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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

  • SECTION 8.1 The Laplace Transform and its Inverse 771

    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 .

  • SECTION 8.1 The Laplace Transform and its Inverse 773

    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 fracti