

d d Idt dt

d tdt

t

= =

=

(b) ( )1/ 2 1/ 21/ 23/ 2

( ) ( )

4 (see problem 32)3

2

d dt I tdt dt

d tdt

t

= =

=

(c) ( )1/ 2 2 21/ 21/ 22

3/ 2 1/ 2 1/ 2

( ) ( )

16 4 2 (see problem 32)15 3

1 8 23

d dat bt c I at bt cdt dt

d t at bt cdt

at bt ct

+ + = + + = + + = + +

Trendy Savings 35. 6( ) 10 .tf t te= (a) 60.08 10 , (0) 0tA A te A = + = (b) .08 6( ) 10t tA t e te=

Investment and Savings 36. .01.04 10000 , (0) 0tA A e A = + =

.04 .01

20 .04(20 ) .01 5

0

( ) 10000

(20) ( )(10000 ) 3.34712 10 or $334,712

t t

w w

A t e e

A e e dw= =

Consistency Check 37. Student Project.


Lake Pollutant 38. .050.1 2 , (0) 0tP P e P = + =

0.1 .05 0.1( ) 0.05 0.05 0.1

0

40 40( ) 2 23 3

tt t t w w tP t e e e e dw e e = = = t

Radioactive Decay Chain 39. 0.001.01 , (0) 0tA A e A = + =

0.01 0.001

0.01( ) 0.001

0

0.001 0.01

( )

90.91 90.91

t t

t t w w

t t

A t e e

e e dw

e e

= ==

Volterra Integral Equation

40. 0

( ) 1 ( )t

y t y w dw=

1 1( ) ( )

1( )1

( ) t

Y s Y ss s

Y ss

y t e

=

= +=

41. 0

( ) ( ) ( )t

y t t t w y w dw=

2 2

2

1 1( ) ( )

1( )1

( ) sin

Y s Y ss s

Y ss

y t t

=

= +=

42. 30

( ) sin( ) ( )t

y t t t w y w dw= +

4 2

2

6 4

3 5

6 1( ) ( )1

6( 1) 6 1 5!( )20

1( )20

Y s Y ss s

sY s 6s s s

y t t t

= + ++= = +

= +


43. 0 0

( ) 1 ( ) ( )t tt w t t wy t e e y w dw e e y w dw = + = +

1 1( ) ( )1 1

1( )

( ) 1

Y s Y ss s

Y ss

y t

= + ==

44. 0

( ) cos sin( ) ( )t

y t t t w y w dw= +

2 2

1( ) ( )1 1

1( )

( ) 1

sY s Y ss s

Y ss

y t

= ++ +==

General Solution of Volterras Equation

45. 0

( ) ( ) ( ) ( )t

y t g t k t w y w dw= +

( ) ( ) ( ) ( )( )( )

1 ( )

Y s G s K s Y sG sY s

K s

= +=

Looking for the Current 46. (a) The integrodifferential equation is

( ) 10 ( ) 25 ( ) ( )I t I t I t dt V t + + = where V(t) = 12 24 step (t 1). (b) Using the initial conditions I(0) = (0) 0I = and taking the Laplace Transform gives

{ }2 ( ) 10 ( ) 25 ( ) { ( )}

( ) (0)

12 24 12

24

s

s

s I s sI s I s V ts V t V

s es se

+ + ==

= =

L

L

where I(s) is the Laplace Transform of I(t).

Solving for I(s) gives

2

1( ) ( 24 ),( 5)

sI s es

= +


which is the product of the transfer function 21( )

( 5)H s

s= + and the Laplace Transform

of . ( )V t

Hence 1 52( ) 24 24 step ( 1) ( 5 )( 5)

st te sI t t e te

s s

= = + L 5 ,

2 25 5because

( 5) ( 5) ( 5)s s

s s s+= + + + 2 .

(c) 5 5 5( 1)12 12 1 12( ) ( 96 120 ) step ( 1)25 5 25 25

t t tI t e e t t e = + + +

Transfer Functions for Circuits

47. 1( ) ( ) ( ) and (0) (0) 0.LQ t Q t V t Q IC

+ = = = Applying Laplace transforms

2

1( ) ( ) { ( )}

1( ) ( ) { ( )}

LQ t Q t V tC

Ls Q s Q s V tC

+ = + =

L L

L

22 2

11( ) 1 11

( ) sin sin

( ) ( ) { ( )}

C C LCH sLCs LCLs s

C L

C t C th tLC LC LC

Q s H s V t

= = =+ + + = =

=L

L

C

Taking the inverse Laplace Transform

{ }1( ) ( ) { ( )}( ) * ( )

sin * ( )

Q t H s V th t V t

C t V tL LC

===

L L


48. ( ) ( ) ( )LI t RI t V t + = Applying Laplace transforms yields

( ) ( ) { ( )},1( ) { ( )},

1 so that ( ) is the transfer function.

LsI s RI s V t

I s V tLs R

H sLs r

+ == += +

L

L

Then { }1 1( ) ( ) { ( )} * ( )R tLI t H s V t e V tL

= =L L

Interesting Convolution 49. siny y + = t

22

2 2

1( )1

1 1( )1 1

( ) sin sin

s Y s Ys

Y ss s

y t t t

+ = += + +=

Duhamels Principle 50. ( ), (0) (0) 0

1, (0) (0) 0ay by cy f t y yaz bz cz z z

+ + = = = + + = = =

2

2

2

2

( ) ( ) ( ) ( )1( ) ( ), and

1( ) ( ) ( )

1( ) , so

( ) ( ) ( )( ) ( ) ( )

as Y s bsY s cY s F s

Y s F sas bs c

as Z s bsZ s cZ ss

sZ sas bs c

Y s sZ s F sy t z t F t

+ + == + +

+ + =

= + +=

=

Using Duhamels Principle 51. ( ), (0) (0) 0y y f t y y = = = 1, (0) (0) 0z z z z = = = has solution

( ) 2t tz t e e= + so

( ) ( 2) ( )t ty t e e f t= +

812 CHAPTER 8 Laplace Transforms Interesting Integral Equation 52.

0( ) ( ) ( )

ty w dw y t y t=

1 ( ) ( ) ( )

1( ) , or ( ) 0

( ) 1 or ( ) 0.

Y s Y s Y ss

Y s Y ss

y t y t

=

= == =

Note y(t) = 0 is easily verified. Also, y(t) = 1 yields 01

tdw t= and 1 * 1 = . 0 (1)(1)t dw t=

Suggested Journal Entry 53. Student Project.

SECTION 8.5 Laplace Transform Solution of Linear Systems 813

8.5 Laplace Transform Solution of Linear Systems

Laplace for Systems 1. , (0) 0

, (0) 1x y xy x y= == =

12

2

2

0 11 0

0 0 1 1 0( ) ( ), so ( ) .

1 0 11 1

11 0 1 01 1( )

1 11 111

sin( )

cos

ss s s s

s

s s sss s ss

st

tt

= = =

+= = = + + =

x x

X X X

X

x

G G

G G G

G

G

2. , (0) 12 4 , (0) 1

x x y xy x y y= = = + =

1

2

2 2

2 2

1 12 4

1 1 1 1 1 1( ) ( ), so ( ) .

2 4 2 41 1

1 1 1 4 1 11( )2 4 21 15 6

4 15 6 5 62 15 6 5 6

ss s s X s

s

s ss

s ss s

ss s s s

ss s s s

= = =

= = + + += + + +

x x

X X

X

G G

G G

G

2

2

12

12

( )t

t

s

se

te

= =

xG

814 CHAPTER 8 Laplace Transforms 3.

4

, (0) 12 3 12 , (0) 1t

x y xy x y e y= == + + =

4

1

2

00 12 3 12

0 11 0 1 1( ) ( ) , so ( ) .12 122 3 2 31 1

4 41 11 31( ) 12 122 3 23 21 1

4 4

te

ss s s s

ss s

s ss

s ss ss s

= +

1

= + = + = = ++ +

x x

X X X

X

G G

G G G

G

2 2

2 2

2

2

2

2

2 4

3 1 1213 2 3 2 4

2 1213 2 3 2 4

6 20 5 6 2( 3 2)( 4) 1 2 4

5 12 86 81 2 4( 3 2)( 4)

5 6 2( )

5

t t t

t

ss s s s s

ss s s s s

s ss s s s s s

s ss s ss s s

e e et

e

+ + + + = + + + + + + + = = + + + +

+= xG

2 412 8t te e

+

4. , (0) 02cos , (0) 0

x y xy x t y= == + =

2 2

0 1 01 0 2cos

0 00 1 1( ) ( ) , so ( ) .2 21 0 1

1 1

t

ss s s ss ss

s s

= + = + = + +

x x

X X X

G G

G G G

1 2 2

2 22 2

2 2

20 0 ( 1)1 11( ) 2 21 11 2

1 1( 1)

sin( )

sin cos

sss s

s s ss ss ss s s

t tt

t t t

+ = = = + + + +

= +

X

x

G

G


5. 3 , (0) 02 3 , (0) 0

tx y e xy x y y= + == + =

3

1

2

2

2

0 12 3 0

1 10 1 1( ) ( ) , so ( ) .3 3

2 3 2 30 0

1 11 3 11( ) 3 32 3 23 20 0

13 22

( 3

te

ss s s ss s

s

s ss s s

s ss s

s s

s s

= + = + =

= +

+= +

x x

X X X

X

G G

G G G

G

2

2 3

1 11 2

1 2 12)( 3) 1 2 3

( )2

t t

t t t

s s

s s s s

e et

e e e

+ = + += +

xG

6. , (0) 03 4 2 4 , (0) 0

x y t xy x y t y= + == + =

2 2

2 2

12 2

2

2 2

0 13 4 2 4

1 10 1 1

( ) ( ) , so ( ) .3 4 3 42 4 2 4

1 11 41( )

3 4 32 4 2 44 3

tt

ss ss s s ss

1

s s s

s ss sss ss s

s s s s

= +

s

= + = = = +

x x

X X X

X

G G

G G G

G

2

2

22 2

3

3

3 1 3 1( 4 3) 2( 1) 2( 3)

1 3 33 2 42( 1) 2( 3)( 4 3)

3 112 2( )3 32 2

t t

t t

s s s s s ss s

s s ss s s

e et

t e e

+ + = = + + + + + = +

xG


7.

1 4 10 0, (0)

1 1 1 0 = + =

G G Gx x x

( ) tGx 1= L { }1 0( ) ( ( )s s +G GI A F x )

11

12

1

1 4 10 011 1 1 0

1 4 1011 1 1( 2 3)

10 141 1 1113 6( 3) 2( 1)

2 1413 .1 113

t

ss s

sss s s

sss s s

e

= + = + = +

= +

L

L

L

8. 3 3 4 1

, (0)2 2 1 0

= + = G G Gx x x

( ) tGx 1= L { }1 0( ) ( ( )s s +G GI A F x )

11

12

21

2

3 3 4 112 2 1 0

2 3 412 3 1( 1)

6 111 11 11

3 1 176 11 .

2 1 12t

ss s

s sss s

s sss s s

e t

= + + + + = + + = + =

L

L

L


9. 5

5

2 1 0, (0)

3 6 1

t

t

ee = + =

G G Gx x x

( )tGx 1= L { }1 0( ) ( ( )s s +G )GI A F x

1

1

3 3

1

51

5

2 1 03 6 1

1 1 11 .1 1 22

t t

s

s

ss

e e

= + = + +

L

10. 0 1 1

, (0)3 4 4 2 1

tt

= + = G G Gx x x

( )tGx 1= L { }1 0( ) ( ( )s s +G )GI A F x

2

2

11

2 21

2 2 2 2

3

1

4 2

1 13 4 1

( 4)( 1) ( 2 4)1( 4 3) 3( 1) ( 2 4)

1 1 11 1 .1 32 2

t t

s

s s

ss

s s s ss s s s s s s

e et

= + + + = + + + = + +

L

L

.

General Solutions of Linear Systems

11. 00

, (0)4 , (0)

x x y x xy x y y y= + == + =

0 0

0 0

1 14 1

1 1 1 1( ) ( ), so ( ) .

4 1 4 1x xs

s s s ssy y

= = =

x x

X X X

G G

G G G


10 0

20 0

2 20

02 2

1 1 1 11( )4 1 4 12 3

1 12 3 2 34 12 3 2 3

1 1 1 12( 1) 2( 3) 4( 1) 4( 3)

1 1 1 1( 1) ( 3) 2( 1) 2( 3)

x xs ss

s sy ys s

sxs s s s

s ys s s s

xs s s s

s s s s

= = =

+ + + + = + + + +

XG

0

0

3 3

0

3 3 0

1 1 1 12 2 4 4( )

1 12 2

t t t t

t t t t

y

e e e e xt

ye e e e

+ + = + + xG

12. 00

4 , (0)1 , (0)

x x y x xy x y y y= == =

0 0

0 0

1 41 1

1 4 1 4( ) ( ), so ( ) .

1 1 1 1x xs

s s s ssy y

= + = = +

x x

X X X

G G

G G G

10 0

20 0

2 22 20 0

0 02 2 2 2

1 4 1 41( )1 1 1 12 5

1 41 4( 1) 4 ( 1) 42 5 2 5

1 1 1 12 5 2 5 ( 1) 4 ( 1) 4

( )

t

x xs ss

s sy ys s

ssx xs ss s s s

s sy ys s s s s s

et

+ + = = + ++ + + + + + + + + + + + = = + + + + + + + + + +

=

X

x

G

G

0

0

cos2 2 sin 21 sin 2 cos22

t

t t

t e t xye t e t


More Complicated Linear System 13. 4 0, (0)

2 0, (0)x x y xx x y y+ + = = + = =

01

1

2

1 1 4 0 01 0 2 1 0

1 1 0 4 0 0 4 1 1 0 1( ) ( ) , so ( )

1 0 2 1 2 1 1 01 0 1

4 1 11( )2 1 2 40 3 4

s ss s s s

s

s s ss

s ss s

+ = + + = = =

+ = =

x x

X X X

X

G G

G G G

G0

s +

2

2

4

4

10

1 115( 1) 5( 4)3 4

2 3 23 4 5( 1) 5( 4)

1 15 5( )3 25 5

t t

t t

s ss ss

s s s s

e et

e e

+ = = + +

= + xG

Higher-Order Systems 14. 1 2 1

2 3 2

3 4 3

4 1 4

, (0) 1, (0) 1, (0) 1, (0)

x x xx x xx x xx x x

= == == == =

1

XG

0 1 0 00 0 1 00 0 0 11 0 0 0

1 10 1 0 0 1 0 01 0 0 1 0 0 1 0 1

( ) ( ), so ( ) .0 0 0 1 0 0 11 11 0 0 0 1 0 01 1

ss

s s s ss

s

= = =

x x

X X

G G

G G


1 3 2

3 2

4 3 2

2 3

3 2

3 2

4 3 2

3 2

2

1 11 0 0 10 1 0 1 11( )0 0 1 1 11 11 0 0 1 11

111

1 11

1 1 12( 1) 2( 1) 1

12(

s s s ss s s ss

s s s s ss s s s

s s ss s s

s s s ss s s

s s s

1

= = + + + + = + + + + +

+ + + +

=

XG

2 2

2 2 2

2 2 2

2 22

11 1

1 11) 2( 1) 1 1 1

1 1 1 12( 1) 2( 1) 1 1 1

11 11 12( 1) 2( 1) 1

1 1 sin2 21 1 cos2 2( )1 1 si2 2

t t

t t

t t

ss s

s ss s s s s

ss s s s s

sss ss s s

e e t

e e tt

e e

+ + + + + + + = + + + + + + + + +

+=

+ xG

cosh sinsinh coscosh sinnsinh cos

1 1 cos2 2

t t

t tt tt ttt t

e e t

+ + =

Finding General Solutions 15. 11 12 1 1

12 22 2 2

( ), (0)( ), (0)

x a x a y f t x cy a x a y f t y c= + + == + + =

1 1

2 2

11 1

2

( ) ( ) ( ), so ( ) ( ) ( ).

( ) ( ) ( ) ( )

( ) ( ) ( ),h p

c cs s s s s s

c c

cs s s s

ct t t

= + = + = +

= + = +

x Ax f

X AX F I A X

X I A I A F

x x x

GG GG G G G

G G

G G G

sFG

where each term is the inverse Laplace transform of the term above. Therefore depends only on A, c

( )h txG

1, and c2, while depends only on A, f( )p txG

1(t) and f2(t).


Drug Metabolism 16. 1 1 1

2 1 2 2

, (0) 1

, (0)

x kx x

x kx kx x

= = = = 0

1

0

1 0 0( ) ( ), so ( )

0 0

kk k

k s ks s s s

k k k s k

= = = +

x x

X X X

GG G

G G G

1

2 2

0 1 0 11( )0 0

1 1

1 1( )( ) 2( ) 2( )

( ) 1 12 2

kt

kt kt

s k s ks

k s k k s ks k

s k s kk

s k s k s k s k

et

e e

+ = = + = = + + =

X

x

G

G

Mass-Spring Systems 17. We apply m1 = m2 = 2 and k1 = k2 = k3 = 1 to the system of DEs in Example 4 to obtain

2 ( ) 22 ( ) 2

x x y x x yy y x y x y= + = += =

so that 2

2

2[ ( ) 1] 2 ( ) ( ) 2[ ( )] ( ) 2 ( )

s X s X s Y ss Y s X s Y s

= +=

2

2

( ) 22 2 1, so

( ) 01 2 2X ssY ss

+ = +

12 2

2 22 2

2

2 24 2

2 24 2

( ) 2 22 2 1 2 2 11( ) 0 04( 1) 11 2 2 1 2 2

1 1142 3 2 14 8 3

1 122 3 2 14 8 3

1 2 3 2 1sin sin2 3 2 2( )

( )

X s s sY s ss s

ss ss s

s ss s

tx ty t

+ + = = + + + + + + ++ + = = + + + + +

+ =

2

1 2 3 2 1sin sin2 3 2 2 2

t

t t

+


)

18. We apply m1 = m2 = 2 and k1 = k2 = k3 = 1 to the system of DEs in Example 4 to obtain 22 (

x y xy y x= =

so that 2

2

2[ ( ) 1] ( ) ( )2 ( ) [ ( ) ( )]

s X s Y s X ss Y s Y s X s

= = , so

2

2

( ) 22 1 1( ) 01 2 1

X ssY ss

+ = +

12 2

4 22 2

2

2 22 2

2 22 2

( ) 2 22 1 1 2 1 11( ) 0 04( )1 2 1 1 2 1

1 11 2 12 2( 1)2 ( 1)1 11

2 2( 1)2 ( 1)

X s s sY s s ss s

ss ss s

s ss s

+ + = = + + + + + ++ = = ++

1 1 sin( ) 2 2( ) 1 1 sin

2 2

t tx ty t t t

+ =

Note that this solution sends the center of mass moving to the right with a constant velocity of 12

in accordance with the laws of classical mechanics.

A Three-Compartment Model

19. 1 1 12 1 21dx

2R ux k x k xdt= + (given)

212 1 21 2 23 2 32 3

323 2 32 3

dx s k x k x k x k xdt

dx k x k xdt

= + +

=


Vibration with a Free End 20. (a) 1 1 1 2 1

2 1 2 2 2

4 2( ), (0) (0)2 2( ), (0) 1, (0) 0x x x x x x

x x x x x= = == = =

0

22

2

12 2

4 22 2

1 0 6 2 00 1 1 1 0

1 0 0 6 2 0 06 2( ) ( ) , so ( ) .

0 1 1 11 0 2 1

0 06 2 1 21( )7 22 1 2 6

ss s s s s

ss

s ss

s ss ss s

+ = + + = = +

+ + = = + + + +

x x

X X X

X

G G

G G G

G

4 2

2

4 2

27 2

( 6)7 2

ss s

s ss s

+ + = + + +

Comparing Laplace

21. 1 0 1 0

, (0)0 2 0 0 = + =

x x xG G G

1

1 11 0 1 0( ) ( ) , so ( ) .

0 2 0 20 0

11 11 0 2 01 ( 1)( ) 10 2 0 1( 1)( 2)0 0 0

( )t

ss s s ss s

s

s ss ss

1 1

0s s s

s ss s

et

+ = + =

s + + += = = = + ++

=

X X X

X

x

G G G

G

G 10

+

22. 2 0 0 0

, (0)0 3 6 0

= + = x x xG G G

1

2

3

0 02 0 2 0( ) ( ) , so ( ) .6 60 3 0 3

00 02 0 3 01( ) 66 60 3 0 25 6( 3) 3

0( )

2 2t

ss s s s

ss s

s ss

s ss ss s

02 2

s s s

te

= + =

s

= = = = + =

X X X

X

x

G G G

G

G


23. 0 1 1 0

, (0)1 0 1 0

= + = x x xG G G

1 2 2

2

2 2

1 10 1 1

( ) ( ) , so ( )1 0 11 1

1 11 11 1 1 ( 1)1 1( )

1 11 1 1 111 ( 1)

ss ss s s ss

s s

s s s s s1 1

1s s ss

s sss s s s s s

= + = + = = = = +

X X X

X

G G G

G s1

1

1( )

1

t

t

s

et

e

=

xG

Suggested Journal Entry I 24. Student Project

Suggested Journal Entry II 25. Student Project

CH08a-new.DOCCH08b-new.DOC

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