nCHAP_6.WXP</title>n<meta http-equiv='Content-Type' content='text/html; charset='UTF-8'>n</head>nn<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 254n</p>n<p><b>Chapter Sixn</b></p>n<p><b>Section 6.1n</b></p>n<p>3.n</p>n<p>The function is 0 >a b <i>continuous</i>.n4.n</p>n<p>The function has a at .0 > > œ "a b <i>jump discontinuityn</i>7. Integration is a linear operation. It follows thatn</p>n<p>( ( (n( (n</p>n<p>! ! !n</p>n<p>E E En=> ,> => ,> =>n</p>n<p>! !n</p>n<p>E En,= >  ,= >n</p>n<p>-9=2 ,> † / .> œ / † / .>  / † / .>n" "n</p>n<p># #n</p>n<p>œ / .>  / .> Þn" "n</p>n<p># #na b a bn</p>n<p>Hencen</p>n<p>( – — – —n!n</p>n<p>En=>n</p>n<p>,= E  ,= En</p>n<p>-9=2 ,> † / .> œ  Þn" "  / " "  /n</p>n<p># =  , # =  ,n</p>n<p>a b a bn</p>n<p>Taking a , as ,<i>limit </i>Ep_</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 255n</p>n<p>( ” • ” •n!n</p>n<p>_n=>n</p>n<p># #n</p>n<p>-9=2 ,> † / .> œ n" " " "n</p>n<p># =  , # =  ,n</p>n<p>œ Þn=n</p>n<p>=  ,n</p>n<p>Note that the above is valid for .=  ,k kn8. Proceeding as in Prob. ,(n</p>n<p>( – — – —n!n</p>n<p>En=>n</p>n<p>,= E  ,= En</p>n<p>=382 ,> † / .> œ  Þn" "  / " "  /n</p>n<p># =  , # =  ,n</p>n<p>a b a bn</p>n<p>Taking a , as ,<i>limit </i>Ep_n</p>n<p>( ” • ” •n!n</p>n<p>_n=>n</p>n<p># #n</p>n<p>=382 ,> † / .> œ n" " " "n</p>n<p># =  , # =  ,n</p>n<p>œ Þn,n</p>n<p>=  ,n</p>n<p>The limit exists as long as .=  ,k kn10. Observe that It follows that/ =382 ,> œ /  / Î# Þ+> Ð+,Ñ> Ð+,Ñ>ˆ ‰n</p>n<p>( – — – —n!n</p>n<p>En+> =>n</p>n<p>+,= E  ,+= En</p>n<p>/ =382 ,> † / .> œ  Þn" "  / " "  /n</p>n<p># =  +  , # =  ,  +n</p>n<p>a b a bn</p>n<p>Taking a , as ,<i>limit </i>Ep_n</p>n<p>( ” • ” •na bn</p>n<p>!n</p>n<p>_n+> =>n</p>n<p># #n</p>n<p>/ =382 ,> † / .> œ n" " " "n</p>n<p># =  +  , # =  ,  +n</p>n<p>œ Þn,n</p>n<p>=  +  ,n</p>n<p>The limit exists as long as .=  +  ,k kn11. Using the of the Laplace transform,<i>linearityn</i></p>n<p>_ _ _c d  ‘  ‘=38 ,> œ /  / Þ" "n#3 #3n</p>n<p>3,> 3,>n</p>n<p>Sincen</p>n<p>(n!n</p>n<p>_n+3, > =>/ / .> œn</p>n<p>"n</p>n<p>=  +  3,na b ,n</p>n<p>we have</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 256n</p>n<p>(n!n</p>n<p>_n„ 3,> =>/ / .> œn</p>n<p>"n</p>n<p>=… 3,n.n</p>n<p>Thereforen</p>n<p>_c d ” •=38 ,> œ " " "n#3 =  3, =  3,n</p>n<p>œn,n</p>n<p>=  ,# #n.n</p>n<p>12. Using the of the Laplace transform,<i>linearityn</i></p>n<p>_ _ _c d  ‘  ‘-9= ,> œ /  / Þ" "n# #n</p>n<p>3,> 3,>n</p>n<p>From Prob. , we have""n</p>n<p>(n!n</p>n<p>_n„ 3,> =>/ / .> œn</p>n<p>"n</p>n<p>=… 3,n.n</p>n<p>Thereforen</p>n<p>_c d ” •-9= ,> œ " " "n# =  3, =  3,n</p>n<p>œn=n</p>n<p>=  ,# #n.n</p>n<p>14. Using the of the Laplace transform,<i>linearityn</i></p>n<p>_ _ _ ‘  ‘  ‘/ -9= ,> œ /  / Þ" "n# #n</p>n<p>+> +3, > +3, >a b a bn</p>n<p>Based on the integration in Prob. ,""n</p>n<p>(n!n</p>n<p>_n+„ 3, > =>/ / .> œ Þn</p>n<p>"n</p>n<p>=  +… 3,na b n</p>n<p>Thereforen</p>n<p>_ ‘ ” •na bn</p>n<p>/ -9= ,> œ n" " "n</p>n<p># =  +  3, =  +  3,n</p>n<p>œn=  +n</p>n<p>=  +  ,n</p>n<p>+>n</p>n<p># #n.n</p>n<p>The above is valid for .=  +n</p>n<p>15. Integrating ,<i>by parts</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 257n</p>n<p>( (ºna bn</p>n<p>a bn! !n</p>n<p>E En+> => += >n</p>n<p>+= > En</p>n<p>!nE += E +=n</p>n<p>#n</p>n<p>>/ † / .> œ   / .>n> / "n</p>n<p>=  + =  +n</p>n<p>œ Þn"  /  E +  = /n</p>n<p>=  +n</p>n<p>a b a bna b a bn</p>n<p>Taking a , as ,<i>limit </i>Ep_n</p>n<p>( a b!n_n</p>n<p>+> =>n#n</p>n<p>>/ † / .> œ Þn"n</p>n<p>=  +n</p>n<p>Note that the limit exists as long as .=  +n</p>n<p>17. Observe that For any value of ,> -9=2 +> œ > /  > / Î# Þ -a b+> +>n( (ºn</p>n<p>a bna bn</p>n<p>! !n</p>n<p>E En-> => -= >n</p>n<p>-= > En</p>n<p>!nE -= E -=n</p>n<p>#n</p>n<p>> / † / .> œ   / .>n> / "n</p>n<p>=  - =  -n</p>n<p>œ Þn"  /  E -  = /n</p>n<p>=  -n</p>n<p>a b a bna b a bn</p>n<p>Taking a , as ,<i>limit </i>Ep_n</p>n<p>( a b!n_n</p>n<p>-> =>n#n</p>n<p>>/ † / .> œ Þn"n</p>n<p>=  -n</p>n<p>Note that the limit exists as long as . Therefore,=  -k kn( – —a b a bn</p>n<p>a b a bn!n</p>n<p>_n=>n</p>n<p># #n</p>n<p># #n</p>n<p># #n</p>n<p>> -9=2 +> † / .> œ n" " "n</p>n<p># =  + =  +n</p>n<p>œn=  +n</p>n<p>=  + =  +n.n</p>n<p>18. Integrating ,<i>by partsn</i></p>n<p>( (ºn(n</p>n<p>! !n</p>n<p>E En8 +> => 8" += >n</p>n<p>8 += > En</p>n<p>!n8  =+ En</p>n<p>!n</p>n<p>En8" += >n</p>n<p>> / † / .> œ   > / .>n> / 8n</p>n<p>=  + =  +n</p>n<p>œ   > / .> ÞnE / 8n</p>n<p>=  + =  +n</p>n<p>a b a bna b a bn</p>n<p>Continuing to integrate by parts, it follows that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 258n</p>n<p>( a bna b a b a bn</p>n<p>ˆ ‰!nEn8 +> =>n</p>n<p>8 += E 8" += En</p>n<p>#n</p>n<p>+= En</p>n<p>$ 8"n</p>n<p>+= En</p>n<p>> / † / .> œ   nE / 8E /n</p>n<p>=  + =  +n</p>n<p> â Þn8xE/n</p>n<p>8  # x =  + =  +n</p>n<p>8x /  "n</p>n<p>a b a bn</p>n<p>a b a bn</p>n<p>That is,n</p>n<p>( a b a b!nEn8 +> => += En</p>n<p>8 8"n> / † / .> œ : E † / n</p>n<p>8xn</p>n<p>=  +na b ,n</p>n<p>in which is a of degree . For any polynomial,: 88a b0 <i>polynomial givenn</i>limnEÄ_n</p>n<p>8n =+ E: E † / œ !a b a b ,n</p>n<p>as long as . Therefore,=  +n</p>n<p>( a b!n_n8 +> =>n</p>n<p>8"n> / † / .> œn</p>n<p>8xn</p>n<p>=  +n.n</p>n<p>20. Observe that Using the result in Prob. ,> =382 +> œ > /  > / Î# Þ ")# # +> # +>a bn( – —a b a bn</p>n<p>a bna bn</p>n<p>!n</p>n<p>_n# =>n</p>n<p>$ $n</p>n<p># #n</p>n<p># # $n</p>n<p>> =382 +> † / .> œ n" #x #xn</p>n<p># =  + =  +n</p>n<p>œn#+ $=  +n</p>n<p>=  +n.n</p>n<p>The above is valid for .=  +k kn22. Integrating ,<i>by partsn</i></p>n<p>( (ºn! !n</p>n<p>E En> > >n</p>n<p>En</p>n<p>!nE En</p>n<p>> / .> œ  > /  / .>n</p>n<p>œ "  /  E/ Þn</p>n<p>Taking a , as ,<i>limit </i>Ep_n</p>n<p>(n!n</p>n<p>_n> E> / .> œ "  / Þn</p>n<p>Hence the integral .<i>convergesn</i></p>n<p>23. Based on a series expansion, note that for ,>  !n</p>n<p>/  "  >  > Î#  > Î#> # # .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 259n</p>n<p>It follows that for ,>  !n</p>n<p>> / n"n</p>n<p>#n# > .n</p>n<p>Hence for any finite ,E  "n</p>n<p>(n"n</p>n<p>En# >> / .> n</p>n<p>E  "n</p>n<p>#n.n</p>n<p>It is evident that the limit as does not exist.Ep_n</p>n<p>24. Using the fact that , and the fact thatk k-9= > Ÿ "n(n!n</p>n<p>_n>/ .> œ " ,n</p>n<p>it follows that the given integral .<i>convergesn</i></p>n<p>25 . Let . Integrating ,a b+ :  ! <i>by partsn</i>( (ºn</p>n<p>(n! !n</p>n<p>E EnB : B : B :"n</p>n<p>En</p>n<p>!n</p>n<p>: E B :"n</p>n<p>!n</p>n<p>En</p>n<p>/ B .B œ  / B  : / B .Bn</p>n<p>œ  E /  : / B .BÞn</p>n<p>Taking a , as ,<i>limit </i>Ep_n</p>n<p>( (n! !n</p>n<p>_ _nB : B :"/ B .B œ : / B .B .n</p>n<p>That is, .> >a b a b:  " œ : :na b, : œ !. Setting ,n</p>n<p>>a b (" œ / .B œ "n!n</p>n<p>_nB .n</p>n<p>a b a b- : œ 8 ,. Let . Using the result in Part ,n> >n</p>n<p>>n</p>n<p>>n</p>n<p>a b a ba b a bna ba b a bn</p>n<p>8  " œ 8 8n</p>n<p>œ 8 8  " 8  "n</p>n<p>ãn</p>n<p>œ 8 8  " 8  # â# † " † " .n</p>n<p>Since , .> >a b a b" œ " 8  " œ 8xna b a b. ,. Using the result in Part ,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 260n</p>n<p>> >n</p>n<p>>n</p>n<p>>n</p>n<p>a b a b a ba ba b a bna ba b a b a bn</p>n<p>:  8 œ :  8  " :  8  "n</p>n<p>œ :  8  " :  8  # :  8  #n</p>n<p>ãn</p>n<p>œ :  8  " :  8  # â :  " : : .n</p>n<p>Hencen</p>n<p>>n</p>n<p>>n</p>n<p>a ba b a ba b a b:  8: œ : :  " :  " â :  8  " ÞnGiven that , it follows that> 1a b È"Î# œn</p>n<p>> >n1Œ  Œ  È$ " "n</p>n<p># # # #nœ œn</p>n<p>andn</p>n<p>> >n1Œ  Œ  È"" * ( & $ $n</p>n<p># # # # # # $#nœ † † † œn</p>n<p>*%&n.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 261n</p>n<p><b>Section 6.2n</b></p>n<p>1. Write the function asn</p>n<p>$ $ #n</p>n<p>=  % # =  %nœn</p>n<p># #n.n</p>n<p>Hence ._" $#c da b] = œ =38 #>n3. Using <i>partial fractions</i>,n</p>n<p># # " "n</p>n<p>=  $=  % & =  " =  %nœ n</p>n<p># ” •.nHence ._" > %>#&c da b ˆ ‰] = œ /  /n5. Note that the denominator is =  #=  &# <i>irreducible</i> over the reals. Completing thensquare, =  #=  & œ =  "  %# #a b . Now convert the function to a <i>rational functionn</i>of the variable . That is,0 œ =  "n</p>n<p>#=  # # =  "n</p>n<p>=  #=  &nœn</p>n<p>#n</p>n<p>a bna b=  "  %# .n</p>n<p>We know thatn</p>n<p>_n0n</p>n<p>0n"n</p>n<p>#” •# % œ # -9= #> .nUsing the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+n</p>n<p>_" >” •#=  #n=  #=  &#n</p>n<p>œ #/ -9= #> .n</p>n<p>6. Using <i>partial fractions</i>,n</p>n<p>#=  $ " " (n</p>n<p>=  % % =  # =  #nœ n</p>n<p># ” •.nHence . Note that we can also write_" #> #>"%c d a ba b] = œ /  (/n</p>n<p>#=  $ = $ #n</p>n<p>=  % =  % # =  %nœ # n</p>n<p># # #n.n</p>n<p>8. ,Using <i>partial fractionsn</i></p>n<p>)=  %=  "# " = #n</p>n<p>= =  % = =  % =  %nœ $  &  #n</p>n<p>#n</p>n<p># # #a b .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 262n</p>n<p>Hence ._"c da b] = œ $  & -9= #>  # =38 #>n9. The denominator is =  %=  &# <i>irreducible</i> over the reals. Completing the square,n=  %=  & œ =  #  "# #a b . Now convert the function to a <i>rational function </i>of thenvariable . That is,0 œ =  #n</p>n<p>"  #= &  # =  #n</p>n<p>=  %=  &nœn</p>n<p>#n</p>n<p>a bna b=  #  "# .n</p>n<p>We find thatn</p>n<p>_n0 0n</p>n<p>0"n# #” •& # "  " œ & =38 >  # -9= > .n</p>n<p>Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+n_" #>” • a b"  #=n</p>n<p>=  %=  &#nœ / & =38 >  # -9= > .n</p>n<p>10. over the reals. CompletingNote that the denominator is =  #=  "!# <i>irreduciblen</i>the square, =  #=  "! œ =  "  *# #a b . Now convert the function to a <i>rationalnfunction </i>of the variable . That is,0 œ =  "n</p>n<p>#=  $ # =  "  &n</p>n<p>=  #=  "!nœn</p>n<p>#n</p>n<p>a bna b=  "  *# .n</p>n<p>We find thatn</p>n<p>_n0n</p>n<p>0 0n"n</p>n<p># #” •# & & *  * $ œ # -9= $>  =38 $> .nUsing the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+n</p>n<p>_" >” • Œ #=  $n=  #=  "!#n</p>n<p>œ / # -9= $>  =38 $>n&n</p>n<p>$n.n</p>n<p>12. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  $ = ] =  C !  #] = œ !# wa b a b a b c d a ba b a b .nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  $= ] =  #] =  =  $ œ !# a b a b a b .nSolving for , the transform of the solution is] =a bn</p>n<p>] = œn=  $n</p>n<p>=  $=  #na bn</p>n<p>#n.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 263n</p>n<p>Using <i>partial fractions</i>,n</p>n<p>=  $n</p>n<p>=  $=  ##nœ n</p>n<p># "n</p>n<p>=  " =  #n.n</p>n<p>Hence .C > œ ] = œ # /  /a b c da b_" > #>n13. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  # = ] =  C !  #] = œ !# wa b a b a b c d a ba b a b .nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  #= ] =  #] =  " œ !# a b a b a b .nSolving for , the transform of the solution is] =a bn</p>n<p>] = œn"n</p>n<p>=  #=  #na bn</p>n<p>#n.n</p>n<p>Since the denominator is , write the transform as a function of .<i>irreducible </i>0 œ =  "nThat is,n</p>n<p>"n</p>n<p>=  #=  ##nœn</p>n<p>"n</p>n<p>=  "  "a b# .nFirst note thatn</p>n<p>_n0n</p>n<p>"n#” •" " œ =38 > .n</p>n<p>Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+n_" >” •"n</p>n<p>=  #=  ##nœ / =38 > .n</p>n<p>Hence .C > œ / =38 >a b >n15. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  # = ] =  C !  #] = œ !# wa b a b a b c d a ba b a b .nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  #= ] =  #] =  #=  % œ !# a b a b a b .nSolving for , the transform of the solution is] =a bn</p>n<p>] = œn#=  %n</p>n<p>=  #=  #na bn</p>n<p>#n.n</p>n<p>Since the denominator is , write the transform as a function of .<i>irreducible </i>0 œ =  "nCompleting the square,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 264n</p>n<p>#=  %n</p>n<p>=  #=  ##nœn# =  "  #n</p>n<p>=  "  $n</p>n<p>a bna b# .n</p>n<p>First note thatn</p>n<p>_n0n</p>n<p>0 0n"n</p>n<p># #” • È ÈÈ# # # $  $ œ # -9=2 $ >  =382 $ >$ .nUsing the fact that , the solution of the IVP is_ _c d c da b a b/ 0 > œ 0 >+> =p=+n</p>n<p>C > œa b _" >” •  È ÈÈ#=  %=  #=  ## œ / # -9=2 $ >  =382 $ >#$ .n</p>n<p>16. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  # = ] =  C !  &] = œ !# wa b a b a b c d a ba b a b .nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  #= ] =  &] =  #=  $ œ !# a b a b a b .nSolving for , the transform of the solution is] =a bn</p>n<p>] = œn#=  $n</p>n<p>=  #=  &na bn</p>n<p>#n.n</p>n<p>Since the denominator is , write the transform as a function of .<i>irreducible </i>0 œ =  "nThat is,n</p>n<p>#=  $n</p>n<p>=  #=  &#nœn# =  "  "n</p>n<p>=  "  %n</p>n<p>a bna b# .n</p>n<p>We know thatn</p>n<p>_n0n</p>n<p>0 0n"n</p>n<p># #” •# " " %  % # œ # -9= #>  =38 #> .nUsing the fact that , the solution of the IVP is_ _c d c da b a b/ 0 > œ 0 >+> =p=+n</p>n<p>C > œa b _" >” • Œ #=  $n=  #=  &#n</p>n<p>œ / # -9= #>  =38 #>n"n</p>n<p>#n.n</p>n<p>17. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  = C !  = C !  C !  % = ] =  = C !  = C !  C ! n</p>n<p> ' = ] =  = C !  C !  % = ] =  C !  ] = œ !n</p>n<p>% $ # w ww www $ # w wwn</p>n<p># wn</p>n<p>a b a b a b a b a b a b a b a b a b ‘ ‘a b a b a b c d a ba b a bnApplying the ,<i>initial conditions</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 265n</p>n<p>= ] =  %= ] =  '= ] =  %= ] =  ] =  =  %=  ( œ !% $ # #a b a b a b a b a b .nSolving for the transform of the solution,n</p>n<p>] = œ œn=  %=  ( =  %=  (n</p>n<p>=  %=  '=  %=  " =  "na b a bn</p>n<p># #n</p>n<p>% $ # %n.n</p>n<p>Using <i>partial fractions</i>,n</p>n<p>=  %=  ( % # "n</p>n<p>=  " =  "nœ  n</p>n<p>=  " =  "n</p>n<p>#n</p>n<p>% % $ #a b a b a b a b .nNote that and . Hence the solution_ _ _c d a b c d c da b a b> œ 8x Î= / 0 > œ 0 >8 8" +> =p=+nof the IVP isn</p>n<p>C > œn=  %=  (n</p>n<p>=  "na b a b_" $ > # > >– —n</p>n<p>#n</p>n<p>%nœ > /  > /  > /n#n</p>n<p>$n.n</p>n<p>18. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  = C !  = C !  C !  ] = œ ! Þ% $ # w ww wwwa b a b a b a b a b a b nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  ] =  =  = œ !% $a b a b .nSolving for the transform of the solution,n</p>n<p>] = œn=n</p>n<p>=  "na bn</p>n<p>#n.n</p>n<p>By inspection, it follows that C > œa b _" ‘ "# œ -9=2 > Þn19. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  = C !  = C !  C !  %] = œ ! Þ% $ # w ww wwwa b a b a b a b a b a b nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  %] =  =  #= œ !% $a b a b .nSolving for the transform of the solution,n</p>n<p>] = œn=n</p>n<p>=  #na bn</p>n<p>#n.n</p>n<p>It follows that C > œa b _" ‘ È ## œ -9= # > Þn20. Taking the Laplace transform of both sides of the ODE, we obtain</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 266n</p>n<p>= ] =  = C !  C !  ] = œn=n</p>n<p>=  %n# w #n</p>n<p>#na b a b a b a b= .n</p>n<p>Applying the ,<i>initial conditionsn</i></p>n<p>= ] =  ] =  = œn=n</p>n<p>=  %n# #n</p>n<p>#na b a b= .n</p>n<p>Solving for , the transform of the solution is] =a bn] = œa b = =n</p>n<p>=  =  % = na ba b# # # # #= = .n</p>n<p>Using <i>partial fractions</i> on the first term,n</p>n<p>= " = =n</p>n<p>=  =  % %  =  =  %nœ a ba b ” •# # # # # # #= = = .n</p>n<p>First note thatn</p>n<p>_ = _n=n</p>n<p>" "n# # #” • ” •=  =  %œ -9= > œ -9= #> and .n</p>n<p>Hence the solution of the IVP isn</p>n<p>C > œ -9= >  -9= #>  -9= >n" "n</p>n<p>%  % n</p>n<p>œ -9= >  -9= #>n&  "n</p>n<p>%  % n</p>n<p>a bn= =n</p>n<p>= =n</p>n<p>=n</p>n<p>= =n=n</p>n<p># #n</p>n<p>#n</p>n<p># #n.n</p>n<p>21. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  # = ] =  C !  #] = œn=n</p>n<p>=  "n# wn</p>n<p>#na b a b a b c d a ba b a b .n</p>n<p>Applying the ,<i>initial conditionsn</i></p>n<p>= ] =  #= ] =  #] =  =  # œn=n</p>n<p>=  "n#n</p>n<p>#na b a b a b .n</p>n<p>Solving for , the transform of the solution is] =a bn] = œa b = =  #n</p>n<p>=  #=  # =  " =  #=  #na ba b# # # .n</p>n<p>Using <i>partial fractions</i> on the first term,n</p>n<p>= " =  # =  %n</p>n<p>=  #=  # =  " & =  " =  #=  #nœ a ba b ” •# # # # .n</p>n<p>Thus we can write</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 267n</p>n<p>] = œn"n</p>n<p>&na b = # " # #=  $n</p>n<p>=  " & =  " & =  #=  #n n</p>n<p># # #n.n</p>n<p>For the , we note that . So that<i>last term </i>=  #=  # œ =  "  "# #a bn#=  $ # =  "  "n</p>n<p>=  #=  #nœn=  "  "# #na bna b .n</p>n<p>We know thatn</p>n<p>_n0n</p>n<p>0 0n"n</p>n<p># #” •# " "  " œ # -9= >  =38 > .nBased on the of the Laplace transform,<i>translation propertyn</i></p>n<p>_" >” • a b#=  $n=  #=  ##n</p>n<p>œ / # -9= >  =38 > .n</p>n<p>Combining the above, the solution of the IVP isn</p>n<p>C > œ -9= >  =38 > n" # #n</p>n<p>& & &na b / # -9= >  =38 > Þ>a bn</p>n<p>23. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  # = ] =  C !  ] = œn%n</p>n<p>=  "n# wa b a b a b c d a ba b a b .n</p>n<p>Applying the ,<i>initial conditionsn</i></p>n<p>= ] =  #= ] =  ] =  #=  $ œn%n</p>n<p>=  "n# a b a b a b .n</p>n<p>Solving for , the transform of the solution is] =a bn] = œa b % #=  $n</p>n<p>=  " =  "na b a b$ # .n</p>n<p>First writen</p>n<p>#=  $ # =  "  " # "n</p>n<p>=  " =  " =  "nœ œ  Þn</p>n<p>=  "a b a b a bna bn</p>n<p># # #n</p>n<p>We note thatn</p>n<p>_n0 0 0n</p>n<p>" #n$ #” •% # "  œ # >  #  > .n</p>n<p>So based on the of the Laplace transform, the solution of the IVP is<i>translation property</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 268n</p>n<p>C > œa b # > /  > /  # / Þ# > > >n25. Let be the on the right-hand-side. Taking the Laplace0 >a b <i>forcing functionn</i>transformnof both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  ] = œ 0 ># wa b a b a b a b c da b_ .nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  ] = œ 0 ># a b a b c da b_ .nBased on the definition of the Laplace transform,n</p>n<p>_c d a ba b (n(n</p>n<p>0 > œ 0 > / .>n</p>n<p>œ > / .>n</p>n<p>œ  n" / /n</p>n<p>= = =n</p>n<p>!n</p>n<p>_n=>n</p>n<p>!n</p>n<p>"n=>n</p>n<p># #n</p>n<p>= =n</p>n<p>.n</p>n<p>Solving for the transform,n</p>n<p>] = œ  / Þn" =  "n</p>n<p>= =  " = =  "na b a b a b# # # #=n</p>n<p>Using <i>partial fractions</i>,n</p>n<p>" " "n</p>n<p>= =  " = =  "nœ n</p>n<p># # # #a bnandn</p>n<p>= " =n</p>n<p>= =  " = =  "nœ  Þn</p>n<p># # #a bnWe find, by inspection, thatn</p>n<p>_"” •"n= =  "# #a b œ >  =38 > .n</p>n<p>Referring to , in Table ,<i>Line </i>"$ 'Þ#Þ"n</p>n<p>_ _c d c da b a b a b? > 0 >  - œ / 0 >- -= .nLetn</p>n<p>_c da b a b1 > œ œ=  "= =  "# # " " = "= = =  " =  "  # # # .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 269n</p>n<p>Then . It follows, therefore, that1 > œ "  >  -9= >  =38 >a bn_" = "” • a bc d/ † œ ? >=  "n</p>n<p>= =  "# #a b "  >  "  -9= >  "  =38 >  "a b a b a b .nCombining the above, the solution of the IVP isn</p>n<p>C > œ "  >  "  -9= >  "  =38 >  "a b a b a b a b>  =38 >  ? >"a bc d .n26. Let be the on the right-hand-side. Taking the Laplace0 >a b <i>forcing functionn</i>transformnof both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  %] = œ 0 ># wa b a b a b a b c da b_ .nApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  %] = œ 0 ># a b a b c da b_ .nBased on the definition of the Laplace transform,n</p>n<p>_c d a ba b (n( (n</p>n<p>0 > œ 0 > / .>n</p>n<p>œ > / .>  / .>n</p>n<p>œ n" /n</p>n<p>= =n</p>n<p>!n</p>n<p>_n=>n</p>n<p>! "n</p>n<p>" _n=> =>n</p>n<p># #n</p>n<p>=n</p>n<p>.n</p>n<p>Solving for the transform,n</p>n<p>] = œ  / Þn" "n</p>n<p>= =  % = =  %na b a b a b# # # #=n</p>n<p>Using <i>partial fractions</i>,n</p>n<p>" " " "n</p>n<p>= =  % % = =  %nœ  Þn</p>n<p># # # #a b ” •nWe find thatn</p>n<p>_"n# #” •a b" " "= =  % % )œ >  =38 > .n</p>n<p>Referring to , in Table ,<i>Line </i>"$ 'Þ#Þ"n</p>n<p>_ _c d c da b a b a b? > 0 >  - œ / 0 >- -= .nIt follows that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 270n</p>n<p>_" =n# # "” • ” •a b a b a b a b/ † œ ? > >  "  =38 >  "" " "= =  % % ) .n</p>n<p>Combining the above, the solution of the IVP isn</p>n<p>C > œ >  =38 >  ? > >  "  =38 >  "n" " " "n</p>n<p>% ) % )na b a b a b a b” •" .n</p>n<p>28 . Assuming that the conditions of Theorem are satisfied,a b+ 'Þ#Þ"nJ = œ / 0 > .>n</p>n<p>.n</p>n<p>.=n</p>n<p>œ / 0 > .>n`n</p>n<p>`=n</p>n<p>œ  > / 0 > .>n</p>n<p>œ /  >0 > .> Þn</p>n<p>w =>n</p>n<p>!n</p>n<p>_n</p>n<p>!n</p>n<p>_n=>n</p>n<p>!n</p>n<p>_n=>n</p>n<p>!n</p>n<p>_n=>n</p>n<p>a b a b(n(  ‘a bn(  ‘a bn( c da bn</p>n<p>a b, 5   ". Using , suppose that for some ,<i>mathematical inductionn</i>J = œ /  > 0 > .>a b5 =>n</p>n<p>!n</p>n<p>_n5a b a b a b( ’ “ .n</p>n<p>Differentiating both sides,n</p>n<p>J = œ /  > 0 > .>n.n</p>n<p>.=n</p>n<p>œ /  > 0 > .>n`n</p>n<p>`=n</p>n<p>œ  > /  > 0 > .>n</p>n<p>œ /  > 0 > .>n</p>n<p>a b5" =>n!n</p>n<p>_n5n</p>n<p>!n</p>n<p>_n=> 5n</p>n<p>!n</p>n<p>_n=> 5n</p>n<p>!n</p>n<p>_n=> 5"n</p>n<p>a b a b a b( ’ “n( ’ “a b a bn( ’ “a b a bn( ’ “a b a b .n</p>n<p>29. We know thatn</p>n<p>_ ‘/ œ "n=  +n</p>n<p>+> .n</p>n<p>Based on Prob. ,#)n</p>n<p>_ ‘ ” • > / œ . "n.= =  +n</p>n<p>+> .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 271n</p>n<p>Therefore,n</p>n<p>_ ‘ a b> / œn"n</p>n<p>=  +n+>n</p>n<p>#n.n</p>n<p>31. Based on Prob. ,#)n</p>n<p>_ _c d c da bn” •n</p>n<p> > œ "n.n</p>n<p>.=n</p>n<p>œ Þn. "n</p>n<p>.= =n</p>n<p>8n8n</p>n<p>8n</p>n<p>8n</p>n<p>8n</p>n<p>Therefore,n</p>n<p>_c d a b a b> œ  "  " 8xn=n</p>n<p>œ Þn8xn</p>n<p>=n</p>n<p>8 8n8n</p>n<p>8"n</p>n<p>8"n</p>n<p>33. Using the of the Laplace transform,<i>translation propertyn</i></p>n<p>_ ‘ a b/ =38 ,> œ Þn,n</p>n<p>=  +  ,n+>n</p>n<p># #n</p>n<p>Therefore,n</p>n<p>_ ‘ – —a ba bna bn</p>n<p>> / =38 ,> œ n. ,n</p>n<p>.= =  +  ,n</p>n<p>œn#, =  +n</p>n<p>=  #+=  +  ,n</p>n<p>+>n# #n</p>n<p># # # #n.n</p>n<p>34. Using the of the Laplace transform,<i>translation propertyn</i></p>n<p>_ ‘ a b/ -9= ,> œ Þn=  +n</p>n<p>=  +  ,n+>n</p>n<p># #n</p>n<p>Therefore,n</p>n<p>_ ‘ – —a bna bn</p>n<p>a bn</p>n<p>> / -9= ,> œ n. =  +n</p>n<p>.= =  +  ,n</p>n<p>œn=  +  ,n</p>n<p>=  #+=  +  ,n</p>n<p>+>n# #n</p>n<p># #n</p>n<p># # # #n.n</p>n<p>35 . Taking the Laplace transform of the given ,a b+ <i>Bessel equation</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 272n</p>n<p>_ _ _c d c d c d> C  C  > C œ !ww w .nUsing the of the transform,<i>differentiation propertyn</i></p>n<p> C  C  C œ !n. .n</p>n<p>.= .=n_ _ _c d c d c dww w .n</p>n<p>That is,n</p>n<p> = ] =  = C !  C !  =] =  C !  ] = œ !n. .n</p>n<p>.= .=n ‘a b a b a b a b a b a b# w .n</p>n<p>It follows thatn</p>n<p>ˆ ‰ a b a b"  = ] =  =] = œ !# w .na b a b, ] =. We obtain a ODE in :<i>first-order linearn</i></p>n<p>] =  ] = œ !n=n</p>n<p>=  "nwn</p>n<p>#na b a b ,n</p>n<p>with <i>integrating factorn</i></p>n<p>.a b Œ ( È= œ /B: .= œ =  "=n=  "#n</p>n<p># .n</p>n<p>The first-order ODE can be written asn</p>n<p>.n</p>n<p>.=n=  " † ] = œ !’ “È a b# ,n</p>n<p>with solutionn</p>n<p>] = œn-n</p>n<p>=  "na b È # .n</p>n<p>a b- =. In order to obtain powers of , first write<i>negativen</i>" " "n</p>n<p>=  "nœ "  Þn= =È ” •# #n</p>n<p>"Î#n</p>n<p>Expanding in a ,Š ‹"  "=# "Î# <i>binomial seriesn</i>" " " † $ " † $ † &n</p>n<p>"  "Î=nœ "  =  =  = ân</p>n<p># # † % # † % † 'È a b# # % ' ,nvalid for . Hence, we can formally express as=  " ] =# a bn</p>n<p>] = œ -    â Þn" " " " † $ " " † $ † & "n</p>n<p>= # = # † % = # † % † ' =na b ” •$ & (</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 273n</p>n<p>Assuming that inversion is valid,<i>term-by-termn</i></p>n<p>C > œ - "    ân" > " † $ > " † $ † & >n</p>n<p># #x # † % %x # † % † ' 'xn</p>n<p>œ - "    â Þn#x > %x > 'x >n</p>n<p># #x # † % %x # † % † ' 'xn</p>n<p>a b ” •n” •n</p>n<p># % 'n</p>n<p># # # # # #n</p>n<p># % 'n</p>n<p>It follows thatn</p>n<p>C > œ - "  >  >  > ân" " "n</p>n<p># # † % # † % † 'n</p>n<p>œ - > Þn "n</p>n<p># 8xn</p>n<p>a b ” •n" a ba bn</p>n<p># # # # # #n# % 'n</p>n<p>8œ!n</p>n<p>_ 8n</p>n<p>#8 #n#8n</p>n<p>The series is evidently the expansion, about , of B œ ! N > Þ!a bn36 . Taking the Laplace transform of the given ,a b, <i>Legendre equationn</i></p>n<p>_ _ _ ! ! _c d c d a b c d ‘C  > C  # > C   " C œ !ww # ww w .nUsing the of the transform,<i>differentiation propertyn</i></p>n<p>_ _ _ ! ! _c d c d c d a b c dC  C  # C   " C œ !. .n.= .=n</p>n<p>ww ww wn#n</p>n<p>#n.n</p>n<p>That is,n</p>n<p> ‘  ‘a b a b a b a b a b a bnc d a b a ba b a bn</p>n<p>= ] =  = C !  C !  = ] =  = C !  C ! n.n</p>n<p>.=n</p>n<p> # = ] =  C !   " ] = œ ! Þn.n</p>n<p>.=n</p>n<p># w # wn#n</p>n<p>#n</p>n<p>! !n</p>n<p>Invoking the , we have<i>initial conditionsn</i></p>n<p>= ] =  "  = ] =  "  # =] =   " ] = œ ! Þn. .n</p>n<p>.= .=n# #n</p>n<p>#n</p>n<p>#na b a b c d a b a b ‘ a b ! !n</p>n<p>After carrying out the differentiation, the equation simplifies ton</p>n<p>. .n</p>n<p>.= .=n= ] =  # = ] =  =   " ] = œ  " Þn</p>n<p>#n</p>n<p>#n# # ‘  ‘a b c d a b a ba b ! !n</p>n<p>That is,n</p>n<p>= ] =  #= ] =  =   " ] = œ  "n. .n</p>n<p>.= .=n# #n#n</p>n<p>#na b a b a b a b ‘! ! .n</p>n<p>37. By definition of the Laplace transform, given the appropriate conditions,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 274n</p>n<p>_ 7 7n</p>n<p>7 7n</p>n<p>c d a ba b ( (” •n( ( a bn</p>n<p>1 > œ / 0 . .>n</p>n<p>œ / 0 . .> Þn</p>n<p>! !n</p>n<p>_ >n=>n</p>n<p>! !n</p>n<p>_ >n=>n</p>n<p>Assuming that the order of integration can be exchanged,n</p>n<p>_ 7 7n</p>n<p>7 7n</p>n<p>c d a ba b ( (” •n( a b” •n</p>n<p>1 > œ 0 / .> .n</p>n<p>œ 0 . Þn/n</p>n<p>=n</p>n<p>!n</p>n<p>_ _n=>n</p>n<p>!n</p>n<p>_ =n7n7n</p>n<p>c da b a bNote the of integration is the area between the lines and <i>region </i>7 7> œ > > œ ! ÞnHencen</p>n<p>_ 7 7n</p>n<p>_n</p>n<p>c d a ba b (nc da bn</p>n<p>1 > œ 0 / .n"n</p>n<p>=n</p>n<p>œ 0 > Þn"n</p>n<p>=n</p>n<p>!n</p>n<p>_n=7</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 275n</p>n<p><b>Section 6.3n</b></p>n<p>1.n</p>n<p>3.n</p>n<p>5.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 276n</p>n<p>6.n</p>n<p>7. Using the Heaviside function, we can writen</p>n<p>0 > œ >  # ? >a b a b a b# # .nThe Laplace transform has the property thatn</p>n<p>_ _c d c da b a b a b? > 0 >  - œ / 0 >- -= .nHencen</p>n<p>_ ‘a b a b>  # ? ># # œ # /n=n</p>n<p> =n</p>n<p>#n</p>n<p>2n.n</p>n<p>9. The function can be expressed asn</p>n<p>0 > œ >  ? >  ? > Þa b a bc da b a b1 1 1#nBefore invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function asn</p>n<p>0 > œ >  ? >  >  # ? >  ? > Þa b a b a b a b a b a b1 1 11 1 1# #nIt follows thatn</p>n<p>_n1c da b0 > œ  / / /n</p>n<p>= = =n</p>n<p> = # = # =n</p>n<p># #n</p>n<p>1 1 1n</p>n<p>.n</p>n<p>10. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform thatn</p>n<p>_c da b0 > œ  #  '/ / /n= = =n</p>n<p>= $= %=n</p>n<p>.n</p>n<p>11. Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function asn</p>n<p>0 > œ >  # ? >  ? >  >  $ ? >  ? > Þa b a b a b a b a b a b a b# # $ $</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 277n</p>n<p>It follows thatn</p>n<p>_c da b0 > œ   / / / /n= = = =n</p>n<p>#= #= $= $=n</p>n<p># #n.n</p>n<p>12. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform thatn</p>n<p>_c da b0 > œ " /n= =# #n</p>n<p>=n</p>n<p>.n</p>n<p>13. Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+n_" $ #>n</p>n<p>%– —a bn$xn</p>n<p>=  #nœ > / .n</p>n<p>15. First consider the functionn</p>n<p>K = œ Þn# =  "n</p>n<p>=  #=  #na b a bn</p>n<p>#n</p>n<p>Completing the square in the denominator,n</p>n<p>K = œ Þn# =  "n</p>n<p>=  "  "na b a ba b#n</p>n<p>It follows thatn</p>n<p>_" >c da bK = œ # / -9= > ÞnHencen</p>n<p>_" #= ># # ‘a b a b a b/ K = œ # / -9= >  # ? > Þa bn16. The of the function is Using <i>inverse transform </i>#Î =  % 0 > œ =382 #> Þa b a b# then><+8=6+>398 :<9:/<>C of the transform,n</p>n<p>_"n#=n</p>n<p># #” • a b a b# /=  % œ =382 # >  # † ? > Þn</p>n<p>17. First consider the functionn</p>n<p>K = œ Þn=  #n</p>n<p>=  %=  $na b a bn</p>n<p>#n</p>n<p>Completing the square in the denominator,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 278n</p>n<p>K = œ Þn=  #n</p>n<p>=  #  "na b a ba b#n</p>n<p>It follows thatn</p>n<p>_" #>c da bK = œ / -9=2 > ÞnHencen</p>n<p>_" # >"n=n</p>n<p># "” •a b a b a b=  # /=  %=  $ œ / -9=2 >  " ? > Þa bn</p>n<p>18. Write the function asn</p>n<p>J = œ    Þn/ / / /n</p>n<p>= = = =na b = #= $= %=n</p>n<p>It follows from the ><+8=6+>398 :<9:/<>C of the transform, thatn</p>n<p>_"n= #= $= %=n</p>n<p>" # $ %” • a b a b a b a b/  /  /  /n=n</p>n<p>œ ? >  ? >  ? >  ? > Þn</p>n<p>19 . By definition of the Laplace transform,a b+n_c d a ba b (0 -> œ / 0 -> .> Þn</p>n<p>!n</p>n<p>_n=>n</p>n<p>Making a change of variable, , we have7 œ ->n</p>n<p>_ 7 7n</p>n<p>7 7n</p>n<p>c d a ba b (n( a bn</p>n<p>0 -> œ / 0 .n"n</p>n<p>-n</p>n<p>œ / 0 . Þn"n</p>n<p>-n</p>n<p>!n</p>n<p>_n= Î-n</p>n<p>!n</p>n<p>_n =Î-n</p>n<p>a bna bn7n</p>n<p>7n</p>n<p>Hence , where ._c da b ˆ ‰0 -> œ J =Î-  +" =- -na b a b, +. Using the result in Part ,n</p>n<p>_” •Œ  a b0 œ 5 J 5= Þ>n5n</p>n<p>Hencen</p>n<p>_"c da b Œ J 5= œ 0 Þ" >n5 5</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 279n</p>n<p>a b a b- ,. From Part ,n_"c da b Œ J += œ 0 Þ" >n</p>n<p>+ +n</p>n<p>Note that . Using the fact that ,+=  , œ + =  ,Î+ / 0 > œ 0 >a b c d c da b a b_ _-> =p=-n_" ,>Î+c dJ +=  , 0" >n</p>n<p>+ +na b Œ œ / .n</p>n<p>20. First writen</p>n<p>J = œn8xa b ˆ ‰=n#n</p>n<p>8" .n</p>n<p>Let Based on the results in Prob. ,K = œ 8xÎ= Þ "*a b 8"n" =n</p>n<p># #nK œ 1 #>_"’ “Š ‹ a b,n</p>n<p>in which . Hence1 > œ >a b 8n_" 8" 88c d a ba bJ = œ # #> œ # > Þn</p>n<p>23. First writen</p>n<p>J = œ Þn/n</p>n<p># =  "Î#na b a bn</p>n<p>% ="Î#a bn</p>n<p>Now considern</p>n<p>K = œn/n</p>n<p>=na b #= .n</p>n<p>Using the result in Prob. ,"* ,a bn_"c da b Œ K #= œ 1" >n</p>n<p># #n,n</p>n<p>in which . Hence . It follows that1 > œ ? > K #= œ ? >Î# œ ? >a b a b c d a b a ba b# # %" " "# #_n_" >Î# %c d a ba bJ = œ / ? >"n</p>n<p>#n.n</p>n<p>24. By definition of the Laplace transform,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 280n</p>n<p>_c d a ba b (0 > œ / ? > .> Þn!n</p>n<p>_n=>n</p>n<p>"n</p>n<p>That is,n</p>n<p>_c da b (0 > œ / .>nœ Þn"  /n</p>n<p>=n</p>n<p>!n</p>n<p>"n=>n</p>n<p>=n</p>n<p>25. First write the function as . It follows that0 > œ ? >  ? >  ? >  ? >a b a b a b a b a b! " # $n_c da b ( (0 > œ / .>  / .> Þn</p>n<p>! #n</p>n<p>" $n=> =>n</p>n<p>That is,n</p>n<p>_c da b0 > œ "  / /  /n= =n</p>n<p>œ Þn"  /  /  /n</p>n<p>=n</p>n<p>= #= $=n</p>n<p>= #= $=n</p>n<p>26. The transform may be computed directly. On the other hand, using the ><+8=6+>398n:<9:/<>C of the transform,n</p>n<p>_c d a ba b "n– —"a bn</p>n<p>a bn</p>n<p>0 > œ   "n" /n</p>n<p>= =n</p>n<p>œ  /n"n</p>n<p>=n</p>n<p>œn" "   /n</p>n<p>= "  /n</p>n<p>5œ"n</p>n<p>#8"n5n5=n</p>n<p>5 œ!n</p>n<p>#8"n= 5n</p>n<p>= #8#n</p>n<p>=n.n</p>n<p>That is,n</p>n<p>_c da b a ba b0 > œ "  /= "  /n#= 8"n</p>n<p>=n.n</p>n<p>29. The given function is , with . Using the result of Prob. ,<i>periodic </i>X œ # #)n</p>n<p>_c d a ba b ( (0 > œ / 0 > .> œ / .>" "n"  / "  /#= #=! !n</p>n<p># "n=> => .n</p>n<p>That is,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 281n</p>n<p>_c da b a bna bn</p>n<p>0 > œn"  /n</p>n<p>= "  /n</p>n<p>œn"n</p>n<p>= "  /n</p>n<p>=n</p>n<p>#=n</p>n<p>=n.n</p>n<p>31. , with . Using the result of Prob. , The function is <i>periodic </i>X œ " #)n</p>n<p>_c da b (0 > œ > / .> Þ"n"  /= !n</p>n<p>"n=>n</p>n<p>It follows thatn</p>n<p>_c da b a ba b0 > œ Þ"  / "  "  /n=n</p>n<p># =n</p>n<p>32. , with . Using the result of Prob. ,The function is <i>periodic </i>X œ #)1n</p>n<p>_c da b (0 > œ =38 > † / .> Þ"n"  / = !n</p>n<p>=>n1n</p>n<p>1n</p>n<p>We first calculaten</p>n<p>(n!n</p>n<p>=>n =n</p>n<p>#n</p>n<p>1 1n</p>n<p>=38 > † / .> œ Þn"  /n</p>n<p>"  =n</p>n<p>Hencen</p>n<p>_c da b a ba b0 > œ Þ"  /"  / "  =n =n</p>n<p> = #n</p>n<p>1n</p>n<p>1n</p>n<p>33 .a b+n</p>n<p>_ _ _c d c d c da b a b0 > œ "  ? >nœ n" /n</p>n<p>= =n</p>n<p>"n=n</p>n<p>.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 282n</p>n<p>a b, .n</p>n<p>Let . ThenJ = œ "  ? >a b c da b_ "n_ 7 7” •( c d a ba bn</p>n<p>!n</p>n<p>>n</p>n<p>"n</p>n<p>=n</p>n<p>#n"  ? . œ J = œ Þn</p>n<p>" "  /n</p>n<p>= =n</p>n<p>a b- .n</p>n<p>Let . ThenK = œ 1 >a b c da b_n_c d a b a ba bn</p>n<p>a bn</p>n<p>2 > œ K =  / K =n</p>n<p>œ  /n"  / "  /n</p>n<p>= =n</p>n<p>œn"  /n</p>n<p>=n</p>n<p>=n</p>n<p>= =n</p>n<p># #n=n</p>n<p>= #n</p>n<p>#n.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 283n</p>n<p>34 .a b+n</p>n<p>a b, X œ # #). , with . Using the result of Prob. ,The given function is <i>periodicn</i>_c d a ba b (0 > œ / : > .> Þ"n</p>n<p>"  /#= !n</p>n<p>#n=>n</p>n<p>Based on the piecewise definition of ,: >a bn( ( (a b a bn</p>n<p>a bn! ! "n</p>n<p># " #n=> => =>n</p>n<p>#n= #n</p>n<p>/ : > .> œ > / .>  #  > / .>n</p>n<p>œ "  / Þn"n</p>n<p>=n</p>n<p>Hencen</p>n<p>_c da b a ba b: > œ Þ"  /= "  /n=n</p>n<p># =n</p>n<p>a b a b- : > 'Þ#Þ". Since satisfies the hypotheses of Theorem ,n_ _c d c d a ba b a b: > œ = : >  : !w .n</p>n<p>Using the result of Prob. ,$!n</p>n<p>_c da b a ba b: > œ "  /= "  /wn=n</p>n<p>=n.n</p>n<p>We note the , hence: ! œ !a bn_c da b ” •a ba b: > œ " "  /= = "  /n</p>n<p>=n</p>n<p>=n.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 284n</p>n<p><b>Section 6.4n</b></p>n<p>2. Let be the on the right-hand-side. Taking the Laplace transform2 >a b <i>forcing functionn</i>of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  # = ] =  C !  #] = œ 2 ># wa b a b a b c d a b c da b a b a b_ .nApplying the initial conditions,n</p>n<p>= ] =  #= ] =  #] =  " œ 2 ># a b a b a b c da b_ .nThe forcing function can be written as Its transform is2 > œ ? >  ? > Þa b a b a b1 1#n</p>n<p>_c da b2 > œ /  /n=n</p>n<p> = # =1 1n</p>n<p>.n</p>n<p>Solving for , the transform of the solution is] =a bn] = œ n</p>n<p>"n</p>n<p>=  #=  # = =  #=  #na b a b# #/  /n</p>n<p> = # =1 1n</p>n<p>.n</p>n<p>First note thatn</p>n<p>"n</p>n<p>=  #=  ##nœn</p>n<p>"n</p>n<p>=  "  "a b# .nUsing partial fractions,n</p>n<p>" " " "n</p>n<p>= =  #=  # # = #nœ a b# a ba bn</p>n<p>=  "  "n</p>n<p>=  "  "nÞ#n</p>n<p>Taking the inverse transform, term-by-term,n</p>n<p>_ _” • – —"=  #=  # œ / =38 ># >œ "=  "  "a b# .nNow letn</p>n<p>K = œa b "n= =  #=  #a b# .n</p>n<p>Thenn</p>n<p>_"c da bK = œ " " "n# # #n</p>n<p>/ -9= >  / =38 >> > .n</p>n<p>Using Theorem ,'Þ$Þ"n</p>n<p>_" -= - -c d a b a ba b/ K = œ ? >  ? >" "n# #n</p>n<p>/ -9= >  -  =38 >  - >-a bc da b a b .nHence the solution of the IVP is</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 285n</p>n<p>C > œ ? >  ? > n" "n</p>n<p># #n</p>n<p> ? >  ? >n" "n</p>n<p># #n</p>n<p>a b a b a bna b a bn</p>n<p>/ =38 >  / -9= >   =38 > n</p>n<p>/ -9= >  #  =38 >  #n</p>n<p>>  >n</p>n<p> >#n</p>n<p>1 1n</p>n<p>1 1n</p>n<p>a bna bn</p>n<p>1n</p>n<p>1n</p>n<p>c da b a bnc da b a bn</p>n<p>1 1n</p>n<p>1 1# # .n</p>n<p>That is,n</p>n<p>C > œ ? >  ? >  ? > n" "n</p>n<p># #n</p>n<p> ? >n"n</p>n<p>#n</p>n<p>a b c d a ba b a bna bn</p>n<p>/ =38 >  / -9= >  =38 >n</p>n<p>/ -9= >  =38 >n</p>n<p>>  >n</p>n<p> >#n</p>n<p>1 1 1n</p>n<p>1n</p>n<p>#n</p>n<p>#n</p>n<p>a bna bn</p>n<p>1n</p>n<p>1n</p>n<p>c dnc d .n</p>n<p>The solution starts out as free oscillation, due to the initial conditions. The amplitudenincreases, as long as the forcing is present. Thereafter, the solution rapidly decays.n</p>n<p>4. Let be the on the right-hand-side. Taking the Laplace transform2 >a b <i>forcing functionn</i>of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  %] = œ 2 ># wa b a b a b a b c da b_ .nApplying the initial conditions,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 286n</p>n<p>= ] =  %] = œ 2 ># a b a b c da b_ .nThe transform of the forcing function isn</p>n<p>_c da b2 > œ " /n=  " =  "# #n</p>n<p> =1n</p>n<p>.n</p>n<p>Solving for , the transform of the solution is] =a bn] = œ n</p>n<p>"n</p>n<p>=  % =  " =  % =  "na b a ba b a ba b# # # #/n</p>n<p> =1n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " " "n</p>n<p>=  % =  " $ =  " =  %nœ a ba b ” •# # # # .n</p>n<p>It follows thatn</p>n<p>_"” • ” •"n=  % =  "a ba b# # œ =38 >  =38 #>" "$ # .n</p>n<p>Based on Theorem ,'Þ$Þ"n</p>n<p>_ 1 1"” • ” •a b a b a b/n=  % =  "n</p>n<p> =n</p>n<p># #n</p>n<p>1n</p>n<p>a ba b œ =38 >   =38 #>  # ? >" "$ # 1 .nHence the solution of the IVP isn</p>n<p>C > œ =38 >  =38 #>  =38 >  =38 #> ? >n" " " "n</p>n<p>$ # $ #na b a b” • ” • 1 .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 287n</p>n<p>Since there is no , the solution follows the forcing function, after which<i>damping termn</i>the response is a steady oscillation about .C œ !n</p>n<p>5. Let be the on the right-hand-side. Taking the Laplace transform0 >a b <i>forcing functionn</i>of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  $ = ] =  C !  #] = œ 0 ># wa b a b a b c d a b c da b a b a b_ .nApplying the initial conditions,n</p>n<p>= ] =  $= ] =  #] = œ 0 ># a b a b a b c da b_ .nThe transform of the forcing function isn</p>n<p>_c da b0 > œ " /n= =n</p>n<p>"!=n</p>n<p>.n</p>n<p>Solving for the transform,n</p>n<p>] = œ n" /n</p>n<p>= =  $=  # = =  $=  #na b a b a b# #n</p>n<p>"!=n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " " " #n</p>n<p>= =  $=  # # = =  # =  "nœ   Þa b ” •#n</p>n<p>Hencen</p>n<p>_" >n#n</p>n<p>#>” •a b" " /= =  $=  # # #œ   / .nBased on Theorem ,'Þ$Þ"n</p>n<p>_" # >"!  >"!n# "!” •a b  ‘ a b/n"!=n</p>n<p>= =  $=  # #nœ "  /  #/ ? >n" a b a b .n</p>n<p>Hence the solution of the IVP is</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 288n</p>n<p>C > œ "  ? >   /  /  #/ ? >n" / "n</p>n<p># # #na b c d a ba b  ‘"! "!#> >  #>#!  >"!a b a b .n</p>n<p>The solution increases to a steady value of . After the forcing ceases,<i>temporary </i>C œ "Î#nthe response decays exponentially to .C œ !n</p>n<p>6. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  $ = ] =  C !  #] = œn/n</p>n<p>=n# wn</p>n<p>#=a b a b a b c d a ba b a b .nApplying the initial conditions,n</p>n<p>= ] =  $= ] =  #] =  " œn/n</p>n<p>=n#n</p>n<p>#=a b a b a b .nSolving for the transform,n</p>n<p>] = œ n" /n</p>n<p>=  $=  # = =  $=  #na b a b# #n</p>n<p>#=n</p>n<p>.n</p>n<p>Using partial fractions,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 289n</p>n<p>" " "n</p>n<p>=  $=  # =  " =  #nœ n</p>n<p>#n</p>n<p>andn</p>n<p>" " " " #n</p>n<p>= =  $=  # # = =  # =  "nœ   Þa b ” •#n</p>n<p>Taking the inverse transform. term-by-term, the solution of the IVP isn</p>n<p>C > œ /  /   /  / ? > Þn" "n</p>n<p># #na b a b” •> #>  ># # ># #a b a bn</p>n<p>Due to the initial conditions, the response has a transient , followed by an<i>overshootn</i>exponential convergence to a steady value of .C œ "Î#=n</p>n<p>7. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  ] = œn/n</p>n<p>=n# wn</p>n<p>$ =a b a b a b a b 1 .nApplying the initial conditions,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 290n</p>n<p>= ] =  ] =  = œn/n</p>n<p>=n#n</p>n<p>$ =a b a b 1 .nSolving for the transform,n</p>n<p>] = œ n= /n</p>n<p>=  " = =  "na b a b# #n</p>n<p>$ =1n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " =n</p>n<p>= =  " = =  "nœ a b# # .n</p>n<p>Hencen</p>n<p>] = œ  / n= " =n</p>n<p>=  " = =  "na b ” •# #$ =1 .n</p>n<p>Taking the inverse transform, the solution of the IVP isn</p>n<p>C > œ -9= >  "  -9= >  $ ? >n</p>n<p>œ -9= >  "  -9= > ? > Þn</p>n<p>a b c d a ba bc d a b1 $$ 11n</p>n<p>Due to initial conditions, the solution temporarily oscillates about . After theC œ !nforcing is applied, the response is a steady oscillation about .C œ "7</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 291n</p>n<p>9. Let be the on the right-hand-side. Taking the Laplace transform1 >a b <i>forcing functionn</i>of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  ] = œ 1 ># wa b a b a b a b c da b_ .nApplying the initial conditions,n</p>n<p>= ] =  ] =  " œ 1 ># a b a b c da b_ .nThe forcing function can be written asn</p>n<p>1 > œ "  ? >  $? >n>n</p>n<p>#n</p>n<p>œ  >  ' ? >n> "n</p>n<p># #n</p>n<p>a b c d a ba bna b a bn' 'n</p>n<p>'n</p>n<p>with Laplace transformn</p>n<p>_c da b1 > œ " /n#= #=# #n</p>n<p>'=n</p>n<p>.n</p>n<p>Solving for the transform,n</p>n<p>] = œ  n" " /n</p>n<p>=  " #= =  " #= =  "na b a b a b# # # # #n</p>n<p>'=n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " " "n</p>n<p>#= =  " # = =  "nœ  Þn</p>n<p># # # #a b ” •nTaking the inverse transform, and using Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ =38 >  >  =38 >  >  '  =38 >  ' ? >n" "n</p>n<p># #n</p>n<p>œ >  =38 >  >  '  =38 >  ' ? > Þn" "n</p>n<p># #n</p>n<p>a b c d c d a ba b a bnc d c d a ba b a bn</p>n<p>'n</p>n<p>'</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 292n</p>n<p>The solution increases, in response to the , and thereafter oscillates about a<i>ramp inputn</i>mean value of .C œ $7n</p>n<p>11. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  %] = œ n/ /n</p>n<p>= =n# wn</p>n<p> = $ =a b a b a b a b 1 1 .nApplying the initial conditions,n</p>n<p>= ] =  %] = œ n/ /n</p>n<p>= =n#n</p>n<p> = $ =a b a b 1 1 .nSolving for the transform,n</p>n<p>] = œ n/ /n</p>n<p>= =  % = =  %na b a b a bn</p>n<p> = $ =n</p>n<p># #n</p>n<p>1 1n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " " =n</p>n<p>= =  % % = =  %nœ  Þa b ” •# #n</p>n<p>Taking the inverse transform, and applying Theorem ,'Þ$Þ"n</p>n<p>C > œ "  -9= #>  # ? >  "  -9= #>  ' ? >n" "n</p>n<p>% %n</p>n<p>œ ? >  ? >  -9= #> † ? >  ? >n" "n</p>n<p>% %n</p>n<p>a b c d a b c d a ba b a bnc d c da b a b a b a bn</p>n<p>1 11 1n</p>n<p>1 1 1 1n</p>n<p>$n</p>n<p>$ $ .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 293n</p>n<p>Since there is no damping term, the solution responds immediately to the forcing input.nThere is a temporary oscillation about C œ "Î% Þn</p>n<p>12. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  = C !  = C !  C !  ] = œ  Þn/ /n</p>n<p>= =n% $ # w ww wwwn</p>n<p>= #=a b a b a b a b a b a bnApplying the ,<i>initial conditionsn</i></p>n<p>= ] =  ] = œ n/ /n</p>n<p>= =n%n</p>n<p>= #=a b a b .nSolving for the transform of the solution,n</p>n<p>] = œ n/ /n</p>n<p>= =  " = =  "na b a b a bn</p>n<p>= #=n</p>n<p>% %n.n</p>n<p>Using partial fractions,n</p>n<p>" " % " " #=n</p>n<p>= =  " % = =  " =  " =  "nœ     Þa b ” •% #n</p>n<p>It follows that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 294n</p>n<p>_" > >n%” •a b  ‘" "= =  " %œ  %  /  /  # -9= > Þn</p>n<p>Based on Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ  ? >  ? >  /  /  # -9= >  " ? > n"n</p>n<p>%n</p>n<p> /  /  # -9= >  # ? >n"n</p>n<p>%n</p>n<p>a b c d a b a ba b a b  ‘n ‘a b a bn</p>n<p>" # "n >" >"n</p>n<p> ># >#n#n</p>n<p>a b a bna b a b .n</p>n<p>The solution increases without bound, exponentially.n</p>n<p>13. Taking the Laplace transform of the ODE, we obtainn</p>n<p>= ] =  = C !  = C !  = C !  C ! n</p>n<p> & = ] =  = C !  C !  %] = œ  Þn" /n</p>n<p>= =n</p>n<p>% $ # w ww wwwn</p>n<p># wn =n</p>n<p>a b a b a b a b a bn ‘a b a b a b a b 1n</p>n<p>Applying the ,<i>initial conditionsn</i></p>n<p>= ] =  &= ] =  %] = œ n" /n</p>n<p>= =n% #n</p>n<p> =a b a b a b 1 .nSolving for the transform of the solution,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 295n</p>n<p>] = œ n" /n</p>n<p>= =  &=  % = =  &=  %na b a b a b% # % #n</p>n<p> =1n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " $ = %=n</p>n<p>= =  &=  % "# = =  % =  "nœ   Þa b ” •% # # #n</p>n<p>It follows thatn</p>n<p>_"n% #” •a b c d" "= =  &=  % "#œ $  -9= #>  % -9= > Þn</p>n<p>Based on Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ "  ? >  -9= #>  % -9= > n" "n</p>n<p>% "#n</p>n<p> -9= # >   % -9= >  ? >n"n</p>n<p>"#n</p>n<p>a b c d c da bnc d a ba b a bn</p>n<p>1n</p>n<p>11 1 .n</p>n<p>That is,n</p>n<p>C > œ "  ? >  -9= #>  % -9= > n" "n</p>n<p>% "#n</p>n<p> -9= #>  % -9= > ? >n"n</p>n<p>"#n</p>n<p>a b c d c da bnc d a bn</p>n<p>1n</p>n<p>1 .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 296n</p>n<p>After an initial transient, the solution oscillates about .C œ !7n</p>n<p>14. The specified function is defined byn</p>n<p>0 > œn</p>n<p>!ß ! Ÿ >  >n</p>n<p>>  > ß > Ÿ >  >  5n</p>n<p>2ß >   >  5n</p>n<p>a bnÚnÛÜ a bn</p>n<p>!n</p>n<p>! ! !n</p>n<p>!n</p>n<p>2n5n</p>n<p>which can conveniently be expressed asn</p>n<p>0 > œ >  > ? >  >  >  5 ? > Þn2 2n</p>n<p>5 5na b a b a b a b a b! !> > 5! !n</p>n<p>15. The function is defined byn</p>n<p>1 > œn</p>n<p>!ß ! Ÿ >  >n</p>n<p>>  > ß > Ÿ >  >  5n</p>n<p> >  >  #5 ß >  5 Ÿ >  >  #5n</p>n<p>!ß >   >  #5n</p>n<p>a bnÚÝÝÛÝÝÜna ba bn</p>n<p>!n</p>n<p>! ! !n</p>n<p>! ! !n</p>n<p>!n</p>n<p>2n5n2n5n</p>n<p>which can also be written asn</p>n<p>1 > œ >  > ? >  >  >  5 ? >  >  >  #5 ? > Þn2 #2 2n</p>n<p>5 5 5na b a b a b a b a b a b a b! ! !> > 5 > #5! ! !n</p>n<p>16 . From Part , the solution isa b a b. -n? > œ %5 ? > 2 >   %5 ? > 2 > n</p>n<p>$ &n</p>n<p># #na b a b a bŒ  Œ $Î# &Î# ,n</p>n<p>wheren</p>n<p>2 > œ  / =38  / -9= Þn" ( $ ( > " $ ( >n</p>n<p>% )% ) % )na b È È È   >Î) >Î)n</p>n<p>Due to the , the solution will decay to . The maximum will occur<i>damping term zero</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 297n</p>n<p>shortly after the forcing ceases. By plotting the various solutions, it appears that thensolution will reach a value of , as long as .C œ # 5  #Þ&"n</p>n<p>a b/ Þn</p>n<p>Based on the graph, and numerical calculation, for .k ka b? >  !Þ" >  #&Þ'(($n17. We consider the initial value problemn</p>n<p>C  %C œ >  & ? >  >  &  5 ? >n"n</p>n<p>5nww c da b a b a b a b& &5 ,n</p>n<p>with .C ! œ C ! œ !a b a bwna b+ . The specified function is defined byn</p>n<p>0 > œ >  & ß & Ÿ >  &  5n</p>n<p>!ß ! Ÿ >  &n</p>n<p>"ß >   &  5n</p>n<p>a b a bnÚnÛÜn"n5n</p>n<p>a b, Þ Taking the Laplace transform of both sides of the ODE, we obtain</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 298n</p>n<p>= ] =  = C !  C !  %] = œ n/ /n</p>n<p>5= 5=n# wn</p>n<p>&=  &5 =n</p>n<p># #na b a b a b a b a b .n</p>n<p>Applying the initial conditions,n</p>n<p>= ] =  %] = œ n/ /n</p>n<p>5= 5=n#n</p>n<p>&=  &5 =n</p>n<p># #na b a b a b .n</p>n<p>Solving for the transform,n</p>n<p>] = œ n/ /n</p>n<p>5= =  % 5= =  %na b a b a bn</p>n<p>&=  &5 =n</p>n<p># # # #n</p>n<p>a bn.n</p>n<p>Using partial fractions,n</p>n<p>" " " "n</p>n<p>= =  % % = =  %nœ  Þn</p>n<p># # # #a b ” •nIt follows thatn</p>n<p>_"n# #” •a b" " "= =  % % )œ >  =38 #> .n</p>n<p>Using Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ 2 >  & ? >  2 >  &  5 ? >n"n</p>n<p>5na b c da b a b a b a b& &5 ,n</p>n<p>in which .2 > œ >  =38 #>a b " "% )na b- >  &  5. Note that for , the solution is given byn</p>n<p>C > œ  =38 #>  "!  =38 #>  "!  #5n" " "n</p>n<p>% )5 )5n</p>n<p>œ  -9= #>  "!  5 Þn" =38 5n</p>n<p>% %5n</p>n<p>a b a b a bna bn</p>n<p>So for , the solution oscillates about , with an amplitude of>  &  5 C œ "Î%7n</p>n<p>E œn=38 5n</p>n<p>%5n</p>n<p>k ka bn.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 299</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 300n</p>n<p>18 .a b+n</p>n<p>a b, . The forcing function can be expressed asn0 > œ ? >  ? > Þn</p>n<p>"n</p>n<p>#5n5 %5 %5a b c da b a bn</p>n<p>Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  =] =  C !  %] = œ n" / /n</p>n<p>$ #5= #5=n# wn</p>n<p> %5 =  %5 =a b a b a b c d a ba b a b a b a b .nApplying the initial conditions,n</p>n<p>= ] =  =] =  %] = œ n" / /n</p>n<p>$ #5= #5=n#n</p>n<p> %5 =  %5 =a b a b a b a b a b .nSolving for the transform,n</p>n<p>] = œ n$ / $ /n</p>n<p>#5= $=  =  "# #5= $=  =  "#na b a b a bn</p>n<p> %5 =  %5 =n</p>n<p># #n</p>n<p>a b a bn.n</p>n<p>Using partial fractions,n</p>n<p>" " " "  $=n</p>n<p>= $=  =  "# "# = $=  =  "#nœ n</p>n<p>œ  Þn" " "n</p>n<p>"# = 'n</p>n<p>"  ' = n</p>n<p>=  n</p>n<p>a b ” •n– —ˆ ‰ˆ ‰n</p>n<p># #n</p>n<p>"n'n</p>n<p>" "%$n' $'n</p>n<p>#n</p>n<p>Letn</p>n<p>L = œ   Þn" "n</p>n<p>)5 = =   =  n</p>n<p>= a b – —ˆ ‰ ˆ ‰n" "n' 'n</p>n<p>" "%$ " "%$n' $' ' $'n</p>n<p># #n</p>n<p>It follows that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 301n</p>n<p>2 > œ L = œ  =38  -9= Þn" / " "%$ > "%$ >n</p>n<p>)5 )5 ' '"%$na b c da b – —È    n</p>n<p>È Èn_"n</p>n<p>>Î'n</p>n<p>Based on Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ 2 >  %  5 ? >  2 >  %  5 ? > Þa b a b a b a b a b%5 %5na b- .n</p>n<p>As the parameter decreases, the solution remains for a longer period of time.5 <i>null</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 302n</p>n<p>Since the of the impulsive force , the initial of the<i>magnitude increases overshootn</i>response also increases. The of the impulse decreases. All solutions eventually<i>durationn</i>decay to C œ ! Þn</p>n<p>19 .a b+n</p>n<p>a b a b- Þ , From Part ,n? > œ "  -9= >  #  " "  -9= >  5 ? > Þa b a b c d a b" a bn</p>n<p>5 œ"n</p>n<p>8n5 1 51</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 303n</p>n<p>21 .a b+n</p>n<p>a b, . Taking the Laplace transform of both sides of the ODE, we obtainn= Y =  =? !  ? !  Y = œ n</p>n<p>"  " /n</p>n<p>= =n# wn</p>n<p>5 œ"n</p>n<p>8 5 5 =a b a b a b a b "a b 1 .nApplying the initial conditions,n</p>n<p>= Y =  Y = œ n"  " /n</p>n<p>= =n#n</p>n<p>5 œ"n</p>n<p>8 5 5 =a b a b "a b 1 .nSolving for the transform,n</p>n<p>Y = œ n"  " /n</p>n<p>= =  " = =  "na b a b a b"a b# #n</p>n<p>5 œ"n</p>n<p>8 5 5 =1n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " =n</p>n<p>= =  " = =  "nœ a b# # .n</p>n<p>Let</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 304n</p>n<p>2 > œ œ "  -9= >n"n</p>n<p>= =  "na b ” •a b_" # .n</p>n<p>Applying Theorem , term-by-term, the solution of the IVP is'Þ$Þ"n</p>n<p>? > œ 2 >   " 2 >  5 ? >a b a b a b a b a b"n5 œ"n</p>n<p>8n5 1 51 .n</p>n<p>Note thatn</p>n<p>2 >  5 œ ? >  5  -9= >  5n</p>n<p>œ ? >   " -9= > Þn</p>n<p>a b a b a bna b a bn</p>n<p>1 1 1!n</p>n<p>51n5n</p>n<p>Hencen</p>n<p>? > œ "  -9= >   " ? >  -9= > ? > Þa b a b a b a b a b" "n5 œ" 5 œ"n</p>n<p>8 8n5n</p>n<p>5 51 1n</p>n<p>a b- .n</p>n<p>The ODE has no . Each interval of forcing adds to the energy of the<i>damping termn</i>system.nHence the amplitude will increase. For , when . Therefore the8 œ "& 1 > œ ! >  "&a b 1noscillation will eventually become , with an amplitude depending on the values of<i>steadyn</i>? "& ? "&a b a b1 1 and .wna b. 8. As increases, the interval of forcing also increases. Hence the amplitude of thentransient will increase with . Eventually, the forcing function will be . In fact,8 <i>constantn</i>for values of ,<i>large </i>>n</p>n<p>1 > œn" ß 8n! ß 8n</p>n<p>a b œ even oddnFurther, for ,>  81</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 305n</p>n<p>? > œ "  -9= >  8 -9= >  Þn"   "n</p>n<p>#na b a b8n</p>n<p>Hence the steady state solution will oscillate about or , depending on , with an! " 8namplitude of .E œ 8  "n</p>n<p>In the limit, as , the forcing function will be a periodic function, with period .8p_ #1nFrom Prob. , in Section ,#( 'Þ$n</p>n<p>_c da b a b1 > œ "= "  /= .nAs increases, the duration and magnitude of the transient will increase without bound.8n</p>n<p>22 . Taking the initial conditions into consideration, the transform of the ODE isa b+n= Y =  !Þ" =Y =  Y = œ n</p>n<p>"  " /n</p>n<p>= =n#n</p>n<p>5 œ"n</p>n<p>8 5 5 =a b a b a b "a b 1 .nSolving for the transform,n</p>n<p>Y = œ n"  " /n</p>n<p>= =  !Þ"=  " = =  !Þ"=  "na b a b a b" a b# #n</p>n<p>5 œ"n</p>n<p>8 5 5 =1n</p>n<p>.n</p>n<p>Using partial fractions,n</p>n<p>" " =  !Þ"n</p>n<p>= =  !Þ"=  " = =  !Þ"=  "nœ  Þa b# #n</p>n<p>Since the denominator in the second term is irreducible, writen</p>n<p>=  !Þ" =  !Þ!&  !Þ!&n</p>n<p>=  !Þ"=  "nœ Þn=  !Þ!&  Ð$**Î%!!Ñ# #na bn</p>n<p>a bnLetn</p>n<p>2 > œ  n" =  !Þ!& !Þ!&n</p>n<p>= =  !Þ!&  Ð$**Î%!!Ñ =  !Þ!&  Ð$**Î%!!Ñn</p>n<p>œ "  / -9= >  =38 > Þn$** " $**n</p>n<p>#! #!$**n</p>n<p>a b – —a ba b a bn– —   È ÈÈn</p>n<p>_"n# #n</p>n<p>>Î#!n</p>n<p>Applying Theorem , term-by-term, the solution of the IVP is'Þ$Þ"n</p>n<p>? > œ 2 >   " 2 >  5 ? >a b a b a b a b a b"n5 œ"n</p>n<p>8n5 1 51 .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 306n</p>n<p>For values of , the solution approaches .<i>odd </i>8 C œ !n</p>n<p>For values of , the solution approaches .<i>even </i>8 C œ "n</p>n<p>a b È, œ $** Î#! ¸ ". The solution is a sum of , each of frequency .<i>damped sinusoids </i>=nEach term has an 'initial' amplitude of approximately For any given , the solution" Þ 8ncontains such terms. Although the amplitude will with , the amplitude8  " 8<i>increasen</i>will also be bounded by .8  "n</p>n<p>a b a b- 1 > œ =38 >. Suppose that the forcing function is replaced by . Based on thenmethodsnin Chapter , the general solution of the differential equation is$n</p>n<p>? > œ / - -9= >  - =38 >  ? > Þn$** $**n</p>n<p>#! #!na b a b– —   È È>Î#! :" #n</p>n<p>Note that . Using the method of ,? > œ E -9= >  F =38 >:a b <i>undetermined coefficientsn</i>E œ  "! F œ ! Þ and Based on the initial conditions, the solution of the IVP is</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 307n</p>n<p>? > œ "! / -9= >  =38 >  "! -9= > Þn$** " $**n</p>n<p>#! #!$**na b – —   È ÈÈ>Î#!n</p>n<p>Observe that both solutions have the same frequency, .= œ $**Î#! ¸ "Èn</p>n<p>23 . Taking the initial conditions into consideration, the transform of the ODE isa b+n= Y =  Y = œ  #n</p>n<p>"  " /n</p>n<p>= =n#n</p>n<p>5 œ"n</p>n<p>8 5  ""5Î% =a b a b "a b a b .nSolving for the transform,n</p>n<p>Y = œ  #n"  " /n</p>n<p>= =  " = =  "na b a b a b"a b# #n</p>n<p>5 œ"n</p>n<p>8 5  ""5Î% =a bn.n</p>n<p>Using partial fractions,n</p>n<p>" " =n</p>n<p>= =  " = =  "nœ a b# # .n</p>n<p>Letn</p>n<p>2 > œ œ "  -9= >n"n</p>n<p>= =  "na b ” •a b_" # .n</p>n<p>Applying Theorem , term-by-term, the solution of the IVP is'Þ$Þ"n</p>n<p>? > œ 2 >  #  " 2 >  ? >n""5n</p>n<p>%na b a b a b a b" Œ n</p>n<p>5 œ"n</p>n<p>8n5n</p>n<p>""5Î% .n</p>n<p>That is,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 308n</p>n<p>? > œ "  -9= >  #  " "  -9= >  ? >n""5n</p>n<p>%na b a b a b" ” •Œ n</p>n<p>5 œ"n</p>n<p>8n5n</p>n<p>""5Î% .n</p>n<p>a b, .n</p>n<p>a b- )). Based on the plot, the ' ' appears to be . The ' ' appears to<i>slow period fast periodn</i>be about . These values correspond to a ' ' of and a ' œ !Þ!("%<i>slow frequency fast</i>n<i>frequency</i>' .=0 œ "Þ!%(#n</p>n<p>a b. œ " Þ. The natural frequency of the system is The forcing function is initially=!nperiodic, with period . Hence the corresponding forcing frequency isX œ ""Î# œ &Þ&nA œ "Þ"%#% $Þ*. Using the results in Section , the ' ' is given by<i>slow frequencyn</i></p>n<p>=n= =n</p>n<p>= œ œ !Þ!("#nn</p>n<p>#n</p>n<p>k k!nand the ' ' is given by<i>fast frequencyn</i></p>n<p>=n= =n</p>n<p>0n!n</p>n<p>œ œ "Þ!("#nn</p>n<p>#n</p>n<p>k kn.n</p>n<p>Based on theses values, the ' ' is predicted as and the ' ' is<i>slow period fast period</i>))Þ#%(ngiven as .&Þ)'&'</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 309n</p>n<p><b>Section 6.5n</b></p>n<p>2. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  %] = œ /  /# w  = # =a b a b a b a b 1 1 .nApplying the initial conditions,n</p>n<p>= ] =  %] = œ /  /#  = # =a b a b 1 1 .nSolving for the transform,n</p>n<p>] = œ œ n/  / / /n</p>n<p>=  % =  % =  %na b  = # =  = # =n</p>n<p># # #n</p>n<p>1 1 1 1n</p>n<p>.n</p>n<p>Applying Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ =38 #>  # ? >  =38 #>  % ? >n" "n</p>n<p># #n</p>n<p>œ =38 #> ? >  ? > Þn"n</p>n<p>#n</p>n<p>a b a b a b a b a bna bc da b a bn</p>n<p>1 11 1n</p>n<p>1 1n</p>n<p>#n</p>n<p>#n</p>n<p>4. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  ] = œ  #! /# w $=a b a b a b a b .nApplying the initial conditions,n</p>n<p>= ] =  ] =  = œ  #! /# $=a b a b .nSolving for the transform,n</p>n<p>] = œ n= #! /n</p>n<p>=  " =  "na bn</p>n<p># #n</p>n<p>$=n</p>n<p>.n</p>n<p>Using a , and Theorem , the solution of the IVP is<i>table of transforms </i>'Þ$Þ"n</p>n<p>C > œ -9=2 >  #! =382 >  $ ? >a b a b a b$ .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 310n</p>n<p>6. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  %] =  =Î# œ /# % =a b a b 1 .nSolving for the transform,n</p>n<p>] = œ n=Î# /n</p>n<p>=  % =  %na bn</p>n<p># #n</p>n<p>% =1n</p>n<p>.n</p>n<p>Using a , and Theorem , the solution of the IVP is<i>table of transforms </i>'Þ$Þ"n</p>n<p>C > œ -9= #>  =38 #>  ) ? >n" "n</p>n<p># #n</p>n<p>œ -9= #>  =38 #> ? > Þn" "n</p>n<p># #n</p>n<p>a b a b a bna b a bn</p>n<p>1 %n</p>n<p>%n</p>n<p>1n</p>n<p>1n</p>n<p>8. Taking the Laplace transform of both sides of the ODE, we obtainn</p>n<p>= ] =  = C !  C !  %] = œ # /# w  Î% =a b a b a b a b a b1 .nApplying the initial conditions,n</p>n<p>= ] =  %] = œ # /#  Î% =a b a b a b1 .nSolving for the transform,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 311n</p>n<p>] = œn# /n</p>n<p>=  %na b  Î% =n</p>n<p>#n</p>n<p>a b1n.n</p>n<p>Applying Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ =38 #>  ? > œ  -9= #> ? > Þn#n</p>n<p>a b a b a b a bŠ ‹1 1 1Î% Î%n</p>n<p>9. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  ] = œ  $ / n/ /n</p>n<p>= =n#  $ Î# =n</p>n<p> Î# = # =a b a b a b a b1 11 .nSolving for the transform,n</p>n<p>] = œ  n/ $ / /n</p>n<p>= =  " =  " = =  "na b a b a bn</p>n<p> Î# =  $ Î# = # =n</p>n<p># # #n</p>n<p>a b a b1 1 1n.n</p>n<p>Using partial fractions,n</p>n<p>" " =n</p>n<p>= =  " = =  "nœ  Þa b# #n</p>n<p>Hencen</p>n<p>] = œa b / = / $ / / = /n= =  " =  " = =  "n</p>n<p>    Þn Î# =  Î# =  $ Î# = # = # =n</p>n<p># # #n</p>n<p>a b a b a b1 1 1 1 1n</p>n<p>Based on Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ ? >  -9= >  ? >  $ =38 >  ? > n# #n</p>n<p>$n</p>n<p> ? >  -9= >  # ? >n</p>n<p>a b a b a b a bŠ ‹ Œ na b a b a bn1 1 1n</p>n<p>1 1n</p>n<p>Î# Î# $ Î#n</p>n<p># #n</p>n<p>1 1n</p>n<p>1 .n</p>n<p>That is,n</p>n<p>C > œ "  =38 > ? >  $ -9= > ? >  "  -9= > ? >a b c d a b a b a b c d a ba b a b1 1 1Î# $ Î# # .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 312n</p>n<p>10. Taking the transform of both sides of the ODE,n</p>n<p>#= ] =  =] =  %] = œ / >  =38 > .>n'n</p>n<p>œ / Þn"n</p>n<p>#n</p>n<p># =>n</p>n<p>!n</p>n<p>_n</p>n<p> Î' =n</p>n<p>a b a b a b ( Š ‹$ 1na b1n</p>n<p>Solving for the transform,n</p>n<p>] = œn/n</p>n<p># #=  =  %na b a bn</p>n<p> Î' =n</p>n<p>#n</p>n<p>a b1n.n</p>n<p>First writen</p>n<p>"n</p>n<p># #=  =  %nœn=  a b ˆ ‰#n</p>n<p>"n%n</p>n<p>" $"n% "'n</p>n<p># .n</p>n<p>It follows thatn</p>n<p>C > œ ] = œ / † =38 >  ? > Þn" $"n</p>n<p>$" % 'na b c d a ba b Èn</p>n<p>È Š ‹_ 1"  > Î' Î%a b1 1Î'</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 313n</p>n<p>11. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  #= ] =  #] = œ  /n=n</p>n<p>=  "n#  Î# =n</p>n<p>#na b a b a b a b1 .n</p>n<p>Solving for the transform,n</p>n<p>] = œ n= /n</p>n<p>=  " =  #=  # =  #=  #na b a ba b# # #n</p>n<p> Î# =a b1n.n</p>n<p>Using partial fractions,n</p>n<p>= " = # =  %n</p>n<p>=  " =  #=  # & =  " =  " =  #=  #nœ   Þa ba b ” •# # # # #n</p>n<p>We can also writen</p>n<p>=  % =  "  $n</p>n<p>=  #=  #nœ Þn=  "  "# #na bna bn</p>n<p>Letn</p>n<p>] = œ Þn=n</p>n<p>=  " =  #=  #n" # #na b a ba bn</p>n<p>Thenn</p>n<p>_" >"c d c da b] = œ -9= >  =38 >  / -9= >  $ =38 > Þ" # "n& & &n</p>n<p>Applying Theorem ,'Þ$Þ"n</p>n<p>_n1"  >n</p>n<p> Î# =n</p>n<p>#– — Š ‹ a b/=  #=  # #œ / =38 >  ? > Þna b ˆ ‰1 1n</p>n<p>#n1Î#n</p>n<p>Hence the solution of the IVP isn</p>n<p>C > œ -9= >  =38 >  / -9= >  $ =38 > n" # "n</p>n<p>& & &n</p>n<p> / -9= > ? > Þn</p>n<p>a b c dna b a bn</p>n<p>>n</p>n<p> >ˆ ‰1#n1Î#</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 314n</p>n<p>12. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  ] = œ /% =a b a b .nSolving for the transform,n</p>n<p>] = œn/n</p>n<p>=  "na b =n</p>n<p>%n.n</p>n<p>Using partial fractions,n</p>n<p>" " " "n</p>n<p>=  " # =  " =  "nœ  Þn</p>n<p>% # #” •nIt follows thatn</p>n<p>_"n%” •" " "=  " # #œ =382 >  =38 > Þn</p>n<p>Applying Theorem , the solution of the IVP is'Þ$Þ"n</p>n<p>C > œ =382 >  "  =38 >  " ? > Þn"n</p>n<p>#na b c d a ba b a b "</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 315n</p>n<p>14 . The Laplace transform of the ODE isa b+n= ] =  =] =  ] = œ /n</p>n<p>"n</p>n<p>#n# =a b a b a b .n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œn/n</p>n<p>=  =Î#  "na b =n</p>n<p>#n.n</p>n<p>First writen</p>n<p>" "n</p>n<p>=  =Î#  "nœ Þn=  n</p>n<p># " "&n% "'n</p>n<p>#ˆ ‰nTaking the inverse transform and applying both ,<i>shifting theoremsn</i></p>n<p>C > œ / =38 >  " ? >n% "&n</p>n<p>"& %na b a b a bÈn</p>n<p>Èn >" Î%a bn</p>n<p>" .n</p>n<p>a b, >  #. As shown on the graph, the maximum is attained at some . Note that for"n>  #,n</p>n<p>C > œ / =38 >  " Þn% "&n</p>n<p>"& %na b a bÈn</p>n<p>Èn >" Î%a bn</p>n<p>Setting , we find that . The maximum value is calculated asC > œ ! > ¸ #Þ$'"$wa b "nC #Þ$'"$ ¸ !Þ(""&$ Þa bna b- œ "Î%. Setting , the transform of the solution is#n</p>n<p>] = œn/n</p>n<p>=  =Î%  "na b =n</p>n<p>#n.n</p>n<p>Following the same steps, it follows that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 316n</p>n<p>C > œ / =38 >  " ? >n) $ (n</p>n<p>$ ( )na b a b a bÈn</p>n<p>Èn >" Î)a bn</p>n<p>" .n</p>n<p>Once again, the maximum is attained at some . Setting , we find that>  # C > œ !" wa bn> ¸ #Þ%&'* C > ¸ !Þ)$$& Þ" ", with a bna b. !   ". Now suppose that . Then the transform of the solution is#n</p>n<p>] = œn/n</p>n<p>=  =  "na b =n</p>n<p># #n.n</p>n<p>First writen</p>n<p>" "n</p>n<p>=  =  "nœ Þn=  Î#  "  Î%# # ## # #a b a bn</p>n<p>It follows thatn</p>n<p>2 > œ œ / =38 "  Î% † > Þn" #n</p>n<p>=  =  " % na b ” • È Š ‹È_ ## #"  >Î## # ##n</p>n<p>Hence the solution isn</p>n<p>C > œ 2 >  " ? >a b a b a b" .nThe solution is nonzero only if , in which case Setting >  " C > œ 2 >  " Þ C > œ !a b a b a bwn,nwe obtainn</p>n<p>>+8 "  Î% † >  " œ % n"’ “È a b È# #n#n</p>n<p># # ,n</p>n<p>that is,n</p>n<p>>+8 "  Î% † >  "n</p>n<p>"  Î%nœn# ‘È a bÈ # # #n</p>n<p>#n</p>n<p>#n.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 317n</p>n<p>As , we obtain the equation . Hence . Setting#p! >+8 >  " œ _ > p" <i>formal </i>a b " 1#n> œ Î# 2 > p ! C p" Þ1 # in , and letting , we find that These conclusions agree witha b "nthe case , for which it is easy to show that the solution is# œ !n</p>n<p>C > œ =38 >  " ? >a b a b a b" .n15 . See Prob. . It follows that the solution of the IVP isa b+ "%n</p>n<p>C > œ / =38 >  " ? >n%5 "&n</p>n<p>"& %na b a b a bÈn</p>n<p>Èn >" Î%a bn</p>n<p>" .n</p>n<p>This function is a of the answer in Prob. . Hence the peak value occurs at<i>multiple </i>"% +a bn> ¸ #Þ$'"$ C #Þ$'"$ ¸ !Þ(""&$ 5 Þ" . The maximum value is calculated as We finda bnthat the appropriate value of is .5 5 œ #Î!Þ(""&$ ¸ #Þ)"!)"n</p>n<p>a b a b, "% -. Based on Prob. , the solution isnC > œ / =38 >  " ? >n</p>n<p>) 5 $ (n</p>n<p>$ ( )na b a b a bÈn</p>n<p>Èn >" Î)a bn</p>n<p>" .n</p>n<p>Since this function is a of the solution in Prob. , we have ,<i>multiple </i>"% - > ¸ #Þ%&'*a b "nwith The solution attains a value of , for ,C > ¸ !Þ)$$& 5 Þ C œ # 5 œ #Î!Þ)$$&a b" "nthat is, .5 ¸ #Þ$**&"n</p>n<p>a b a b- "% . !   ". Similar to Prob. , for , the solution is#nC > œ 2 >  " ? >a b a b a b" ,n</p>n<p>in whichn</p>n<p>2 > œ / =38 "  Î% † > Þn# 5n</p>n<p>% na b È Š ‹È# ##  >Î# ##n</p>n<p>It follows that . Setting in , and letting , we find that>  "p Î# > œ Î# 2 > p !" 1 1 #a bnC p5 Þ C œ # 5" Requiring that the remains at , the limiting value of is<i>peak valuen</i>5 œ # œ !" . These conclusions agree with the case , for which it is easy to show#nthat the solution isn</p>n<p>C > œ 5 =38 >  " ? >a b a b a b" .n16 . Taking the initial conditions into consideration, the transformation of the ODE isa b+n</p>n<p>= ] =  ] = œ n" / /n</p>n<p>#5 = =n#n</p>n<p> %5 =  %5 =a b a b – —a b a b .nSolving for the transform of the solution,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 318n</p>n<p>] = œ  Þn" / /n</p>n<p>#5 = =  " = =  "na b – —a b a bn</p>n<p> %5 =  %5 =n</p>n<p># #n</p>n<p>a b a bn</p>n<p>Using partial fractions,n</p>n<p>" " =n</p>n<p>= =  " = =  "nœ  Þa b# #n</p>n<p>Now letn</p>n<p>2 > œ œ "  -9= > Þn"n</p>n<p>= =  "na b ” •a b_" #n</p>n<p>Applying Theorem , the solution is'Þ$Þ"n</p>n<p>9a b c da b a b a b a b> ß 5 œ 2 >  %  5 ? >  2 >  %  5 ? > Þ"n#5n</p>n<p>%5 %5n</p>n<p>That is,n</p>n<p>9a b c da b a bnc da b a b a b a bn</p>n<p>> ß 5 œ ? >  ? > n"n</p>n<p>#5n</p>n<p> -9= >  %  5 ? >  -9= >  %  5 ? > Þn"n</p>n<p>#5n</p>n<p>%5 %5n</p>n<p>%5 %5n</p>n<p>a b a b, > >  % > ß 5 œ !. Consider various values of . For any fixed , , as long as9n%  5  >Þ >   % %  5  > If , then for ,n</p>n<p>9a b c da b a b> ß 5 œ  -9= >  %  5  -9= >  %  5 Þ"n#5n</p>n<p>It follows thatn</p>n<p>lim limn5Ä! 5Ä!n9a b a b a bn</p>n<p>a bn> ß 5 œ n</p>n<p>-9= >  %  5  -9= >  %  5n</p>n<p>#5nœ =38 >  % Þn</p>n<p>Hencen</p>n<p>limn5Ä!n9a b a b a b> ß 5 œ =38 >  % ? >% .n</p>n<p>a b- . The Laplace transform of the differential equationnC  C œ >  %ww $a b,n</p>n<p>with , isC ! œ C ! œ !a b a bwn= ] =  ] = œ /# %=a b a b .n</p>n<p>Solving for the transform of the solution,</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 319n</p>n<p>] = œ Þn/n</p>n<p>=  "na b %=n</p>n<p>#n</p>n<p>It follows that the solution isn</p>n<p>9! %a b a b a b> œ =38 >  % ? > .na b. .n</p>n<p>18 . The transform of the ODE given the specified initial conditions isa b a b,n= ] =  ] = œ  " /# 5 =n</p>n<p>5 œ"n</p>n<p>#!n5"a b a b a b" 1 .n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œ  " /n"n</p>n<p>=  "na b a b"n</p>n<p>#n5 œ"n</p>n<p>#!n5" 5 =1 .n</p>n<p>Applying Theorem , term-by-term,'Þ$Þ"n</p>n<p>C > œ  " =38 >  5 ? >n</p>n<p>œ  =38 > † ? > Þn</p>n<p>a b a b a b a b"na b a b"n</p>n<p>5 œ"n</p>n<p>#!n5"n</p>n<p>5 œ"n</p>n<p>#!n</p>n<p>1 5n</p>n<p>5n</p>n<p>1n</p>n<p>1</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 320n</p>n<p>a b- .n</p>n<p>19 . Taking the initial conditions into consideration, the transform of the ODE isa b,n= ] =  ] = œ /#  5 Î# =n</p>n<p>5 œ"n</p>n<p>#!a b a b " a b1 .nSolving for the transform of the solution,n</p>n<p>] = œ /n"n</p>n<p>=  "na b "n</p>n<p>#n5 œ"n</p>n<p>#!n 5 Î# =a b1 .n</p>n<p>Applying Theorem , term-by-term,'Þ$Þ"n</p>n<p>C > œ =38 >  ? > Þn5n</p>n<p>#na b a b" Œ n</p>n<p>5 œ"n</p>n<p>#! 1n5 Î#1n</p>n<p>a b- .</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 321n</p>n<p>20 . The transform of the ODE given the specified initial conditions isa b a b,n= ] =  ] = œ  " /#  5 Î# =n</p>n<p>5 œ"n</p>n<p>#!n5a b a b a b" +1 a b1 .n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œ  "n/n</p>n<p>=  "na b a b"n</p>n<p>5 œ"n</p>n<p>#!n5n</p>n<p> 5 Î# =n</p>n<p>#n+1n</p>n<p>a b1n.n</p>n<p>Applying Theorem , term-by-term,'Þ$Þ"n</p>n<p>C > œ  " =38 >  ? > Þn5n</p>n<p>#na b a b a b" Œ n</p>n<p>5 œ"n</p>n<p>#!n5" 1n</p>n<p>5 Î#1n</p>n<p>a b- .n</p>n<p>22 . Taking the initial conditions into consideration, the transform of the ODE isa b,n= ] =  ] = œ  " /#  ""5Î% =n</p>n<p>5 œ"n</p>n<p>%!n5"a b a b a b" a b .n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œ  "n/n</p>n<p>=  "na b a b"n</p>n<p>5 œ"n</p>n<p>%!n5"n</p>n<p> ""5Î% =n</p>n<p>#n</p>n<p>a bn.n</p>n<p>Applying Theorem , term-by-term,'Þ$Þ"</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 322n</p>n<p>C > œ  " =38 >  ? > Þn""5n</p>n<p>%na b a b a b" Œ n</p>n<p>5 œ"n</p>n<p>%!n5"n</p>n<p>""5Î%n</p>n<p>a b- .n</p>n<p>23 The transform of the ODE given the specified initial conditions isa b a b, Þn= ] =  !Þ"= ] =  ] = œ  " /# 5 =n</p>n<p>5 œ"n</p>n<p>#!n5"a b a b a b a b" 1 .n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œn/n</p>n<p>=  !Þ"=  "na b "n</p>n<p>5 œ"n</p>n<p>#! 5 =n</p>n<p>#n</p>n<p>1n</p>n<p>.n</p>n<p>First writen</p>n<p>" "n</p>n<p>=  !Þ"=  "nœn=  n</p>n<p># " $**n#! %!!n</p>n<p>#ˆ ‰ .nIt follows thatn</p>n<p>_" >Î#!n#” • È  n</p>n<p>È" #! $**n=  !Þ"=  " #!n</p>n<p>œ / =38 > Þn$**n</p>n<p>Applying Theorem , term-by-term,'Þ$Þ"</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 323n</p>n<p>C > œ  " 2 >  5 ? >a b a b a b a b"n5 œ"n</p>n<p>#!n5" 1 51 ,n</p>n<p>in whichn</p>n<p>2 > œ / =38 > Þn#! $**n</p>n<p>$** #!na b È  n</p>n<p>Èn>Î#!n</p>n<p>a b- .n</p>n<p>24 Taking the initial conditions into consideration, the transform of the ODE isa b, Þn= ] =  !Þ"= ] =  ] = œ /#  #5" =n</p>n<p>5 œ"n</p>n<p>"&a b a b a b " a b1 .nSolving for the transform of the solution,n</p>n<p>] = œn/n</p>n<p>=  !Þ"=  "na b "n</p>n<p>5 œ"n</p>n<p>"&  #5" =n</p>n<p>#n</p>n<p>a b1n.n</p>n<p>As shown in Prob. ,#$n</p>n<p>_" >Î#!n#” • È  n</p>n<p>È" #! $**n=  !Þ"=  " #!n</p>n<p>œ / =38 > Þn$**n</p>n<p>Applying Theorem , term-by-term,'Þ$Þ"n</p>n<p>C > œ 2 >  #5  " ? >a b c d a b" a bn5 œ"n</p>n<p>"&n</p>n<p>1 a b#5" 1 ,n</p>n<p>in which</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 324n</p>n<p>2 > œ / =38 > Þn#! $**n</p>n<p>$** #!na b È  n</p>n<p>Èn>Î#!n</p>n<p>a b- .n</p>n<p>25 . A fundamental set of solutions is and .a b a b a b+ C > œ / -9= > C > œ / =38 >" #> >nBased on Prob. , in Section , a particular solution is given by## $Þ(n</p>n<p>C > œ 0 = .=nC = C >  C > C =n</p>n<p>[ C ß C =n:n</p>n<p>" # " #n</p>n<p>" #n</p>n<p>a b a b( a b a b a b a ba ba b!n>n</p>n<p>.n</p>n<p>In the given problem,n</p>n<p>C > œ 0 = .=n/ -9= = =38 >  =38 = -9= >n</p>n<p>/B:  #=n</p>n<p>œ / =38 >  = 0 = .=n</p>n<p>:a b a b( c da b a b a b a ba bn( a b a bn!n</p>n<p>> =>n</p>n<p>!n</p>n<p>>n >=a b .n</p>n<p>.n</p>n<p>Given the specified initial conditions,n</p>n<p>C > œ / =38 >  = 0 = .=a b a b a b(n!n</p>n<p>>n >=a b .n</p>n<p>a b a b a b a b, Þ 0 > œ >  >  C > œ ! >  Let . It is easy to see that if , . If ,$ 1 1 1n( a b a b a bn!n</p>n<p>>n >=  >/ =38 >  = =  .= œ / =38 > a b a b$ 1 11 .n</p>n<p>Setting , and letting , we find that . Hence> œ  p! C œ !1 & & 1a b</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 325n</p>n<p>C > œ / =38 >  ? > Þa b a b a b >a b1 11na b- . The Laplace transform of the solution isn</p>n<p>] = œn/n</p>n<p>=  #=  #n</p>n<p>œ Þn/n</p>n<p>=  "  "n</p>n<p>a bna bn</p>n<p> =n</p>n<p>#n</p>n<p> =n</p>n<p>#n</p>n<p>1n</p>n<p>1n</p>n<p>Hence the solutions agree.</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 326n</p>n<p><b>Section 6.6n</b></p>n<p>1 . The a b+ <i>convolution integral</i> is defined asn0 ‡ 1 > œ 0 >  1 . Þa b a b a b(n</p>n<p>!n</p>n<p>>n</p>n<p>7 7 7n</p>n<p>Consider the change of variable . It follows that? œ >  7n</p>n<p>( (a b a b a b a ba bn( a b a bn</p>n<p>a bn</p>n<p>! >n</p>n<p>> !n</p>n<p>!n</p>n<p>>n</p>n<p>0 >  1 . œ 0 ? 1 >  ?  .?n</p>n<p>œ 1 >  ? 0 ? .?n</p>n<p>œ 1 ‡ 0 > Þn</p>n<p>7 7 7n</p>n<p>a b, . Based on the distributive property of the , the convolution is also<i>real numbersn</i>distributive.n</p>n<p>a b- . By definition,n0 ‡ 1 ‡ 2 > œ 0 >  1 ‡ 2 .n</p>n<p>œ 0 >  1  2 . .n</p>n<p>œ 0 >  1  2 . . Þn</p>n<p>a ba b a bc d( a bn( (a b a b a b” •n( ( a b a b a bn</p>n<p>!n</p>n<p>>n</p>n<p>! !n</p>n<p>>n</p>n<p>! !n</p>n<p>>n</p>n<p>7 7 7n</p>n<p>7 7 ( ( ( 7n</p>n<p>7 7 ( ( ( 7n</p>n<p>7n</p>n<p>7n</p>n<p>The region of integration, in the double integral is the area between the straight linesn( ( 7 7œ ! œ œ > Þ, and Interchanging the order of integration,n</p>n<p>( ( ( (a b a b a b a b a b a bn( (” •a b a b a bn</p>n<p>! ! !n</p>n<p>> > >n</p>n<p>!n</p>n<p>> >n</p>n<p>7n</p>n<p>(n</p>n<p>(n</p>n<p>0 >  1  2 . . œ 0 >  1  2 . .n</p>n<p>œ 0 >  1  . 2 . Þn</p>n<p>7 7 ( ( ( 7 7 7 ( ( 7 (n</p>n<p>7 7 ( 7 ( (n</p>n<p>Now let . Then7 ( œ ?n</p>n<p>( (a b a b a b a bna b(n(> >n</p>n<p>!n</p>n<p>0 >  1  . œ 0 >   ? 1 ? .?n</p>n<p>œ 0 ‡ 1 > n</p>n<p>7 7 ( 7 (n</p>n<p>( .n</p>n<p>Hencen</p>n<p>( (a bc d c d a ba b a bn! !n</p>n<p>> >n</p>n<p>0 >  1 ‡ 2 . œ 0 ‡ 1 >  2 . Þ7 7 7 7 7 7</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 327n</p>n<p>2. Let . Then0 > œ /a b >n0 ‡ " > œ / † " .n</p>n<p>œ / / .n</p>n<p>œ /  "n</p>n<p>a b (n(n!n</p>n<p>>n>n</p>n<p>> n</p>n<p>!n</p>n<p>>n</p>n<p>>n</p>n<p>7n</p>n<p>7n</p>n<p>7n</p>n<p>7n</p>n<p>.n</p>n<p>3. It follows directly thatn</p>n<p>0 ‡ 0 > œ =38 >  =38 .n</p>n<p>œ -9= >  #  -9= > .n"n</p>n<p>#n</p>n<p>œ =38 >  > -9= >n"n</p>n<p>#n</p>n<p>a b a b a b(n( c da b a bnc da b a bn</p>n<p>!n</p>n<p>>n</p>n<p>!n</p>n<p>>n</p>n<p>7 7 7n</p>n<p>7 7n</p>n<p>.n</p>n<p>The of the resulting function is <i>range </i>‘ Þn</p>n<p>5. We have and Based on Theorem ,_ _c d a b c d/ œ "Î =  " =38 > œ "Î Þ 'Þ'Þ"> a b=  "#n_ 7 7” •( a bn</p>n<p>a ba bn!n</p>n<p>>n >n</p>n<p>#n</p>n<p>#n</p>n<p>/ =38 . œ †n" "n</p>n<p>=  " =  "n</p>n<p>œ Þn"n</p>n<p>=  " =  "n</p>n<p>a b7n</p>n<p>6. Let and . Then . Applying Theorem ,1 > œ > 2 > œ / 0 > œ 1 ‡ 2 > 'Þ'Þ"a b a b a b a b>n_ 7 7 7” •( a b a bn</p>n<p>a bn!n</p>n<p>>n</p>n<p>#n</p>n<p>#n</p>n<p>1 >  2 . œ †n" "n</p>n<p>= =  "n</p>n<p>œ Þn"n</p>n<p>= =  "n</p>n<p>7. We have , in which and . The transform0 > œ 1 ‡ 2 > 1 > œ =38 > 2 > œ -9= >a b a b a b a bnof the convolution integral isn</p>n<p>_ 7 7 7” •( a b a bna bn</p>n<p>!n</p>n<p>>n</p>n<p># #n</p>n<p># #n</p>n<p>1 >  2 . œ †n" =n</p>n<p>=  " =  "n</p>n<p>œ Þn=n</p>n<p>=  "n</p>n<p>9. It is easy to see that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 328n</p>n<p>_ _" > "n#” • ” •"  " =  %œ / œ -9= #> and .n</p>n<p>Applying Theorem ,'Þ'Þ"n</p>n<p>_ 7 7"  >n#n</p>n<p>!n</p>n<p>>” •a ba b (  " =  % œ / -9= # . Þa b7n</p>n<p>10. We first note thatn</p>n<p>_ _" > "n# #– —a b ” •n</p>n<p>" " "n</p>n<p>=  "nœ > / œ =38 #>n</p>n<p>=  % #n and .n</p>n<p>Based on the ,<i>convolution theoremn</i></p>n<p>_ 7 7 7n</p>n<p>7 7 7n</p>n<p>"  >n# #n</p>n<p>!n</p>n<p>>n</p>n<p>!n</p>n<p>>nn</p>n<p>– —a b a b ( a bn( a bn</p>n<p>" "n</p>n<p>=  " =  %nœ >  / =38 # .n#n</p>n<p>œ / =38 #>  # . Þn"n</p>n<p>#n</p>n<p>a b7n</p>n<p>7n</p>n<p>11. Let . Since , the inverse transform of1 > œ K = "Î œ =38 >a b c d c da b_ _" " a b=  "#nthe product isn</p>n<p>_ 7 7 7n</p>n<p>7 7 7n</p>n<p>"n#n</p>n<p>!n</p>n<p>>n</p>n<p>!n</p>n<p>>n</p>n<p>” •a b ( a bn( a b a bn</p>n<p>K =n</p>n<p>=  "nœ 1 >  =38 .n</p>n<p>œ =38 >  1 . Þn</p>n<p>12. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  "  ] = œ K =# #a b a b a b= .nSolving for the transform of the solution,n</p>n<p>] = œ n" K =n</p>n<p>=  = na b a bn</p>n<p># # # #= =n.n</p>n<p>As shown in a related situation, Prob. ,""n</p>n<p>_ = 7 7 7n= =n</p>n<p>"n# #n</p>n<p>!n</p>n<p>>” •a b ( a b a bK = "n= n</p>n<p>œ =38 >  1 . Þn</p>n<p>Hence the solution of the IVP is</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 329n</p>n<p>C > œ =38 >  =38 >  1 . Þn" "a b a b a b(n= =n</p>n<p>= = 7 7 7n!n</p>n<p>>n</p>n<p>14. The transform of the ODE given the specified initial conditions isa bn%= ] =  %= ] =  "(] = œ K =# a b a b a b a b.n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œnK =n</p>n<p>%=  %=  "(na b a bn</p>n<p>#n.n</p>n<p>First writen</p>n<p>"n</p>n<p>%=  %=  "(nœ Þn=   %n</p>n<p>#n</p>n<p>"n%n</p>n<p>"n#n</p>n<p>#ˆ ‰nBased on the elementary properties of the Laplace transform,n</p>n<p>_" >Î#n#” •" "%=  %=  "( )œ / =38 #> Þn</p>n<p>Applying the , the solution of the IVP is<i>convolution theoremn</i></p>n<p>C > œ / =38 # >  1 . Þn"n</p>n<p>)na b a b a b(n</p>n<p>!n</p>n<p>>n > Î#a b7 7 7 7n</p>n<p>16. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  #=  $  % =] =  #  %] = œ K =# a b c d a b a ba b .nSolving for the transform of the solution,n</p>n<p>] = œ n#=  & K =n</p>n<p>=  # =  #na b a b a bn</p>n<p>a bn# # .n</p>n<p>We can writen</p>n<p>#=  & # "n</p>n<p>=  # =  #nœ  Þn=  #a b a b# #n</p>n<p>It follows thatn</p>n<p>_ _" #> " #>n#” • – —a bn</p>n<p># "n</p>n<p>=  #nœ #/ œ > /n</p>n<p>=  #n and .n</p>n<p>Based on the , the solution of the IVP is<i>convolution theorem</i></p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 330n</p>n<p>C > œ #/  > /  >  / 1 . Þa b a b a b(#> #> # >n!n</p>n<p>>n</p>n<p>7 7 7a b7n</p>n<p>18. The transform of the ODE given the specified initial conditions isa bn= ] =  ] = œ K =% a b a b a b.n</p>n<p>Solving for the transform of the solution,n</p>n<p>] = œnK =n</p>n<p>=  "na b a bn</p>n<p>%n.n</p>n<p>First writen</p>n<p>" " " "n</p>n<p>=  " # =  " =  "nœ  Þn</p>n<p>% # #” •nIt follows thatn</p>n<p>_"” • c d" "n=  " #n</p>n<p>œ =382 >  =38 > Þ4n</p>n<p>Based on the , the solution of the IVP is<i>convolution theoremn</i></p>n<p>C > œ =382 >   =38 >  1 . Þn"n</p>n<p>#na b c d a b( a b a bn</p>n<p>!n</p>n<p>>n</p>n<p>7 7 7 7n</p>n<p>19. Taking the initial conditions into consideration, the transform of the ODE isn</p>n<p>= ] =  =  &= ] =  &=  %] = œ K =% $ #a b a b a b a b.nSolving for the transform of the solution,n</p>n<p>] = œ n=  &= K =n</p>n<p>=  " =  % =  " =  %na b a ba b a ba ba bn</p>n<p>$n</p>n<p># # # #n.n</p>n<p>Using partial fractions, we find thatn</p>n<p>=  &= " %= =n</p>n<p>=  " =  % $ =  " =  %nœ n</p>n<p>$n</p>n<p># # # #a ba b ” •,nandn</p>n<p>" " " "n</p>n<p>=  " =  % $ =  " =  %nœ  Þa ba b ” •# # # #n</p>n<p>It follows that</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 331n</p>n<p>_"n#n</p>n<p># #” •a ba ba b= =  & % "=  " =  % $ $œ -9= >  -9= #>,nandn</p>n<p>_"n# #” •a ba b" " "=  " =  % $ 'œ =38 >  =38 #> Þn</p>n<p>Based on the , the solution of the IVP is<i>convolution theoremn</i></p>n<p>C > œ -9= >  -9= #>  # =38 >   =38 # >  1 . Þn% " "n</p>n<p>$ $ 'na b c d a b( a b a bn</p>n<p>!n</p>n<p>>n</p>n<p>7 7 7 7n</p>n<p>21 . Let . Substitution into the results ina b a b a b+ > œ ? >9 ww <i>integral equationn</i>? >  >  ? . œ =38 #> Þww wwn</p>n<p>!n</p>n<p>>a b a b a b( 0 0 0nIntegrating by parts,n</p>n<p>( (a b a b a b a b a bºna b a b a b! !n</p>n<p>> >nww w wn</p>n<p>œ!n</p>n<p>œ>n</p>n<p>wn</p>n<p>>  ? . œ >  ?  ? .n</p>n<p>œ  >? !  ? >  ? ! Þn</p>n<p>0 0 0 0 0 0 0n0n</p>n<p>0n</p>n<p>Hencen</p>n<p>? >  ? >  > ? !  ? ! œ =38 #> Þww wa b a b a b a bna b a b, ? >. Substituting the given for the function ,<i>initial conditionsn</i></p>n<p>? >  ? > œ =38 #> Þwwa b a bnHence the solution of the IVP is equivalent to solving the integral equation in Part .a b+na b a b c da b- = œ >. Taking the Laplace transform of the integral equation, with ,F _ 9n</p>n<p>F Fa b a b=  † = œ Þ" #n= =  %# #n</p>n<p>Note that the was applied. Solving for the transform ,<i>convolution theorem </i>Fa b=nFa b a ba b= œ Þ#  " =  %n</p>n<p>#n</p>n<p># #n</p>n<p>Using partial fractions, we can write</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 332n</p>n<p>#= # % "n</p>n<p>=  " =  % $ =  % =  "nœ  Þn</p>n<p>#n</p>n<p># # # #a ba b ” •nTherefore the solution of the is<i>integral equationn</i></p>n<p>9a b> œ =38 #>  =38 >% #n$ $n</p>n<p>.n</p>n<p>a b a b c da b. Y = œ ? >. Taking the Laplace transform of the ODE, with ,_n= Y =  Y = œ Þn</p>n<p>#n</p>n<p>=  %n#n</p>n<p>#na b a bn</p>n<p>Solving for the transform of the solution,n</p>n<p>Y = œ Þn#n</p>n<p>=  " =  %na b a ba b# #n</p>n<p>Using partial fractions, we can writen</p>n<p># " # #n</p>n<p>=  " =  % $ =  " =  %nœ  Þa ba b ” •# # # #n</p>n<p>It follows that the solution of the IVP isn</p>n<p>? > œ =38 >  =38 #> Þn# "n</p>n<p>$ $na bn</p>n<p>We find that ? > œ  =38 >  =38 #> Þww # %$ $a bn22 . First note thata b+n</p>n<p>( a bÈ  È a b!n, 0 C "n</p>n<p>,  Cn.C œ ‡ 0 , Þn</p>n<p>Cn</p>n<p>Take the Laplace transformation of both sides of the equation. Using the <i>convolutionntheorem</i>, with ,J = œ 0 Ca b c da b_n</p>n<p>X " "n</p>n<p>= Cnœ J = † Þn</p>n<p>#1n</p>n<p>! È a b – —È_nIt was shown in Prob. , Section , that#( - 'Þ"a bn</p>n<p>_n1– —È Ê"C =œ Þn</p>n<p>Hence</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 333n</p>n<p>X "n</p>n<p>= =nœ J = †n</p>n<p>#1n</p>n<p>! È a b Ê1 ,nwithn</p>n<p>J = œ † Þn#1 Xn</p>n<p>=na b Ê È1 !n</p>n<p>Taking the inverse transform, we obtainn</p>n<p>0 C œ ÞnX #1n</p>n<p>Cna b Ë!1n</p>n<p>a b a b a b, 3 3@. Combining equations and ,n#1 X .Bn</p>n<p>C .Cnœ "  Þ!n</p>n<p>#n</p>n<p>#n</p>n<p>#n</p>n<p>1nŒ n</p>n<p>Solving for the derivative ,.BÎ.Cn</p>n<p>.B #  Cn</p>n<p>.C CnœË ! ,n</p>n<p>in which ! 1œ 1X Î Þ!# #n</p>n<p>a b a b- C œ # =38 Î# Þ. Consider the Using the chain rule,<i>change of variable </i>! )#n.C .n</p>n<p>.B .Bnœ # =38 Î# -9= Î# †! ) )n</p>n<p>)a b a bnandn</p>n<p>.B " .Bn</p>n<p>.C # =38 Î# -9= Î# .nœ †n! ) ) )a b a b .n</p>n<p>It follows thatn</p>n<p>.B -9= Î#n</p>n<p>. =38 Î#nœ # =38 Î# -9= Î#n</p>n<p>œ # -9= Î#n</p>n<p>œ  -9=n</p>n<p>) )n! ) )n</p>n<p>)n</p>n<p>! )n</p>n<p>! ! )n</p>n<p>a b a bË a ba bna bn</p>n<p>#n</p>n<p>#n</p>n<p>#n</p>n<p>.n</p>n<p>Direct integration results inn</p>n<p>B œ  =38  Ga b) ! ) ! ) .nSince the curve passes through the , we require Hence ,<i>origin </i>C ! œ B ! œ ! Þ G œ !a b a bnand . We also haveB œ  =38a b) ! ) ! )</p>nn</div></div>n<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. n</p>n<p><b>________________________________________________________________________n</b> page 334n</p>n<p>C œ # =38 Î#n</p>n<p>œ  -9= Þn</p>n<p>a b a b) ! )n! ! )n</p>n<p>#</p>nn</div></div>n</body></html>','canEdit':false,'canDelete':false,'canReport':false,'userVote':null,'previewLimit':3,'advEnabled':true,'totalVotes':0,'title':'Respostas Capu00edtulo 06, Manuais, Projetos, Pesquisas de Engenharia Biolu00f3gica','isPremiumEnabled':false,'hasQuizcardSet':null}'><div><div><div><div><main><div><div><div><span><span>Pré-visualização</span><span>3 páginas / 81</span></span></div><div><div><div><div><div><div></div><div></div></div></div></div></div></div></div></div><div><div><div></div><div></div><div></div></div></div><div><div><div><div><div><div>CHAP_6.WXP<div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 254</p><p><b>Chapter Six</b></p><p><b>Section 6.1</b></p><p>3.</p><p>The function is 0 >a b <i>continuous</i>.4.</p><p>The function has a at .0 > > œ 'a b <i>jump discontinuity</i>7. Integration is a linear operation. It follows that</p><p>( ( (( (</p><p>! ! !</p><p>E E E=> ,> => ,> =></p><p>! !</p><p>E E,= > ,= ></p><p>-9=2 ,> † / .> œ / † / .> / † / .>' '</p><p># #</p><p>œ / .> / .> Þ' '</p><p># #a b a b</p><p>Hence</p><p>( – — – —!</p><p>E=></p><p>,= E ,= E</p><p>-9=2 ,> † / .> œ Þ' ' / ' ' /</p><p># = , # = ,</p><p>a b a b</p><p>Taking a , as ,<i>limit </i>Ep_</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 255</p><p>( ” • ” •!</p><p>_=></p><p># #</p><p>-9=2 ,> † / .> œ ' ' ' '</p><p># = , # = ,</p><p>œ Þ=</p><p>= ,</p><p>Note that the above is valid for .= ,k k8. Proceeding as in Prob. ,(</p><p>( – — – —!</p><p>E=></p><p>,= E ,= E</p><p>=382 ,> † / .> œ Þ' ' / ' ' /</p><p># = , # = ,</p><p>a b a b</p><p>Taking a , as ,<i>limit </i>Ep_</p><p>( ” • ” •!</p><p>_=></p><p># #</p><p>=382 ,> † / .> œ ' ' ' '</p><p># = , # = ,</p><p>œ Þ,</p><p>= ,</p><p>The limit exists as long as .= ,k k10. Observe that It follows that/ =382 ,> œ / / Î# Þ+> Ð+,Ñ> Ð+,Ñ>ˆ ‰</p><p>( – — – —!</p><p>E+> =></p><p>+,= E ,+= E</p><p>/ =382 ,> † / .> œ Þ' ' / ' ' /</p><p># = + , # = , +</p><p>a b a b</p><p>Taking a , as ,<i>limit </i>Ep_</p><p>( ” • ” •a b</p><p>!</p><p>_+> =></p><p># #</p><p>/ =382 ,> † / .> œ ' ' ' '</p><p># = + , # = , +</p><p>œ Þ,</p><p>= + ,</p><p>The limit exists as long as .= + ,k k11. Using the of the Laplace transform,<i>linearity</i></p><p>_ _ _c d ‘ ‘=38 ,> œ / / Þ' '#3 #3</p><p>3,> 3,></p><p>Since</p><p>(!</p><p>_+3, > =>/ / .> œ</p><p>'</p><p>= + 3,a b ,</p><p>we have</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 256</p><p>(!</p><p>_„ 3,> =>/ / .> œ</p><p>'</p><p>=… 3,.</p><p>Therefore</p><p>_c d ” •=38 ,> œ ' ' '#3 = 3, = 3,</p><p>œ,</p><p>= ,# #.</p><p>12. Using the of the Laplace transform,<i>linearity</i></p><p>_ _ _c d ‘ ‘-9= ,> œ / / Þ' '# #</p><p>3,> 3,></p><p>From Prob. , we have'</p><p>(!</p><p>_„ 3,> =>/ / .> œ</p><p>'</p><p>=… 3,.</p><p>Therefore</p><p>_c d ” •-9= ,> œ ' ' '# = 3, = 3,</p><p>œ=</p><p>= ,# #.</p><p>14. Using the of the Laplace transform,<i>linearity</i></p><p>_ _ _ ‘ ‘ ‘/ -9= ,> œ / / Þ' '# #</p><p>+> +3, > +3, >a b a b</p><p>Based on the integration in Prob. ,'</p><p>(!</p><p>_+„ 3, > =>/ / .> œ Þ</p><p>'</p><p>= +… 3,a b </p><p>Therefore</p><p>_ ‘ ” •a b</p><p>/ -9= ,> œ ' ' '</p><p># = + 3, = + 3,</p><p>œ= +</p><p>= + ,</p><p>+></p><p># #.</p><p>The above is valid for .= +</p><p>15. Integrating ,<i>by parts</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 257</p><p>( (ºa b</p><p>a b! !</p><p>E E+> => += ></p><p>+= > E</p><p>!E += E +=</p><p>#</p><p>>/ † / .> œ / .>> / '</p><p>= + = +</p><p>œ Þ' / E + = /</p><p>= +</p><p>a b a ba b a b</p><p>Taking a , as ,<i>limit </i>Ep_</p><p>( a b!_</p><p>+> =>#</p><p>>/ † / .> œ Þ'</p><p>= +</p><p>Note that the limit exists as long as .= +</p><p>17. Observe that For any value of ,> -9=2 +> œ > / > / Î# Þ -a b+> +>( (º</p><p>a ba b</p><p>! !</p><p>E E-> => -= ></p><p>-= > E</p><p>!E -= E -=</p><p>#</p><p>> / † / .> œ / .>> / '</p><p>= - = -</p><p>œ Þ' / E - = /</p><p>= -</p><p>a b a ba b a b</p><p>Taking a , as ,<i>limit </i>Ep_</p><p>( a b!_</p><p>-> =>#</p><p>>/ † / .> œ Þ'</p><p>= -</p><p>Note that the limit exists as long as . Therefore,= -k k( – —a b a b</p><p>a b a b!</p><p>_=></p><p># #</p><p># #</p><p># #</p><p>> -9=2 +> † / .> œ ' ' '</p><p># = + = +</p><p>œ= +</p><p>= + = +.</p><p>18. Integrating ,<i>by parts</i></p><p>( (º(</p><p>! !</p><p>E E8 +> => 8' += ></p><p>8 += > E</p><p>!8 =+ E</p><p>!</p><p>E8' += ></p><p>> / † / .> œ > / .>> / 8</p><p>= + = +</p><p>œ > / .> ÞE / 8</p><p>= + = +</p><p>a b a ba b a b</p><p>Continuing to integrate by parts, it follows that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 258</p><p>( a ba b a b a b</p><p>ˆ ‰!E8 +> =></p><p>8 += E 8' += E</p><p>#</p><p>+= E</p><p>$ 8'</p><p>+= E</p><p>> / † / .> œ E / 8E /</p><p>= + = +</p><p> â Þ8xE/</p><p>8 # x = + = +</p><p>8x / '</p><p>a b a b</p><p>a b a b</p><p>That is,</p><p>( a b a b!E8 +> => += E</p><p>8 8'> / † / .> œ : E † / </p><p>8x</p><p>= +a b ,</p><p>in which is a of degree . For any polynomial,: 88a b0 <i>polynomial given</i>limEÄ_</p><p>8 =+ E: E † / œ !a b a b ,</p><p>as long as . Therefore,= +</p><p>( a b!_8 +> =></p><p>8'> / † / .> œ</p><p>8x</p><p>= +.</p><p>20. Observe that Using the result in Prob. ,> =382 +> œ > / > / Î# Þ ')# # +> # +>a b( – —a b a b</p><p>a ba b</p><p>!</p><p>_# =></p><p>$ $</p><p># #</p><p># # $</p><p>> =382 +> † / .> œ ' #x #x</p><p># = + = +</p><p>œ#+ $= +</p><p>= +.</p><p>The above is valid for .= +k k22. Integrating ,<i>by parts</i></p><p>( (º! !</p><p>E E> > ></p><p>E</p><p>!E E</p><p>> / .> œ > / / .></p><p>œ ' / E/ Þ</p><p>Taking a , as ,<i>limit </i>Ep_</p><p>(!</p><p>_> E> / .> œ ' / Þ</p><p>Hence the integral .<i>converges</i></p><p>23. Based on a series expansion, note that for ,> !</p><p>/ ' > > Î# > Î#> # # .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 259</p><p>It follows that for ,> !</p><p>> / '</p><p>## > .</p><p>Hence for any finite ,E '</p><p>('</p><p>E# >> / .> </p><p>E '</p><p>#.</p><p>It is evident that the limit as does not exist.Ep_</p><p>24. Using the fact that , and the fact thatk k-9= > Ÿ '(!</p><p>_>/ .> œ ' ,</p><p>it follows that the given integral .<i>converges</i></p><p>25 . Let . Integrating ,a b+ : ! <i>by parts</i>( (º</p><p>(! !</p><p>E EB : B : B :'</p><p>E</p><p>!</p><p>: E B :'</p><p>!</p><p>E</p><p>/ B .B œ / B : / B .B</p><p>œ E / : / B .BÞ</p><p>Taking a , as ,<i>limit </i>Ep_</p><p>( (! !</p><p>_ _B : B :'/ B .B œ : / B .B .</p><p>That is, .> >a b a b: ' œ : :a b, : œ !. Setting ,</p><p>>a b (' œ / .B œ '!</p><p>_B .</p><p>a b a b- : œ 8 ,. Let . Using the result in Part ,> ></p><p>></p><p>></p><p>a b a ba b a ba ba b a b</p><p>8 ' œ 8 8</p><p>œ 8 8 ' 8 '</p><p>ã</p><p>œ 8 8 ' 8 # â# † ' † ' .</p><p>Since , .> >a b a b' œ ' 8 ' œ 8xa b a b. ,. Using the result in Part ,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 260</p><p>> ></p><p>></p><p>></p><p>a b a b a ba ba b a ba ba b a b a b</p><p>: 8 œ : 8 ' : 8 '</p><p>œ : 8 ' : 8 # : 8 #</p><p>ã</p><p>œ : 8 ' : 8 # â : ' : : .</p><p>Hence</p><p>></p><p>></p><p>a ba b a ba b a b: 8: œ : : ' : ' â : 8 ' ÞGiven that , it follows that> 1a b È'Î# œ</p><p>> >1Œ Œ È$ ' '</p><p># # # #œ œ</p><p>and</p><p>> >1Œ Œ È' * ( & $ $</p><p># # # # # # $#œ † † † œ</p><p>*%&.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 261</p><p><b>Section 6.2</b></p><p>1. Write the function as</p><p>$ $ #</p><p>= % # = %œ</p><p># #.</p><p>Hence ._' $#c da b] = œ =38 #>3. Using <i>partial fractions</i>,</p><p># # ' '</p><p>= $= % & = ' = %œ </p><p># ” •.Hence ._' > %>#&c da b ˆ ‰] = œ / /5. Note that the denominator is = #= &# <i>irreducible</i> over the reals. Completing thesquare, = #= & œ = ' %# #a b . Now convert the function to a <i>rational function</i>of the variable . That is,0 œ = '</p><p>#= # # = '</p><p>= #= &œ</p><p>#</p><p>a ba b= ' %# .</p><p>We know that</p><p>_0</p><p>0'</p><p>#” •# % œ # -9= #> .Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+</p><p>_' >” •#= #= #= &#</p><p>œ #/ -9= #> .</p><p>6. Using <i>partial fractions</i>,</p><p>#= $ ' ' (</p><p>= % % = # = #œ </p><p># ” •.Hence . Note that we can also write_' #> #>'%c d a ba b] = œ / (/</p><p>#= $ = $ #</p><p>= % = % # = %œ # </p><p># # #.</p><p>8. ,Using <i>partial fractions</i></p><p>)= %= '# ' = #</p><p>= = % = = % = %œ $ & #</p><p>#</p><p># # #a b .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 262</p><p>Hence ._'c da b] = œ $ & -9= #> # =38 #>9. The denominator is = %= &# <i>irreducible</i> over the reals. Completing the square,= %= & œ = # '# #a b . Now convert the function to a <i>rational function </i>of thevariable . That is,0 œ = #</p><p>' #= & # = #</p><p>= %= &œ</p><p>#</p><p>a ba b= # '# .</p><p>We find that</p><p>_0 0</p><p>0'# #” •& # ' ' œ & =38 > # -9= > .</p><p>Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+_' #>” • a b' #=</p><p>= %= &#œ / & =38 > # -9= > .</p><p>10. over the reals. CompletingNote that the denominator is = #= '!# <i>irreducible</i>the square, = #= '! œ = ' *# #a b . Now convert the function to a <i>rationalfunction </i>of the variable . That is,0 œ = '</p><p>#= $ # = ' &</p><p>= #= '!œ</p><p>#</p><p>a ba b= ' *# .</p><p>We find that</p><p>_0</p><p>0 0'</p><p># #” •# & & * * $ œ # -9= $> =38 $> .Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+</p><p>_' >” • Œ #= $= #= '!#</p><p>œ / # -9= $> =38 $>&</p><p>$.</p><p>12. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! C ! $ = ] = C ! #] = œ !# wa b a b a b c d a ba b a b .Applying the ,<i>initial conditions</i></p><p>= ] = $= ] = #] = = $ œ !# a b a b a b .Solving for , the transform of the solution is] =a b</p><p>] = œ= $</p><p>= $= #a b</p><p>#.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 263</p><p>Using <i>partial fractions</i>,</p><p>= $</p><p>= $= ##œ </p><p># '</p><p>= ' = #.</p><p>Hence .C > œ ] = œ # / /a b c da b_' > #>13. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! C ! # = ] = C ! #] = œ !# wa b a b a b c d a ba b a b .Applying the ,<i>initial conditions</i></p><p>= ] = #= ] = #] = ' œ !# a b a b a b .Solving for , the transform of the solution is] =a b</p><p>] = œ'</p><p>= #= #a b</p><p>#.</p><p>Since the denominator is , write the transform as a function of .<i>irreducible </i>0 œ = 'That is,</p><p>'</p><p>= #= ##œ</p><p>'</p><p>= ' 'a b# .First note that</p><p>_0</p><p>'#” •' ' œ =38 > .</p><p>Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+_' >” •'</p><p>= #= ##œ / =38 > .</p><p>Hence .C > œ / =38 >a b >15. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! C ! # = ] = C ! #] = œ !# wa b a b a b c d a ba b a b .Applying the ,<i>initial conditions</i></p><p>= ] = #= ] = #] = #= % œ !# a b a b a b .Solving for , the transform of the solution is] =a b</p><p>] = œ#= %</p><p>= #= #a b</p><p>#.</p><p>Since the denominator is , write the transform as a function of .<i>irreducible </i>0 œ = 'Completing the square,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 264</p><p>#= %</p><p>= #= ##œ# = ' #</p><p>= ' $</p><p>a ba b# .</p><p>First note that</p><p>_0</p><p>0 0'</p><p># #” • È ÈÈ# # # $ $ œ # -9=2 $ > =382 $ >$ .Using the fact that , the solution of the IVP is_ _c d c da b a b/ 0 > œ 0 >+> =p=+</p><p>C > œa b _' >” • È ÈÈ#= %= #= ## œ / # -9=2 $ > =382 $ >#$ .</p><p>16. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! C ! # = ] = C ! &] = œ !# wa b a b a b c d a ba b a b .Applying the ,<i>initial conditions</i></p><p>= ] = #= ] = &] = #= $ œ !# a b a b a b .Solving for , the transform of the solution is] =a b</p><p>] = œ#= $</p><p>= #= &a b</p><p>#.</p><p>Since the denominator is , write the transform as a function of .<i>irreducible </i>0 œ = 'That is,</p><p>#= $</p><p>= #= &#œ# = ' '</p><p>= ' %</p><p>a ba b# .</p><p>We know that</p><p>_0</p><p>0 0'</p><p># #” •# ' ' % % # œ # -9= #> =38 #> .Using the fact that , the solution of the IVP is_ _c d c da b a b/ 0 > œ 0 >+> =p=+</p><p>C > œa b _' >” • Œ #= $= #= &#</p><p>œ / # -9= #> =38 #>'</p><p>#.</p><p>17. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! = C ! = C ! C ! % = ] = = C ! = C ! C ! </p><p> ' = ] = = C ! C ! % = ] = C ! ] = œ !</p><p>% $ # w ww www $ # w ww</p><p># w</p><p>a b a b a b a b a b a b a b a b a b ‘ ‘a b a b a b c d a ba b a bApplying the ,<i>initial conditions</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 265</p><p>= ] = %= ] = '= ] = %= ] = ] = = %= ( œ !% $ # #a b a b a b a b a b .Solving for the transform of the solution,</p><p>] = œ œ= %= ( = %= (</p><p>= %= '= %= ' = 'a b a b</p><p># #</p><p>% $ # %.</p><p>Using <i>partial fractions</i>,</p><p>= %= ( % # '</p><p>= ' = 'œ </p><p>= ' = '</p><p>#</p><p>% % $ #a b a b a b a b .Note that and . Hence the solution_ _ _c d a b c d c da b a b> œ 8x Î= / 0 > œ 0 >8 8' +> =p=+of the IVP is</p><p>C > œ= %= (</p><p>= 'a b a b_' $ > # > >– —</p><p>#</p><p>%œ > / > / > /#</p><p>$.</p><p>18. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! = C ! = C ! C ! ] = œ ! Þ% $ # w ww wwwa b a b a b a b a b a b Applying the ,<i>initial conditions</i></p><p>= ] = ] = = = œ !% $a b a b .Solving for the transform of the solution,</p><p>] = œ=</p><p>= 'a b</p><p>#.</p><p>By inspection, it follows that C > œa b _' ‘ '# œ -9=2 > Þ19. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! = C ! = C ! C ! %] = œ ! Þ% $ # w ww wwwa b a b a b a b a b a b Applying the ,<i>initial conditions</i></p><p>= ] = %] = = #= œ !% $a b a b .Solving for the transform of the solution,</p><p>] = œ=</p><p>= #a b</p><p>#.</p><p>It follows that C > œa b _' ‘ È ## œ -9= # > Þ20. Taking the Laplace transform of both sides of the ODE, we obtain</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 266</p><p>= ] = = C ! C ! ] = œ=</p><p>= %# w #</p><p>#a b a b a b a b= .</p><p>Applying the ,<i>initial conditions</i></p><p>= ] = ] = = œ=</p><p>= %# #</p><p>#a b a b= .</p><p>Solving for , the transform of the solution is] =a b] = œa b = =</p><p>= = % = a ba b# # # # #= = .</p><p>Using <i>partial fractions</i> on the first term,</p><p>= ' = =</p><p>= = % % = = %œ a ba b ” •# # # # # # #= = = .</p><p>First note that</p><p>_ = _=</p><p>' '# # #” • ” •= = %œ -9= > œ -9= #> and .</p><p>Hence the solution of the IVP is</p><p>C > œ -9= > -9= #> -9= >' '</p><p>% % </p><p>œ -9= > -9= #>& '</p><p>% % </p><p>a b= =</p><p>= =</p><p>=</p><p>= </p><p># #</p><p>#</p><p># #.</p><p>21. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! # = ] = C ! #] = œ=</p><p>= '# w</p><p>#a b a b a b c d a ba b a b .</p><p>Applying the ,<i>initial conditions</i></p><p>= ] = #= ] = #] = = # œ=</p><p>= '#</p><p>#a b a b a b .</p><p>Solving for , the transform of the solution is] =a b] = œa b = = #</p><p>= #= # = ' = #= #a ba b# # # .</p><p>Using <i>partial fractions</i> on the first term,</p><p>= ' = # = %</p><p>= #= # = ' & = ' = #= #œ a ba b ” •# # # # .</p><p>Thus we can write</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 267</p><p>] = œ'</p><p>&a b = # ' # #= $</p><p>= ' & = ' & = #= # </p><p># # #.</p><p>For the , we note that . So that<i>last term </i>= #= # œ = ' '# #a b#= $ # = ' '</p><p>= #= #œ= ' '# #a ba b .</p><p>We know that</p><p>_0</p><p>0 0'</p><p># #” •# ' ' ' œ # -9= > =38 > .Based on the of the Laplace transform,<i>translation property</i></p><p>_' >” • a b#= $= #= ##</p><p>œ / # -9= > =38 > .</p><p>Combining the above, the solution of the IVP is</p><p>C > œ -9= > =38 > ' # #</p><p>& & &a b / # -9= > =38 > Þ>a b</p><p>23. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! # = ] = C ! ] = œ%</p><p>= '# wa b a b a b c d a ba b a b .</p><p>Applying the ,<i>initial conditions</i></p><p>= ] = #= ] = ] = #= $ œ%</p><p>= '# a b a b a b .</p><p>Solving for , the transform of the solution is] =a b] = œa b % #= $</p><p>= ' = 'a b a b$ # .</p><p>First write</p><p>#= $ # = ' ' # '</p><p>= ' = ' = 'œ œ Þ</p><p>= 'a b a b a ba b</p><p># # #</p><p>We note that</p><p>_0 0 0</p><p>' #$ #” •% # ' œ # > # > .</p><p>So based on the of the Laplace transform, the solution of the IVP is<i>translation property</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 268</p><p>C > œa b # > / > / # / Þ# > > >25. Let be the on the right-hand-side. Taking the Laplace0 >a b <i>forcing function</i>transformof both sides of the ODE, we obtain</p><p>= ] = = C ! C ! ] = œ 0 ># wa b a b a b a b c da b_ .Applying the ,<i>initial conditions</i></p><p>= ] = ] = œ 0 ># a b a b c da b_ .Based on the definition of the Laplace transform,</p><p>_c d a ba b ((</p><p>0 > œ 0 > / .></p><p>œ > / .></p><p>œ ' / /</p><p>= = =</p><p>!</p><p>_=></p><p>!</p><p>'=></p><p># #</p><p>= =</p><p>.</p><p>Solving for the transform,</p><p>] = œ / Þ' = '</p><p>= = ' = = 'a b a b a b# # # #=</p><p>Using <i>partial fractions</i>,</p><p>' ' '</p><p>= = ' = = 'œ </p><p># # # #a band</p><p>= ' =</p><p>= = ' = = 'œ Þ</p><p># # #a bWe find, by inspection, that</p><p>_'” •'= = '# #a b œ > =38 > .</p><p>Referring to , in Table ,<i>Line </i>'$ 'Þ#Þ'</p><p>_ _c d c da b a b a b? > 0 > - œ / 0 >- -= .Let</p><p>_c da b a b1 > œ œ= '= = '# # ' ' = '= = = ' = ' # # # .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 269</p><p>Then . It follows, therefore, that1 > œ ' > -9= > =38 >a b_' = '” • a bc d/ † œ ? >= '</p><p>= = '# #a b ' > ' -9= > ' =38 > 'a b a b a b .Combining the above, the solution of the IVP is</p><p>C > œ ' > ' -9= > ' =38 > 'a b a b a b a b> =38 > ? >'a bc d .26. Let be the on the right-hand-side. Taking the Laplace0 >a b <i>forcing function</i>transformof both sides of the ODE, we obtain</p><p>= ] = = C ! C ! %] = œ 0 ># wa b a b a b a b c da b_ .Applying the ,<i>initial conditions</i></p><p>= ] = %] = œ 0 ># a b a b c da b_ .Based on the definition of the Laplace transform,</p><p>_c d a ba b (( (</p><p>0 > œ 0 > / .></p><p>œ > / .> / .></p><p>œ ' /</p><p>= =</p><p>!</p><p>_=></p><p>! '</p><p>' _=> =></p><p># #</p><p>=</p><p>.</p><p>Solving for the transform,</p><p>] = œ / Þ' '</p><p>= = % = = %a b a b a b# # # #=</p><p>Using <i>partial fractions</i>,</p><p>' ' ' '</p><p>= = % % = = %œ Þ</p><p># # # #a b ” •We find that</p><p>_'# #” •a b' ' '= = % % )œ > =38 > .</p><p>Referring to , in Table ,<i>Line </i>'$ 'Þ#Þ'</p><p>_ _c d c da b a b a b? > 0 > - œ / 0 >- -= .It follows that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 270</p><p>_' =# # '” • ” •a b a b a b a b/ † œ ? > > ' =38 > ' ' '= = % % ) .</p><p>Combining the above, the solution of the IVP is</p><p>C > œ > =38 > ? > > ' =38 > ' ' ' '</p><p>% ) % )a b a b a b a b” •' .</p><p>28 . Assuming that the conditions of Theorem are satisfied,a b+ 'Þ#Þ'J = œ / 0 > .></p><p>.</p><p>.=</p><p>œ / 0 > .>`</p><p>`=</p><p>œ > / 0 > .></p><p>œ / >0 > .> Þ</p><p>w =></p><p>!</p><p>_</p><p>!</p><p>_=></p><p>!</p><p>_=></p><p>!</p><p>_=></p><p>a b a b(( ‘a b( ‘a b( c da b</p><p>a b, 5 '. Using , suppose that for some ,<i>mathematical induction</i>J = œ / > 0 > .>a b5 =></p><p>!</p><p>_5a b a b a b( ’ “ .</p><p>Differentiating both sides,</p><p>J = œ / > 0 > .>.</p><p>.=</p><p>œ / > 0 > .>`</p><p>`=</p><p>œ > / > 0 > .></p><p>œ / > 0 > .></p><p>a b5' =>!</p><p>_5</p><p>!</p><p>_=> 5</p><p>!</p><p>_=> 5</p><p>!</p><p>_=> 5'</p><p>a b a b a b( ’ “( ’ “a b a b( ’ “a b a b( ’ “a b a b .</p><p>29. We know that</p><p>_ ‘/ œ '= +</p><p>+> .</p><p>Based on Prob. ,#)</p><p>_ ‘ ” • > / œ . '.= = +</p><p>+> .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 271</p><p>Therefore,</p><p>_ ‘ a b> / œ'</p><p>= ++></p><p>#.</p><p>31. Based on Prob. ,#)</p><p>_ _c d c da b” •</p><p> > œ '.</p><p>.=</p><p>œ Þ. '</p><p>.= =</p><p>88</p><p>8</p><p>8</p><p>8</p><p>Therefore,</p><p>_c d a b a b> œ ' ' 8x=</p><p>œ Þ8x</p><p>=</p><p>8 88</p><p>8'</p><p>8'</p><p>33. Using the of the Laplace transform,<i>translation property</i></p><p>_ ‘ a b/ =38 ,> œ Þ,</p><p>= + ,+></p><p># #</p><p>Therefore,</p><p>_ ‘ – —a ba ba b</p><p>> / =38 ,> œ . ,</p><p>.= = + ,</p><p>œ#, = +</p><p>= #+= + ,</p><p>+># #</p><p># # # #.</p><p>34. Using the of the Laplace transform,<i>translation property</i></p><p>_ ‘ a b/ -9= ,> œ Þ= +</p><p>= + ,+></p><p># #</p><p>Therefore,</p><p>_ ‘ – —a ba b</p><p>a b</p><p>> / -9= ,> œ . = +</p><p>.= = + ,</p><p>œ= + ,</p><p>= #+= + ,</p><p>+># #</p><p># #</p><p># # # #.</p><p>35 . Taking the Laplace transform of the given ,a b+ <i>Bessel equation</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 272</p><p>_ _ _c d c d c d> C C > C œ !ww w .Using the of the transform,<i>differentiation property</i></p><p> C C C œ !. .</p><p>.= .=_ _ _c d c d c dww w .</p><p>That is,</p><p> = ] = = C ! C ! =] = C ! ] = œ !. .</p><p>.= .= ‘a b a b a b a b a b a b# w .</p><p>It follows that</p><p>ˆ ‰ a b a b' = ] = =] = œ !# w .a b a b, ] =. We obtain a ODE in :<i>first-order linear</i></p><p>] = ] = œ !=</p><p>= 'w</p><p>#a b a b ,</p><p>with <i>integrating factor</i></p><p>.a b Œ ( È= œ /B: .= œ = ' '#</p><p># .</p><p>The first-order ODE can be written as</p><p>.</p><p>. ' † ] = œ !’ “È a b# ,</p><p>with solution</p><p>] = œ-</p><p>= 'a b È # .</p><p>a b- =. In order to obtain powers of , first write<i>negative</i>' ' '</p><p>= 'œ ' Þ= =È ” •# #</p><p>'Î#</p><p>Expanding in a ,Š ‹' '=# 'Î# <i>binomial series</i>' ' ' † $ ' † $ † &</p><p>' 'Î=œ ' = = = â</p><p># # † % # † % † 'È a b# # % ' ,valid for . Hence, we can formally express as= ' ] =# a b</p><p>] = œ - â Þ' ' ' ' † $ ' ' † $ † & '</p><p>= # = # † % = # † % † ' =a b ” •$ & (</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 273</p><p>Assuming that inversion is valid,<i>term-by-term</i></p><p>C > œ - ' â' > ' † $ > ' † $ † & ></p><p># #x # † % %x # † % † ' 'x</p><p>œ - ' â Þ#x > %x > 'x ></p><p># #x # † % %x # † % † ' 'x</p><p>a b ” •” •</p><p># % '</p><p># # # # # #</p><p># % '</p><p>It follows that</p><p>C > œ - ' > > > â' ' '</p><p># # † % # † % † '</p><p>œ - > Þ '</p><p># 8x</p><p>a b ” •' a ba b</p><p># # # # # ## % '</p><p>8œ!</p><p>_ 8</p><p>#8 ##8</p><p>The series is evidently the expansion, about , of B œ ! N > Þ!a b36 . Taking the Laplace transform of the given ,a b, <i>Legendre equation</i></p><p>_ _ _ ! ! _c d c d a b c d ‘C > C # > C ' C œ !ww # ww w .Using the of the transform,<i>differentiation property</i></p><p>_ _ _ ! ! _c d c d c d a b c dC C # C ' C œ !. ..= .=</p><p>ww ww w#</p><p>#.</p><p>That is,</p><p> ‘ ‘a b a b a b a b a b a bc d a b a ba b a b</p><p>= ] = = C ! C ! = ] = = C ! C ! .</p><p>.=</p><p> # = ] = C ! ' ] = œ ! Þ.</p><p>.=</p><p># w # w#</p><p>#</p><p>! !</p><p>Invoking the , we have<i>initial conditions</i></p><p>= ] = ' = ] = ' # =] = ' ] = œ ! Þ. .</p><p>.= .=# #</p><p>#</p><p>#a b a b c d a b a b ‘ a b ! !</p><p>After carrying out the differentiation, the equation simplifies to</p><p>. .</p><p>.= . ] = # = ] = = ' ] = œ ' Þ</p><p>#</p><p>## # ‘ ‘a b c d a b a ba b ! !</p><p>That is,</p><p>= ] = #= ] = = ' ] = œ '. .</p><p>.= .=# ##</p><p>#a b a b a b a b ‘! ! .</p><p>37. By definition of the Laplace transform, given the appropriate conditions,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 274</p><p>_ 7 7</p><p>7 7</p><p>c d a ba b ( (” •( ( a b</p><p>1 > œ / 0 . .></p><p>œ / 0 . .> Þ</p><p>! !</p><p>_ >=></p><p>! !</p><p>_ >=></p><p>Assuming that the order of integration can be exchanged,</p><p>_ 7 7</p><p>7 7</p><p>c d a ba b ( (” •( a b” •</p><p>1 > œ 0 / .> .</p><p>œ 0 . Þ/</p><p>=</p><p>!</p><p>_ _=></p><p>!</p><p>_ =77</p><p>c da b a bNote the of integration is the area between the lines and <i>region </i>7 7> œ > > œ ! ÞHence</p><p>_ 7 7</p><p>_</p><p>c d a ba b (c da b</p><p>1 > œ 0 / .'</p><p>=</p><p>œ 0 > Þ'</p><p>=</p><p>!</p><p>_=7</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 275</p><p><b>Section 6.3</b></p><p>1.</p><p>3.</p><p>5.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 276</p><p>6.</p><p>7. Using the Heaviside function, we can write</p><p>0 > œ > # ? >a b a b a b# # .The Laplace transform has the property that</p><p>_ _c d c da b a b a b? > 0 > - œ / 0 >- -= .Hence</p><p>_ ‘a b a b> # ? ># # œ # /=</p><p> =</p><p>#</p><p>2.</p><p>9. The function can be expressed as</p><p>0 > œ > ? > ? > Þa b a bc da b a b1 1 1#Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function as</p><p>0 > œ > ? > > # ? > ? > Þa b a b a b a b a b a b1 1 11 1 1# #It follows that</p><p>_1c da b0 > œ / / /</p><p>= = =</p><p> = # = # =</p><p># #</p><p>1 1 1</p><p>.</p><p>10. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform that</p><p>_c da b0 > œ # '/ / /= = =</p><p>= $= %=</p><p>.</p><p>11. Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function as</p><p>0 > œ > # ? > ? > > $ ? > ? > Þa b a b a b a b a b a b a b# # $ $</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 277</p><p>It follows that</p><p>_c da b0 > œ / / / /= = = =</p><p>#= #= $= $=</p><p># #.</p><p>12. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform that</p><p>_c da b0 > œ ' /= =# #</p><p>=</p><p>.</p><p>13. Using the fact that ,_ _c d c da b a b/ 0 > œ 0 >+> =p=+_' $ #></p><p>%– —a b$x</p><p>= #œ > / .</p><p>15. First consider the function</p><p>K = œ Þ# = '</p><p>= #= #a b a b</p><p>#</p><p>Completing the square in the denominator,</p><p>K = œ Þ# = '</p><p>= ' 'a b a ba b#</p><p>It follows that</p><p>_' >c da bK = œ # / -9= > ÞHence</p><p>_' #= ># # ‘a b a b a b/ K = œ # / -9= > # ? > Þa b16. The of the function is Using <i>inverse transform </i>#Î = % 0 > œ =382 #> Þa b a b# the><+8=6+>398 :<9:/<>C of the transform,</p><p>_'#=</p><p># #” • a b a b# /= % œ =382 # > # † ? > Þ</p><p>17. First consider the function</p><p>K = œ Þ= #</p><p>= %= $a b a b</p><p>#</p><p>Completing the square in the denominator,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 278</p><p>K = œ Þ= #</p><p>= # 'a b a ba b#</p><p>It follows that</p><p>_' #>c da bK = œ / -9=2 > ÞHence</p><p>_' # >'=</p><p># '” •a b a b a b= # /= %= $ œ / -9=2 > ' ? > Þa b</p><p>18. Write the function as</p><p>J = œ Þ/ / / /</p><p>= = = =a b = #= $= %=</p><p>It follows from the ><+8=6+>398 :<9:/<>C of the transform, that</p><p>_'= #= $= %=</p><p>' # $ %” • a b a b a b a b/ / / /=</p><p>œ ? > ? > ? > ? > Þ</p><p>19 . By definition of the Laplace transform,a b+_c d a ba b (0 -> œ / 0 -> .> Þ</p><p>!</p><p>_=></p><p>Making a change of variable, , we have7 œ -></p><p>_ 7 7</p><p>7 7</p><p>c d a ba b (( a b</p><p>0 -> œ / 0 .'</p><p>-</p><p>œ / 0 . Þ'</p><p>-</p><p>!</p><p>_= Î-</p><p>!</p><p>_ =Î-</p><p>a ba b7</p><p>7</p><p>Hence , where ._c da b ˆ ‰0 -> œ J =Î- +' =- -a b a b, +. Using the result in Part ,</p><p>_” •Œ a b0 œ 5 J 5= Þ>5</p><p>Hence</p><p>_'c da b Œ J 5= œ 0 Þ' >5 5</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 279</p><p>a b a b- ,. From Part ,_'c da b Œ J += œ 0 Þ' ></p><p>+ +</p><p>Note that . Using the fact that ,+= , œ + = ,Î+ / 0 > œ 0 >a b c d c da b a b_ _-> =p=-_' ,>Î+c dJ += , 0' ></p><p>+ +a b Œ œ / .</p><p>20. First write</p><p>J = œ8xa b ˆ ‰=#</p><p>8' .</p><p>Let Based on the results in Prob. ,K = œ 8xÎ= Þ '*a b 8' =</p><p># #K œ 1 #>_'’ “Š ‹ a b,</p><p>in which . Hence1 > œ >a b 8_' 8' 88c d a ba bJ = œ # #> œ # > Þ</p><p>23. First write</p><p>J = œ Þ/</p><p># = 'Î#a b a b</p><p>% ='Î#a b</p><p>Now consider</p><p>K = œ/</p><p>=a b #= .</p><p>Using the result in Prob. ,'* ,a b_'c da b Œ K #= œ 1' ></p><p># #,</p><p>in which . Hence . It follows that1 > œ ? > K #= œ ? >Î# œ ? >a b a b c d a b a ba b# # %' ' '# #__' >Î# %c d a ba bJ = œ / ? >'</p><p>#.</p><p>24. By definition of the Laplace transform,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 280</p><p>_c d a ba b (0 > œ / ? > .> Þ!</p><p>_=></p><p>'</p><p>That is,</p><p>_c da b (0 > œ / .>œ Þ' /</p><p>=</p><p>!</p><p>'=></p><p>=</p><p>25. First write the function as . It follows that0 > œ ? > ? > ? > ? >a b a b a b a b a b! ' # $_c da b ( (0 > œ / .> / .> Þ</p><p>! #</p><p>' $=> =></p><p>That is,</p><p>_c da b0 > œ ' / / /= =</p><p>œ Þ' / / /</p><p>=</p><p>= #= $=</p><p>= #= $=</p><p>26. The transform may be computed directly. On the other hand, using the ><+8=6+>398:<9:/<>C of the transform,</p><p>_c d a ba b '– —'a b</p><p>a b</p><p>0 > œ ' /</p><p>= =</p><p>œ /'</p><p>=</p><p>œ' ' /</p><p>= ' /</p><p>5œ'</p><p>#8'55=</p><p>5 œ!</p><p>#8'= 5</p><p>= #8#</p><p>=.</p><p>That is,</p><p>_c da b a ba b0 > œ ' /= ' /#= 8'</p><p>=.</p><p>29. The given function is , with . Using the result of Prob. ,<i>periodic </i>X œ # #)</p><p>_c d a ba b ( (0 > œ / 0 > .> œ / .>' ' / ' /#= #=! !</p><p># '=> => .</p><p>That is,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 281</p><p>_c da b a ba b</p><p>0 > œ' /</p><p>= ' /</p><p>œ'</p><p>= ' /</p><p>=</p><p>#=</p><p>=.</p><p>31. , with . Using the result of Prob. , The function is <i>periodic </i>X œ ' #)</p><p>_c da b (0 > œ > / .> Þ' /= !</p><p>'=></p><p>It follows that</p><p>_c da b a ba b0 > œ Þ' / ' ' /=</p><p># =</p><p>32. , with . Using the result of Prob. ,The function is <i>periodic </i>X œ #)1</p><p>_c da b (0 > œ =38 > † / .> Þ' / = !</p><p>=>1</p><p>1</p><p>We first calculate</p><p>(!</p><p>=> =</p><p>#</p><p>1 1</p><p>=38 > † / .> œ Þ' /</p><p>' =</p><p>Hence</p><p>_c da b a ba b0 > œ Þ' /' / ' = =</p><p> = #</p><p>1</p><p>1</p><p>33 .a b+</p><p>_ _ _c d c d c da b a b0 > œ ' ? >œ ' /</p><p>= =</p><p>'=</p><p>.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 282</p><p>a b, .</p><p>Let . ThenJ = œ ' ? >a b c da b_ '_ 7 7” •( c d a ba b</p><p>!</p><p>></p><p>'</p><p>=</p><p>#' ? . œ J = œ Þ</p><p>' ' /</p><p>= =</p><p>a b- .</p><p>Let . ThenK = œ 1 >a b c da b__c d a b a ba b</p><p>a b</p><p>2 > œ K = / K =</p><p>œ /' / ' /</p><p>= =</p><p>œ' /</p><p>=</p><p>=</p><p>= =</p><p># #=</p><p>= #</p><p>#.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 283</p><p>34 .a b+</p><p>a b, X œ # #). , with . Using the result of Prob. ,The given function is <i>periodic</i>_c d a ba b (0 > œ / : > .> Þ'</p><p>' /#= !</p><p>#=></p><p>Based on the piecewise definition of ,: >a b( ( (a b a b</p><p>a b! ! '</p><p># ' #=> => =></p><p>#= #</p><p>/ : > .> œ > / .> # > / .></p><p>œ ' / Þ'</p><p>=</p><p>Hence</p><p>_c da b a ba b: > œ Þ' /= ' /=</p><p># =</p><p>a b a b- : > 'Þ#Þ'. Since satisfies the hypotheses of Theorem ,_ _c d c d a ba b a b: > œ = : > : !w .</p><p>Using the result of Prob. ,$!</p><p>_c da b a ba b: > œ ' /= ' /w=</p><p>=.</p><p>We note the , hence: ! œ !a b_c da b ” •a ba b: > œ ' ' /= = ' /</p><p>=</p><p>=.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 284</p><p><b>Section 6.4</b></p><p>2. Let be the on the right-hand-side. Taking the Laplace transform2 >a b <i>forcing function</i>of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! # = ] = C ! #] = œ 2 ># wa b a b a b c d a b c da b a b a b_ .Applying the initial conditions,</p><p>= ] = #= ] = #] = ' œ 2 ># a b a b a b c da b_ .The forcing function can be written as Its transform is2 > œ ? > ? > Þa b a b a b1 1#</p><p>_c da b2 > œ / /=</p><p> = # =1 1</p><p>.</p><p>Solving for , the transform of the solution is] =a b] = œ </p><p>'</p><p>= #= # = = #= #a b a b# #/ /</p><p> = # =1 1</p><p>.</p><p>First note that</p><p>'</p><p>= #= ##œ</p><p>'</p><p>= ' 'a b# .Using partial fractions,</p><p>' ' ' '</p><p>= = #= # # = #œ a b# a ba b</p><p>= ' '</p><p>= ' 'Þ#</p><p>Taking the inverse transform, term-by-term,</p><p>_ _” • – —'= #= # œ / =38 ># >œ '= ' 'a b# .Now let</p><p>K = œa b '= = #= #a b# .</p><p>Then</p><p>_'c da bK = œ ' ' '# # #</p><p>/ -9= > / =38 >> > .</p><p>Using Theorem ,'Þ$Þ'</p><p>_' -= - -c d a b a ba b/ K = œ ? > ? >' '# #</p><p>/ -9= > - =38 > - >-a bc da b a b .Hence the solution of the IVP is</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 285</p><p>C > œ ? > ? > ' '</p><p># #</p><p> ? > ? >' '</p><p># #</p><p>a b a b a ba b a b</p><p>/ =38 > / -9= > =38 > </p><p>/ -9= > # =38 > #</p><p>> ></p><p> >#</p><p>1 1</p><p>1 1</p><p>a ba b</p><p>1</p><p>1</p><p>c da b a bc da b a b</p><p>1 1</p><p>1 1# # .</p><p>That is,</p><p>C > œ ? > ? > ? > ' '</p><p># #</p><p> ? >'</p><p>#</p><p>a b c d a ba b a ba b</p><p>/ =38 > / -9= > =38 ></p><p>/ -9= > =38 ></p><p>> ></p><p> >#</p><p>1 1 1</p><p>1</p><p>#</p><p>#</p><p>a ba b</p><p>1</p><p>1</p><p>c dc d .</p><p>The solution starts out as free oscillation, due to the initial conditions. The amplitudeincreases, as long as the forcing is present. Thereafter, the solution rapidly decays.</p><p>4. Let be the on the right-hand-side. Taking the Laplace transform2 >a b <i>forcing function</i>of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! %] = œ 2 ># wa b a b a b a b c da b_ .Applying the initial conditions,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 286</p><p>= ] = %] = œ 2 ># a b a b c da b_ .The transform of the forcing function is</p><p>_c da b2 > œ ' /= ' = '# #</p><p> =1</p><p>.</p><p>Solving for , the transform of the solution is] =a b] = œ </p><p>'</p><p>= % = ' = % = 'a b a ba b a ba b# # # #/</p><p> =1</p><p>.</p><p>Using partial fractions,</p><p>' ' ' '</p><p>= % = ' $ = ' = %œ a ba b ” •# # # # .</p><p>It follows that</p><p>_'” • ” •'= % = 'a ba b# # œ =38 > =38 #>' '$ # .</p><p>Based on Theorem ,'Þ$Þ'</p><p>_ 1 1'” • ” •a b a b a b/= % = '</p><p> =</p><p># #</p><p>1</p><p>a ba b œ =38 > =38 #> # ? >' '$ # 1 .Hence the solution of the IVP is</p><p>C > œ =38 > =38 #> =38 > =38 #> ? >' ' ' '</p><p>$ # $ #a b a b” • ” • 1 .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 287</p><p>Since there is no , the solution follows the forcing function, after which<i>damping term</i>the response is a steady oscillation about .C œ !</p><p>5. Let be the on the right-hand-side. Taking the Laplace transform0 >a b <i>forcing function</i>of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! $ = ] = C ! #] = œ 0 ># wa b a b a b c d a b c da b a b a b_ .Applying the initial conditions,</p><p>= ] = $= ] = #] = œ 0 ># a b a b a b c da b_ .The transform of the forcing function is</p><p>_c da b0 > œ ' /= =</p><p>'!=</p><p>.</p><p>Solving for the transform,</p><p>] = œ ' /</p><p>= = $= # = = $= #a b a b a b# #</p><p>'!=</p><p>.</p><p>Using partial fractions,</p><p>' ' ' ' #</p><p>= = $= # # = = # = 'œ Þa b ” •#</p><p>Hence</p><p>_' >#</p><p>#>” •a b' ' /= = $= # # #œ / .Based on Theorem ,'Þ$Þ'</p><p>_' # >'! >'!# '!” •a b ‘ a b/'!=</p><p>= = $= # #œ ' / #/ ? >' a b a b .</p><p>Hence the solution of the IVP is</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 288</p><p>C > œ ' ? > / / #/ ? >' / '</p><p># # #a b c d a ba b ‘'! '!#> > #>#! >'!a b a b .</p><p>The solution increases to a steady value of . After the forcing ceases,<i>temporary </i>C œ 'Î#the response decays exponentially to .C œ !</p><p>6. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! $ = ] = C ! #] = œ/</p><p>=# w</p><p>#=a b a b a b c d a ba b a b .Applying the initial conditions,</p><p>= ] = $= ] = #] = ' œ/</p><p>=#</p><p>#=a b a b a b .Solving for the transform,</p><p>] = œ ' /</p><p>= $= # = = $= #a b a b# #</p><p>#=</p><p>.</p><p>Using partial fractions,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 289</p><p>' ' '</p><p>= $= # = ' = #œ </p><p>#</p><p>and</p><p>' ' ' ' #</p><p>= = $= # # = = # = 'œ Þa b ” •#</p><p>Taking the inverse transform. term-by-term, the solution of the IVP is</p><p>C > œ / / / / ? > Þ' '</p><p># #a b a b” •> #> ># # ># #a b a b</p><p>Due to the initial conditions, the response has a transient , followed by an<i>overshoot</i>exponential convergence to a steady value of .C œ 'Î#=</p><p>7. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! ] = œ/</p><p>=# w</p><p>$ =a b a b a b a b 1 .Applying the initial conditions,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 290</p><p>= ] = ] = = œ/</p><p>=#</p><p>$ =a b a b 1 .Solving for the transform,</p><p>] = œ = /</p><p>= ' = = 'a b a b# #</p><p>$ =1</p><p>.</p><p>Using partial fractions,</p><p>' ' =</p><p>= = ' = = 'œ a b# # .</p><p>Hence</p><p>] = œ / = ' =</p><p>= ' = = 'a b ” •# #$ =1 .</p><p>Taking the inverse transform, the solution of the IVP is</p><p>C > œ -9= > ' -9= > $ ? ></p><p>œ -9= > ' -9= > ? > Þ</p><p>a b c d a ba bc d a b1 $$ 11</p><p>Due to initial conditions, the solution temporarily oscillates about . After theC œ !forcing is applied, the response is a steady oscillation about .C œ '7</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 291</p><p>9. Let be the on the right-hand-side. Taking the Laplace transform1 >a b <i>forcing function</i>of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! ] = œ 1 ># wa b a b a b a b c da b_ .Applying the initial conditions,</p><p>= ] = ] = ' œ 1 ># a b a b c da b_ .The forcing function can be written as</p><p>1 > œ ' ? > $? >></p><p>#</p><p>œ > ' ? >> '</p><p># #</p><p>a b c d a ba ba b a b' '</p><p>'</p><p>with Laplace transform</p><p>_c da b1 > œ ' /#= #=# #</p><p>'=</p><p>.</p><p>Solving for the transform,</p><p>] = œ ' ' /</p><p>= ' #= = ' #= = 'a b a b a b# # # # #</p><p>'=</p><p>.</p><p>Using partial fractions,</p><p>' ' ' '</p><p>#= = ' # = = 'œ Þ</p><p># # # #a b ” •Taking the inverse transform, and using Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ =38 > > =38 > > ' =38 > ' ? >' '</p><p># #</p><p>œ > =38 > > ' =38 > ' ? > Þ' '</p><p># #</p><p>a b c d c d a ba b a bc d c d a ba b a b</p><p>'</p><p>'</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 292</p><p>The solution increases, in response to the , and thereafter oscillates about a<i>ramp input</i>mean value of .C œ $7</p><p>11. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! %] = œ / /</p><p>= =# w</p><p> = $ =a b a b a b a b 1 1 .Applying the initial conditions,</p><p>= ] = %] = œ / /</p><p>= =#</p><p> = $ =a b a b 1 1 .Solving for the transform,</p><p>] = œ / /</p><p>= = % = = %a b a b a b</p><p> = $ =</p><p># #</p><p>1 1</p><p>.</p><p>Using partial fractions,</p><p>' ' ' =</p><p>= = % % = = %œ Þa b ” •# #</p><p>Taking the inverse transform, and applying Theorem ,'Þ$Þ'</p><p>C > œ ' -9= #> # ? > ' -9= #> ' ? >' '</p><p>% %</p><p>œ ? > ? > -9= #> † ? > ? >' '</p><p>% %</p><p>a b c d a b c d a ba b a bc d c da b a b a b a b</p><p>1 11 1</p><p>1 1 1 1</p><p>$</p><p>$ $ .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 293</p><p>Since there is no damping term, the solution responds immediately to the forcing input.There is a temporary oscillation about C œ 'Î% Þ</p><p>12. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! = C ! = C ! C ! ] = œ Þ/ /</p><p>= =% $ # w ww www</p><p>= #=a b a b a b a b a b a bApplying the ,<i>initial conditions</i></p><p>= ] = ] = œ / /</p><p>= =%</p><p>= #=a b a b .Solving for the transform of the solution,</p><p>] = œ / /</p><p>= = ' = = 'a b a b a b</p><p>= #=</p><p>% %.</p><p>Using partial fractions,</p><p>' ' % ' ' #=</p><p>= = ' % = = ' = ' = 'œ Þa b ” •% #</p><p>It follows that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 294</p><p>_' > >%” •a b ‘' '= = ' %œ % / / # -9= > Þ</p><p>Based on Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ ? > ? > / / # -9= > ' ? > '</p><p>%</p><p> / / # -9= > # ? >'</p><p>%</p><p>a b c d a b a ba b a b ‘ ‘a b a b</p><p>' # ' >' >'</p><p> ># >##</p><p>a b a ba b a b .</p><p>The solution increases without bound, exponentially.</p><p>13. Taking the Laplace transform of the ODE, we obtain</p><p>= ] = = C ! = C ! = C ! C ! </p><p> & = ] = = C ! C ! %] = œ Þ' /</p><p>= =</p><p>% $ # w ww www</p><p># w =</p><p>a b a b a b a b a b ‘a b a b a b a b 1</p><p>Applying the ,<i>initial conditions</i></p><p>= ] = &= ] = %] = œ ' /</p><p>= =% #</p><p> =a b a b a b 1 .Solving for the transform of the solution,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 295</p><p>] = œ ' /</p><p>= = &= % = = &= %a b a b a b% # % #</p><p> =1</p><p>.</p><p>Using partial fractions,</p><p>' ' $ = %=</p><p>= = &= % '# = = % = 'œ Þa b ” •% # # #</p><p>It follows that</p><p>_'% #” •a b c d' '= = &= % '#œ $ -9= #> % -9= > Þ</p><p>Based on Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ ' ? > -9= #> % -9= > ' '</p><p>% '#</p><p> -9= # > % -9= > ? >'</p><p>'#</p><p>a b c d c da bc d a ba b a b</p><p>1</p><p>11 1 .</p><p>That is,</p><p>C > œ ' ? > -9= #> % -9= > ' '</p><p>% '#</p><p> -9= #> % -9= > ? >'</p><p>'#</p><p>a b c d c da bc d a b</p><p>1</p><p>1 .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 296</p><p>After an initial transient, the solution oscillates about .C œ !7</p><p>14. The specified function is defined by</p><p>0 > œ</p><p>!ß ! Ÿ > ></p><p>> > ß > Ÿ > > 5</p><p>2ß > > 5</p><p>a bÚÛÜ a b</p><p>!</p><p>! ! !</p><p>!</p><p>25</p><p>which can conveniently be expressed as</p><p>0 > œ > > ? > > > 5 ? > Þ2 2</p><p>5 5a b a b a b a b a b! !> > 5! !</p><p>15. The function is defined by</p><p>1 > œ</p><p>!ß ! Ÿ > ></p><p>> > ß > Ÿ > > 5</p><p> > > #5 ß > 5 Ÿ > > #5</p><p>!ß > > #5</p><p>a bÚÝÝÛÝÝÜa ba b</p><p>!</p><p>! ! !</p><p>! ! !</p><p>!</p><p>2525</p><p>which can also be written as</p><p>1 > œ > > ? > > > 5 ? > > > #5 ? > Þ2 #2 2</p><p>5 5 5a b a b a b a b a b a b a b! ! !> > 5 > #5! ! !</p><p>16 . From Part , the solution isa b a b. -? > œ %5 ? > 2 > %5 ? > 2 > </p><p>$ &</p><p># #a b a b a bŒ Œ $Î# &Î# ,</p><p>where</p><p>2 > œ / =38 / -9= Þ' ( $ ( > ' $ ( ></p><p>% )% ) % )a b È È È >Î) >Î)</p><p>Due to the , the solution will decay to . The maximum will occur<i>damping term zero</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 297</p><p>shortly after the forcing ceases. By plotting the various solutions, it appears that thesolution will reach a value of , as long as .C œ # 5 #Þ&'</p><p>a b/ Þ</p><p>Based on the graph, and numerical calculation, for .k ka b? > !Þ' > #&Þ'(($17. We consider the initial value problem</p><p>C %C œ > & ? > > & 5 ? >'</p><p>5ww c da b a b a b a b& &5 ,</p><p>with .C ! œ C ! œ !a b a bwa b+ . The specified function is defined by</p><p>0 > œ > & ß & Ÿ > & 5</p><p>!ß ! Ÿ > &</p><p>'ß > & 5</p><p>a b a bÚÛÜ'5</p><p>a b, Þ Taking the Laplace transform of both sides of the ODE, we obtain</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 298</p><p>= ] = = C ! C ! %] = œ / /</p><p>5= 5=# w</p><p>&= &5 =</p><p># #a b a b a b a b a b .</p><p>Applying the initial conditions,</p><p>= ] = %] = œ / /</p><p>5= 5=#</p><p>&= &5 =</p><p># #a b a b a b .</p><p>Solving for the transform,</p><p>] = œ / /</p><p>5= = % 5= = %a b a b a b</p><p>&= &5 =</p><p># # # #</p><p>a b.</p><p>Using partial fractions,</p><p>' ' ' '</p><p>= = % % = = %œ Þ</p><p># # # #a b ” •It follows that</p><p>_'# #” •a b' ' '= = % % )œ > =38 #> .</p><p>Using Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ 2 > & ? > 2 > & 5 ? >'</p><p>5a b c da b a b a b a b& &5 ,</p><p>in which .2 > œ > =38 #>a b ' '% )a b- > & 5. Note that for , the solution is given by</p><p>C > œ =38 #> '! =38 #> '! #5' ' '</p><p>% )5 )5</p><p>œ -9= #> '! 5 Þ' =38 5</p><p>% %5</p><p>a b a b a ba b</p><p>So for , the solution oscillates about , with an amplitude of> & 5 C œ 'Î%7</p><p>E œ=38 5</p><p>%5</p><p>k ka b.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 299</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 300</p><p>18 .a b+</p><p>a b, . The forcing function can be expressed as0 > œ ? > ? > Þ</p><p>'</p><p>#55 %5 %5a b c da b a b</p><p>Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! =] = C ! %] = œ ' / /</p><p>$ #5= #5=# w</p><p> %5 = %5 =a b a b a b c d a ba b a b a b a b .Applying the initial conditions,</p><p>= ] = =] = %] = œ ' / /</p><p>$ #5= #5=#</p><p> %5 = %5 =a b a b a b a b a b .Solving for the transform,</p><p>] = œ $ / $ /</p><p>#5= $= = '# #5= $= = '#a b a b a b</p><p> %5 = %5 =</p><p># #</p><p>a b a b.</p><p>Using partial fractions,</p><p>' ' ' ' $=</p><p>= $= = '# '# = $= = '#œ </p><p>œ Þ' ' '</p><p>'# = '</p><p>' ' = </p><p>= </p><p>a b ” •– —ˆ ‰ˆ ‰</p><p># #</p><p>'</p><p>' '%$' $'</p><p>#</p><p>Let</p><p>L = œ Þ' '</p><p>)5 = = = </p><p>= a b – —ˆ ‰ ˆ ‰' ' '</p><p>' '%$ ' '%$' $' ' $'</p><p># #</p><p>It follows that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 301</p><p>2 > œ L = œ =38 -9= Þ' / ' '%$ > '%$ ></p><p>)5 )5 ' '%$a b c da b – —È </p><p>È È_'</p><p>>Î'</p><p>Based on Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ 2 > % 5 ? > 2 > % 5 ? > Þa b a b a b a b a b%5 %5a b- .</p><p>As the parameter decreases, the solution remains for a longer period of time.5 <i>null</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 302</p><p>Since the of the impulsive force , the initial of the<i>magnitude increases overshoot</i>response also increases. The of the impulse decreases. All solutions eventually<i>duration</i>decay to C œ ! Þ</p><p>19 .a b+</p><p>a b a b- Þ , From Part ,? > œ ' -9= > # ' ' -9= > 5 ? > Þa b a b c d a b' a b</p><p>5 œ'</p><p>85 1 51</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 303</p><p>21 .a b+</p><p>a b, . Taking the Laplace transform of both sides of the ODE, we obtain= Y = =? ! ? ! Y = œ </p><p>' ' /</p><p>= =# w</p><p>5 œ'</p><p>8 5 5 =a b a b a b a b 'a b 1 .Applying the initial conditions,</p><p>= Y = Y = œ ' ' /</p><p>= =#</p><p>5 œ'</p><p>8 5 5 =a b a b 'a b 1 .Solving for the transform,</p><p>Y = œ ' ' /</p><p>= = ' = = 'a b a b a b'a b# #</p><p>5 œ'</p><p>8 5 5 =1</p><p>.</p><p>Using partial fractions,</p><p>' ' =</p><p>= = ' = = 'œ a b# # .</p><p>Let</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 304</p><p>2 > œ œ ' -9= >'</p><p>= = 'a b ” •a b_' # .</p><p>Applying Theorem , term-by-term, the solution of the IVP is'Þ$Þ'</p><p>? > œ 2 > ' 2 > 5 ? >a b a b a b a b a b'5 œ'</p><p>85 1 51 .</p><p>Note that</p><p>2 > 5 œ ? > 5 -9= > 5</p><p>œ ? > ' -9= > Þ</p><p>a b a b a ba b a b</p><p>1 1 1!</p><p>515</p><p>Hence</p><p>? > œ ' -9= > ' ? > -9= > ? > Þa b a b a b a b a b' '5 œ' 5 œ'</p><p>8 85</p><p>5 51 1</p><p>a b- .</p><p>The ODE has no . Each interval of forcing adds to the energy of the<i>damping term</i>system.Hence the amplitude will increase. For , when . Therefore the8 œ '& 1 > œ ! > '&a b 1oscillation will eventually become , with an amplitude depending on the values of<i>steady</i>? '& ? '&a b a b1 1 and .wa b. 8. As increases, the interval of forcing also increases. Hence the amplitude of thetransient will increase with . Eventually, the forcing function will be . In fact,8 <i>constant</i>for values of ,<i>large </i>></p><p>1 > œ' ß 8! ß 8</p><p>a b œ even oddFurther, for ,> 81</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 305</p><p>? > œ ' -9= > 8 -9= > Þ' '</p><p>#a b a b8</p><p>Hence the steady state solution will oscillate about or , depending on , with an! ' 8amplitude of .E œ 8 '</p><p>In the limit, as , the forcing function will be a periodic function, with period .8p_ #1From Prob. , in Section ,#( 'Þ$</p><p>_c da b a b1 > œ '= ' /= .As increases, the duration and magnitude of the transient will increase without bound.8</p><p>22 . Taking the initial conditions into consideration, the transform of the ODE isa b+= Y = !Þ' =Y = Y = œ </p><p>' ' /</p><p>= =#</p><p>5 œ'</p><p>8 5 5 =a b a b a b 'a b 1 .Solving for the transform,</p><p>Y = œ ' ' /</p><p>= = !Þ'= ' = = !Þ'= 'a b a b a b' a b# #</p><p>5 œ'</p><p>8 5 5 =1</p><p>.</p><p>Using partial fractions,</p><p>' ' = !Þ'</p><p>= = !Þ'= ' = = !Þ'= 'œ Þa b# #</p><p>Since the denominator in the second term is irreducible, write</p><p>= !Þ' = !Þ!& !Þ!&</p><p>= !Þ'= 'œ Þ= !Þ!& Ð$**Î%!!Ñ# #a b</p><p>a bLet</p><p>2 > œ ' = !Þ!& !Þ!&</p><p>= = !Þ!& Ð$**Î%!!Ñ = !Þ!& Ð$**Î%!!Ñ</p><p>œ ' / -9= > =38 > Þ$** ' $**</p><p>#! #!$**</p><p>a b – —a ba b a b– — È ÈÈ</p><p>_'# #</p><p>>Î#!</p><p>Applying Theorem , term-by-term, the solution of the IVP is'Þ$Þ'</p><p>? > œ 2 > ' 2 > 5 ? >a b a b a b a b a b'5 œ'</p><p>85 1 51 .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 306</p><p>For values of , the solution approaches .<i>odd </i>8 C œ !</p><p>For values of , the solution approaches .<i>even </i>8 C œ '</p><p>a b È, œ $** Î#! ¸ '. The solution is a sum of , each of frequency .<i>damped sinusoids </i>=Each term has an 'initial' amplitude of approximately For any given , the solution' Þ 8contains such terms. Although the amplitude will with , the amplitude8 ' 8<i>increase</i>will also be bounded by .8 '</p><p>a b a b- 1 > œ =38 >. Suppose that the forcing function is replaced by . Based on themethodsin Chapter , the general solution of the differential equation is$</p><p>? > œ / - -9= > - =38 > ? > Þ$** $**</p><p>#! #!a b a b– — È È>Î#! :' #</p><p>Note that . Using the method of ,? > œ E -9= > F =38 >:a b <i>undetermined coefficients</i>E œ '! F œ ! Þ and Based on the initial conditions, the solution of the IVP is</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 307</p><p>? > œ '! / -9= > =38 > '! -9= > Þ$** ' $**</p><p>#! #!$**a b – — È ÈÈ>Î#!</p><p>Observe that both solutions have the same frequency, .= œ $**Î#! ¸ 'È</p><p>23 . Taking the initial conditions into consideration, the transform of the ODE isa b+= Y = Y = œ #</p><p>' ' /</p><p>= =#</p><p>5 œ'</p><p>8 5 '5Î% =a b a b 'a b a b .Solving for the transform,</p><p>Y = œ #' ' /</p><p>= = ' = = 'a b a b a b'a b# #</p><p>5 œ'</p><p>8 5 '5Î% =a b.</p><p>Using partial fractions,</p><p>' ' =</p><p>= = ' = = 'œ a b# # .</p><p>Let</p><p>2 > œ œ ' -9= >'</p><p>= = 'a b ” •a b_' # .</p><p>Applying Theorem , term-by-term, the solution of the IVP is'Þ$Þ'</p><p>? > œ 2 > # ' 2 > ? >'5</p><p>%a b a b a b a b' Œ </p><p>5 œ'</p><p>85</p><p>'5Î% .</p><p>That is,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 308</p><p>? > œ ' -9= > # ' ' -9= > ? >'5</p><p>%a b a b a b' ” •Œ </p><p>5 œ'</p><p>85</p><p>'5Î% .</p><p>a b, .</p><p>a b- )). Based on the plot, the ' ' appears to be . The ' ' appears to<i>slow period fast period</i>be about . These values correspond to a ' ' of and a ' œ !Þ!('%<i>slow frequency fast</i><i>frequency</i>' .=0 œ 'Þ!%(#</p><p>a b. œ ' Þ. The natural frequency of the system is The forcing function is initially=!periodic, with period . Hence the corresponding forcing frequency isX œ 'Î# œ &Þ&A œ 'Þ'%#% $Þ*. Using the results in Section , the ' ' is given by<i>slow frequency</i></p><p> =</p><p>= œ œ !Þ!('#</p><p>#</p><p>k k!and the ' ' is given by<i>fast frequency</i></p><p> =</p><p>0!</p><p>œ œ 'Þ!('#</p><p>#</p><p>k k.</p><p>Based on theses values, the ' ' is predicted as and the ' ' is<i>slow period fast period</i>))Þ#%(given as .&Þ)'&'</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 309</p><p><b>Section 6.5</b></p><p>2. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! %] = œ / /# w = # =a b a b a b a b 1 1 .Applying the initial conditions,</p><p>= ] = %] = œ / /# = # =a b a b 1 1 .Solving for the transform,</p><p>] = œ œ / / / /</p><p>= % = % = %a b = # = = # =</p><p># # #</p><p>1 1 1 1</p><p>.</p><p>Applying Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ =38 #> # ? > =38 #> % ? >' '</p><p># #</p><p>œ =38 #> ? > ? > Þ'</p><p>#</p><p>a b a b a b a b a ba bc da b a b</p><p>1 11 1</p><p>1 1</p><p>#</p><p>#</p><p>4. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! ] = œ #! /# w $=a b a b a b a b .Applying the initial conditions,</p><p>= ] = ] = = œ #! /# $=a b a b .Solving for the transform,</p><p>] = œ = #! /</p><p>= ' = 'a b</p><p># #</p><p>$=</p><p>.</p><p>Using a , and Theorem , the solution of the IVP is<i>table of transforms </i>'Þ$Þ'</p><p>C > œ -9=2 > #! =382 > $ ? >a b a b a b$ .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 310</p><p>6. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = %] = =Î# œ /# % =a b a b 1 .Solving for the transform,</p><p>] = œ =Î# /</p><p>= % = %a b</p><p># #</p><p>% =1</p><p>.</p><p>Using a , and Theorem , the solution of the IVP is<i>table of transforms </i>'Þ$Þ'</p><p>C > œ -9= #> =38 #> ) ? >' '</p><p># #</p><p>œ -9= #> =38 #> ? > Þ' '</p><p># #</p><p>a b a b a ba b a b</p><p>1 %</p><p>%</p><p>1</p><p>1</p><p>8. Taking the Laplace transform of both sides of the ODE, we obtain</p><p>= ] = = C ! C ! %] = œ # /# w Î% =a b a b a b a b a b1 .Applying the initial conditions,</p><p>= ] = %] = œ # /# Î% =a b a b a b1 .Solving for the transform,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 311</p><p>] = œ# /</p><p>= %a b Î% =</p><p>#</p><p>a b1.</p><p>Applying Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ =38 #> ? > œ -9= #> ? > Þ#</p><p>a b a b a b a bŠ ‹1 1 1Î% Î%</p><p>9. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = ] = œ $ / / /</p><p>= =# $ Î# =</p><p> Î# = # =a b a b a b a b1 11 .Solving for the transform,</p><p>] = œ / $ / /</p><p>= = ' = ' = = 'a b a b a b</p><p> Î# = $ Î# = # =</p><p># # #</p><p>a b a b1 1 1.</p><p>Using partial fractions,</p><p>' ' =</p><p>= = ' = = 'œ Þa b# #</p><p>Hence</p><p>] = œa b / = / $ / / = /= = ' = ' = = '</p><p> Þ Î# = Î# = $ Î# = # = # =</p><p># # #</p><p>a b a b a b1 1 1 1 1</p><p>Based on Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ ? > -9= > ? > $ =38 > ? > # #</p><p>$</p><p> ? > -9= > # ? ></p><p>a b a b a b a bŠ ‹ Œ a b a b a b1 1 1</p><p>1 1</p><p>Î# Î# $ Î#</p><p># #</p><p>1 1</p><p>1 .</p><p>That is,</p><p>C > œ ' =38 > ? > $ -9= > ? > ' -9= > ? >a b c d a b a b a b c d a ba b a b1 1 1Î# $ Î# # .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 312</p><p>10. Taking the transform of both sides of the ODE,</p><p>#= ] = =] = %] = œ / > =38 > .>'</p><p>œ / Þ'</p><p>#</p><p># =></p><p>!</p><p>_</p><p> Î' =</p><p>a b a b a b ( Š ‹$ 1a b1</p><p>Solving for the transform,</p><p>] = œ/</p><p># #= = %a b a b</p><p> Î' =</p><p>#</p><p>a b1.</p><p>First write</p><p>'</p><p># #= = %œ= a b ˆ ‰#</p><p>'%</p><p>' $'% '</p><p># .</p><p>It follows that</p><p>C > œ ] = œ / † =38 > ? > Þ' $'</p><p>$' % 'a b c d a ba b È</p><p>È Š ‹_ 1' > Î' Î%a b1 1Î'</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 313</p><p>11. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = #= ] = #] = œ /=</p><p>= '# Î# =</p><p>#a b a b a b a b1 .</p><p>Solving for the transform,</p><p>] = œ = /</p><p>= ' = #= # = #= #a b a ba b# # #</p><p> Î# =a b1.</p><p>Using partial fractions,</p><p>= ' = # = %</p><p>= ' = #= # & = ' = ' = #= #œ Þa ba b ” •# # # # #</p><p>We can also write</p><p>= % = ' $</p><p>= #= #œ Þ= ' '# #a ba b</p><p>Let</p><p>] = œ Þ=</p><p>= ' = #= #' # #a b a ba b</p><p>Then</p><p>_' >'c d c da b] = œ -9= > =38 > / -9= > $ =38 > Þ' # '& & &</p><p>Applying Theorem ,'Þ$Þ'</p><p>_1' ></p><p> Î# =</p><p>#– — Š ‹ a b/= #= # #œ / =38 > ? > Þa b ˆ ‰1 1</p><p>#1Î#</p><p>Hence the solution of the IVP is</p><p>C > œ -9= > =38 > / -9= > $ =38 > ' # '</p><p>& & &</p><p> / -9= > ? > Þ</p><p>a b c da b a b</p><p>></p><p> >ˆ ‰1#1Î#</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 314</p><p>12. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = ] = œ /% =a b a b .Solving for the transform,</p><p>] = œ/</p><p>= 'a b =</p><p>%.</p><p>Using partial fractions,</p><p>' ' ' '</p><p>= ' # = ' = 'œ Þ</p><p>% # #” •It follows that</p><p>_'%” •' ' '= ' # #œ =382 > =38 > Þ</p><p>Applying Theorem , the solution of the IVP is'Þ$Þ'</p><p>C > œ =382 > ' =38 > ' ? > Þ'</p><p>#a b c d a ba b a b '</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 315</p><p>14 . The Laplace transform of the ODE isa b+= ] = =] = ] = œ /</p><p>'</p><p>## =a b a b a b .</p><p>Solving for the transform of the solution,</p><p>] = œ/</p><p>= =Î# 'a b =</p><p>#.</p><p>First write</p><p>' '</p><p>= =Î# 'œ Þ= </p><p># ' '&% '</p><p>#ˆ ‰Taking the inverse transform and applying both ,<i>shifting theorems</i></p><p>C > œ / =38 > ' ? >% '&</p><p>'& %a b a b a bÈ</p><p>È >' Î%a b</p><p>' .</p><p>a b, > #. As shown on the graph, the maximum is attained at some . Note that for'> #,</p><p>C > œ / =38 > ' Þ% '&</p><p>'& %a b a bÈ</p><p>È >' Î%a b</p><p>Setting , we find that . The maximum value is calculated asC > œ ! > ¸ #Þ$'$wa b 'C #Þ$'$ ¸ !Þ('&$ Þa ba b- œ 'Î%. Setting , the transform of the solution is#</p><p>] = œ/</p><p>= =Î% 'a b =</p><p>#.</p><p>Following the same steps, it follows that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 316</p><p>C > œ / =38 > ' ? >) $ (</p><p>$ ( )a b a b a bÈ</p><p>È >' Î)a b</p><p>' .</p><p>Once again, the maximum is attained at some . Setting , we find that> # C > œ !' wa b> ¸ #Þ%&'* C > ¸ !Þ)$$& Þ' ', with a ba b. ! '. Now suppose that . Then the transform of the solution is#</p><p>] = œ/</p><p>= = 'a b =</p><p># #.</p><p>First write</p><p>' '</p><p>= = 'œ Þ= Î# ' Î%# # ## # #a b a b</p><p>It follows that</p><p>2 > œ œ / =38 ' Î% † > Þ' #</p><p>= = ' % a b ” • È Š ‹È_ ## #' >Î## # ##</p><p>Hence the solution is</p><p>C > œ 2 > ' ? >a b a b a b' .The solution is nonzero only if , in which case Setting > ' C > œ 2 > ' Þ C > œ !a b a b a bw,we obtain</p><p>>+8 ' Î% † > ' œ % '’ “È a b È# ##</p><p># # ,</p><p>that is,</p><p>>+8 ' Î% † > '</p><p>' Î%œ# ‘È a bÈ # # #</p><p>#</p><p>#.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 317</p><p>As , we obtain the equation . Hence . Setting#p! >+8 > ' œ _ > p' <i>formal </i>a b ' 1#> œ Î# 2 > p ! C p' Þ1 # in , and letting , we find that These conclusions agree witha b 'the case , for which it is easy to show that the solution is# œ !</p><p>C > œ =38 > ' ? >a b a b a b' .15 . See Prob. . It follows that the solution of the IVP isa b+ '%</p><p>C > œ / =38 > ' ? >%5 '&</p><p>'& %a b a b a bÈ</p><p>È >' Î%a b</p><p>' .</p><p>This function is a of the answer in Prob. . Hence the peak value occurs at<i>multiple </i>'% +a b> ¸ #Þ$'$ C #Þ$'$ ¸ !Þ('&$ 5 Þ' . The maximum value is calculated as We finda bthat the appropriate value of is .5 5 œ #Î!Þ('&$ ¸ #Þ)'!)'</p><p>a b a b, '% -. Based on Prob. , the solution isC > œ / =38 > ' ? ></p><p>) 5 $ (</p><p>$ ( )a b a b a bÈ</p><p>È >' Î)a b</p><p>' .</p><p>Since this function is a of the solution in Prob. , we have ,<i>multiple </i>'% - > ¸ #Þ%&'*a b 'with The solution attains a value of , for ,C > ¸ !Þ)$$& 5 Þ C œ # 5 œ #Î!Þ)$$&a b' 'that is, .5 ¸ #Þ$**&'</p><p>a b a b- '% . ! '. Similar to Prob. , for , the solution is#C > œ 2 > ' ? >a b a b a b' ,</p><p>in which</p><p>2 > œ / =38 ' Î% † > Þ# 5</p><p>% a b È Š ‹È# ## >Î# ##</p><p>It follows that . Setting in , and letting , we find that> 'p Î# > œ Î# 2 > p !' 1 1 #a bC p5 Þ C œ # 5' Requiring that the remains at , the limiting value of is<i>peak value</i>5 œ # œ !' . These conclusions agree with the case , for which it is easy to show#that the solution is</p><p>C > œ 5 =38 > ' ? >a b a b a b' .16 . Taking the initial conditions into consideration, the transformation of the ODE isa b+</p><p>= ] = ] = œ ' / /</p><p>#5 = =#</p><p> %5 = %5 =a b a b – —a b a b .Solving for the transform of the solution,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 318</p><p>] = œ Þ' / /</p><p>#5 = = ' = = 'a b – —a b a b</p><p> %5 = %5 =</p><p># #</p><p>a b a b</p><p>Using partial fractions,</p><p>' ' =</p><p>= = ' = = 'œ Þa b# #</p><p>Now let</p><p>2 > œ œ ' -9= > Þ'</p><p>= = 'a b ” •a b_' #</p><p>Applying Theorem , the solution is'Þ$Þ'</p><p>9a b c da b a b a b a b> ß 5 œ 2 > % 5 ? > 2 > % 5 ? > Þ'#5</p><p>%5 %5</p><p>That is,</p><p>9a b c da b a bc da b a b a b a b</p><p>> ß 5 œ ? > ? > '</p><p>#5</p><p> -9= > % 5 ? > -9= > % 5 ? > Þ'</p><p>#5</p><p>%5 %5</p><p>%5 %5</p><p>a b a b, > > % > ß 5 œ !. Consider various values of . For any fixed , , as long as9% 5 >Þ > % % 5 > If , then for ,</p><p>9a b c da b a b> ß 5 œ -9= > % 5 -9= > % 5 Þ'#5</p><p>It follows that</p><p>lim lim5Ä! 5Ä!9a b a b a b</p><p>a b> ß 5 œ </p><p>-9= > % 5 -9= > % 5</p><p>#5œ =38 > % Þ</p><p>Hence</p><p>lim5Ä!9a b a b a b> ß 5 œ =38 > % ? >% .</p><p>a b- . The Laplace transform of the differential equationC C œ > %ww $a b,</p><p>with , isC ! œ C ! œ !a b a bw= ] = ] = œ /# %=a b a b .</p><p>Solving for the transform of the solution,</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 319</p><p>] = œ Þ/</p><p>= 'a b %=</p><p>#</p><p>It follows that the solution is</p><p>9! %a b a b a b> œ =38 > % ? > .a b. .</p><p>18 . The transform of the ODE given the specified initial conditions isa b a b,= ] = ] = œ ' /# 5 =</p><p>5 œ'</p><p>#!5'a b a b a b' 1 .</p><p>Solving for the transform of the solution,</p><p>] = œ ' /'</p><p>= 'a b a b'</p><p>#5 œ'</p><p>#!5' 5 =1 .</p><p>Applying Theorem , term-by-term,'Þ$Þ'</p><p>C > œ ' =38 > 5 ? ></p><p>œ =38 > † ? > Þ</p><p>a b a b a b a b'a b a b'</p><p>5 œ'</p><p>#!5'</p><p>5 œ'</p><p>#!</p><p>1 5</p><p>5</p><p>1</p><p>1</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 320</p><p>a b- .</p><p>19 . Taking the initial conditions into consideration, the transform of the ODE isa b,= ] = ] = œ /# 5 Î# =</p><p>5 œ'</p><p>#!a b a b ' a b1 .Solving for the transform of the solution,</p><p>] = œ /'</p><p>= 'a b '</p><p>#5 œ'</p><p>#! 5 Î# =a b1 .</p><p>Applying Theorem , term-by-term,'Þ$Þ'</p><p>C > œ =38 > ? > Þ5</p><p>#a b a b' Œ </p><p>5 œ'</p><p>#! 15 Î#1</p><p>a b- .</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 321</p><p>20 . The transform of the ODE given the specified initial conditions isa b a b,= ] = ] = œ ' /# 5 Î# =</p><p>5 œ'</p><p>#!5a b a b a b' +1 a b1 .</p><p>Solving for the transform of the solution,</p><p>] = œ '/</p><p>= 'a b a b'</p><p>5 œ'</p><p>#!5</p><p> 5 Î# =</p><p>#+1</p><p>a b1.</p><p>Applying Theorem , term-by-term,'Þ$Þ'</p><p>C > œ ' =38 > ? > Þ5</p><p>#a b a b a b' Œ </p><p>5 œ'</p><p>#!5' 1</p><p>5 Î#1</p><p>a b- .</p><p>22 . Taking the initial conditions into consideration, the transform of the ODE isa b,= ] = ] = œ ' /# '5Î% =</p><p>5 œ'</p><p>%!5'a b a b a b' a b .</p><p>Solving for the transform of the solution,</p><p>] = œ '/</p><p>= 'a b a b'</p><p>5 œ'</p><p>%!5'</p><p> '5Î% =</p><p>#</p><p>a b.</p><p>Applying Theorem , term-by-term,'Þ$Þ'</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 322</p><p>C > œ ' =38 > ? > Þ'5</p><p>%a b a b a b' Œ </p><p>5 œ'</p><p>%!5'</p><p>'5Î%</p><p>a b- .</p><p>23 The transform of the ODE given the specified initial conditions isa b a b, Þ= ] = !Þ'= ] = ] = œ ' /# 5 =</p><p>5 œ'</p><p>#!5'a b a b a b a b' 1 .</p><p>Solving for the transform of the solution,</p><p>] = œ/</p><p>= !Þ'= 'a b '</p><p>5 œ'</p><p>#! 5 =</p><p>#</p><p>1</p><p>.</p><p>First write</p><p>' '</p><p>= !Þ'= 'œ= </p><p># ' $**#! %!!</p><p>#ˆ ‰ .It follows that</p><p>_' >Î#!#” • È </p><p>È' #! $**= !Þ'= ' #!</p><p>œ / =38 > Þ$**</p><p>Applying Theorem , term-by-term,'Þ$Þ'</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 323</p><p>C > œ ' 2 > 5 ? >a b a b a b a b'5 œ'</p><p>#!5' 1 51 ,</p><p>in which</p><p>2 > œ / =38 > Þ#! $**</p><p>$** #!a b È </p><p>È>Î#!</p><p>a b- .</p><p>24 Taking the initial conditions into consideration, the transform of the ODE isa b, Þ= ] = !Þ'= ] = ] = œ /# #5' =</p><p>5 œ'</p><p>'&a b a b a b ' a b1 .Solving for the transform of the solution,</p><p>] = œ/</p><p>= !Þ'= 'a b '</p><p>5 œ'</p><p>'& #5' =</p><p>#</p><p>a b1.</p><p>As shown in Prob. ,#$</p><p>_' >Î#!#” • È </p><p>È' #! $**= !Þ'= ' #!</p><p>œ / =38 > Þ$**</p><p>Applying Theorem , term-by-term,'Þ$Þ'</p><p>C > œ 2 > #5 ' ? >a b c d a b' a b5 œ'</p><p>'&</p><p>1 a b#5' 1 ,</p><p>in which</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 324</p><p>2 > œ / =38 > Þ#! $**</p><p>$** #!a b È </p><p>È>Î#!</p><p>a b- .</p><p>25 . A fundamental set of solutions is and .a b a b a b+ C > œ / -9= > C > œ / =38 >' #> >Based on Prob. , in Section , a particular solution is given by## $Þ(</p><p>C > œ 0 = .=C = C > C > C =</p><p>[ C ß C =:</p><p>' # ' #</p><p>' #</p><p>a b a b( a b a b a b a ba ba b!></p><p>.</p><p>In the given problem,</p><p>C > œ 0 = .=/ -9= = =38 > =38 = -9= ></p><p>/B: #=</p><p>œ / =38 > = 0 = .=</p><p>:a b a b( c da b a b a b a ba b( a b a b!</p><p>> =></p><p>!</p><p>> >=a b .</p><p>.</p><p>Given the specified initial conditions,</p><p>C > œ / =38 > = 0 = .=a b a b a b(!</p><p>> >=a b .</p><p>a b a b a b a b, Þ 0 > œ > > C > œ ! > Let . It is easy to see that if , . If ,$ 1 1 1( a b a b a b!</p><p>> >= >/ =38 > = = .= œ / =38 > a b a b$ 1 11 .</p><p>Setting , and letting , we find that . Hence> œ p! C œ !1 & & 1a b</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 325</p><p>C > œ / =38 > ? > Þa b a b a b >a b1 11a b- . The Laplace transform of the solution is</p><p>] = œ/</p><p>= #= #</p><p>œ Þ/</p><p>= ' '</p><p>a ba b</p><p> =</p><p>#</p><p> =</p><p>#</p><p>1</p><p>1</p><p>Hence the solutions agree.</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 326</p><p><b>Section 6.6</b></p><p>1 . The a b+ <i>convolution integral</i> is defined as0 ‡ 1 > œ 0 > 1 . Þa b a b a b(</p><p>!</p><p>></p><p>7 7 7</p><p>Consider the change of variable . It follows that? œ > 7</p><p>( (a b a b a b a ba b( a b a b</p><p>a b</p><p>! ></p><p>> !</p><p>!</p><p>></p><p>0 > 1 . œ 0 ? 1 > ? .?</p><p>œ 1 > ? 0 ? .?</p><p>œ 1 ‡ 0 > Þ</p><p>7 7 7</p><p>a b, . Based on the distributive property of the , the convolution is also<i>real numbers</i>distributive.</p><p>a b- . By definition,0 ‡ 1 ‡ 2 > œ 0 > 1 ‡ 2 .</p><p>œ 0 > 1 2 . .</p><p>œ 0 > 1 2 . . Þ</p><p>a ba b a bc d( a b( (a b a b a b” •( ( a b a b a b</p><p>!</p><p>></p><p>! !</p><p>></p><p>! !</p><p>></p><p>7 7 7</p><p>7 7 ( ( ( 7</p><p>7 7 ( ( ( 7</p><p>7</p><p>7</p><p>The region of integration, in the double integral is the area between the straight lines( ( 7 7œ ! œ œ > Þ, and Interchanging the order of integration,</p><p>( ( ( (a b a b a b a b a b a b( (” •a b a b a b</p><p>! ! !</p><p>> > ></p><p>!</p><p>> ></p><p>7</p><p>(</p><p>(</p><p>0 > 1 2 . . œ 0 > 1 2 . .</p><p>œ 0 > 1 . 2 . Þ</p><p>7 7 ( ( ( 7 7 7 ( ( 7 (</p><p>7 7 ( 7 ( (</p><p>Now let . Then7 ( œ ?</p><p>( (a b a b a b a ba b((> ></p><p>!</p><p>0 > 1 . œ 0 > ? 1 ? .?</p><p>œ 0 ‡ 1 > </p><p>7 7 ( 7 (</p><p>( .</p><p>Hence</p><p>( (a bc d c d a ba b a b! !</p><p>> ></p><p>0 > 1 ‡ 2 . œ 0 ‡ 1 > 2 . Þ7 7 7 7 7 7</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 327</p><p>2. Let . Then0 > œ /a b >0 ‡ ' > œ / † ' .</p><p>œ / / .</p><p>œ / '</p><p>a b ((!</p><p>>></p><p>> </p><p>!</p><p>></p><p>></p><p>7</p><p>7</p><p>7</p><p>7</p><p>.</p><p>3. It follows directly that</p><p>0 ‡ 0 > œ =38 > =38 .</p><p>œ -9= > # -9= > .'</p><p>#</p><p>œ =38 > > -9= >'</p><p>#</p><p>a b a b a b(( c da b a bc da b a b</p><p>!</p><p>></p><p>!</p><p>></p><p>7 7 7</p><p>7 7</p><p>.</p><p>The of the resulting function is <i>range </i>‘ Þ</p><p>5. We have and Based on Theorem ,_ _c d a b c d/ œ 'Î = ' =38 > œ 'Î Þ 'Þ'Þ'> a b= '#_ 7 7” •( a b</p><p>a ba b!</p><p>> ></p><p>#</p><p>#</p><p>/ =38 . œ †' '</p><p>= ' = '</p><p>œ Þ'</p><p>= ' = '</p><p>a b7</p><p>6. Let and . Then . Applying Theorem ,1 > œ > 2 > œ / 0 > œ 1 ‡ 2 > 'Þ'Þ'a b a b a b a b>_ 7 7 7” •( a b a b</p><p>a b!</p><p>></p><p>#</p><p>#</p><p>1 > 2 . œ †' '</p><p>= = '</p><p>œ Þ'</p><p>= = '</p><p>7. We have , in which and . The transform0 > œ 1 ‡ 2 > 1 > œ =38 > 2 > œ -9= >a b a b a b a bof the convolution integral is</p><p>_ 7 7 7” •( a b a ba b</p><p>!</p><p>></p><p># #</p><p># #</p><p>1 > 2 . œ †' =</p><p>= ' = '</p><p>œ Þ=</p><p>= '</p><p>9. It is easy to see that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 328</p><p>_ _' > '#” • ” •' ' = %œ / œ -9= #> and .</p><p>Applying Theorem ,'Þ'Þ'</p><p>_ 7 7' >#</p><p>!</p><p>>” •a ba b ( ' = % œ / -9= # . Þa b7</p><p>10. We first note that</p><p>_ _' > '# #– —a b ” •</p><p>' ' '</p><p>= 'œ > / œ =38 #></p><p>= % # and .</p><p>Based on the ,<i>convolution theorem</i></p><p>_ 7 7 7</p><p>7 7 7</p><p>' ># #</p><p>!</p><p>></p><p>!</p><p>></p><p>– —a b a b ( a b( a b</p><p>' '</p><p>= ' = %œ > / =38 # .#</p><p>œ / =38 #> # . Þ'</p><p>#</p><p>a b7</p><p>7</p><p>11. Let . Since , the inverse transform of1 > œ K = 'Î œ =38 >a b c d c da b_ _' ' a b= '#the product is</p><p>_ 7 7 7</p><p>7 7 7</p><p>'#</p><p>!</p><p>></p><p>!</p><p>></p><p>” •a b ( a b( a b a b</p><p>K =</p><p>= 'œ 1 > =38 .</p><p>œ =38 > 1 . Þ</p><p>12. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = ' ] = œ K =# #a b a b a b= .Solving for the transform of the solution,</p><p>] = œ ' K =</p><p>= = a b a b</p><p># # # #= =.</p><p>As shown in a related situation, Prob. ,'</p><p>_ = 7 7 7= =</p><p>'# #</p><p>!</p><p>>” •a b ( a b a bK = '= </p><p>œ =38 > 1 . Þ</p><p>Hence the solution of the IVP is</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 329</p><p>C > œ =38 > =38 > 1 . Þ' 'a b a b a b(= =</p><p>= = 7 7 7!</p><p>></p><p>14. The transform of the ODE given the specified initial conditions isa b%= ] = %= ] = '(] = œ K =# a b a b a b a b.</p><p>Solving for the transform of the solution,</p><p>] = œK =</p><p>%= %= '(a b a b</p><p>#.</p><p>First write</p><p>'</p><p>%= %= '(œ Þ= %</p><p>#</p><p>'%</p><p>'#</p><p>#ˆ ‰Based on the elementary properties of the Laplace transform,</p><p>_' >Î##” •' '%= %= '( )œ / =38 #> Þ</p><p>Applying the , the solution of the IVP is<i>convolution theorem</i></p><p>C > œ / =38 # > 1 . Þ'</p><p>)a b a b a b(</p><p>!</p><p>> > Î#a b7 7 7 7</p><p>16. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = #= $ % =] = # %] = œ K =# a b c d a b a ba b .Solving for the transform of the solution,</p><p>] = œ #= & K =</p><p>= # = #a b a b a b</p><p>a b# # .</p><p>We can write</p><p>#= & # '</p><p>= # = #œ Þ= #a b a b# #</p><p>It follows that</p><p>_ _' #> ' #>#” • – —a b</p><p># '</p><p>= #œ #/ œ > /</p><p>= # and .</p><p>Based on the , the solution of the IVP is<i>convolution theorem</i></p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 330</p><p>C > œ #/ > / > / 1 . Þa b a b a b(#> #> # >!</p><p>></p><p>7 7 7a b7</p><p>18. The transform of the ODE given the specified initial conditions isa b= ] = ] = œ K =% a b a b a b.</p><p>Solving for the transform of the solution,</p><p>] = œK =</p><p>= 'a b a b</p><p>%.</p><p>First write</p><p>' ' ' '</p><p>= ' # = ' = 'œ Þ</p><p>% # #” •It follows that</p><p>_'” • c d' '= ' #</p><p>œ =382 > =38 > Þ4</p><p>Based on the , the solution of the IVP is<i>convolution theorem</i></p><p>C > œ =382 > =38 > 1 . Þ'</p><p>#a b c d a b( a b a b</p><p>!</p><p>></p><p>7 7 7 7</p><p>19. Taking the initial conditions into consideration, the transform of the ODE is</p><p>= ] = = &= ] = &= %] = œ K =% $ #a b a b a b a b.Solving for the transform of the solution,</p><p>] = œ = &= K =</p><p>= ' = % = ' = %a b a ba b a ba ba b</p><p>$</p><p># # # #.</p><p>Using partial fractions, we find that</p><p>= &= ' %= =</p><p>= ' = % $ = ' = %œ </p><p>$</p><p># # # #a ba b ” •,and</p><p>' ' ' '</p><p>= ' = % $ = ' = %œ Þa ba b ” •# # # #</p><p>It follows that</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 331</p><p>_'#</p><p># #” •a ba ba b= = & % '= ' = % $ $œ -9= > -9= #>,and</p><p>_'# #” •a ba b' ' '= ' = % $ 'œ =38 > =38 #> Þ</p><p>Based on the , the solution of the IVP is<i>convolution theorem</i></p><p>C > œ -9= > -9= #> # =38 > =38 # > 1 . Þ% ' '</p><p>$ $ 'a b c d a b( a b a b</p><p>!</p><p>></p><p>7 7 7 7</p><p>21 . Let . Substitution into the results ina b a b a b+ > œ ? >9 ww <i>integral equation</i>? > > ? . œ =38 #> Þww ww</p><p>!</p><p>>a b a b a b( 0 0 0Integrating by parts,</p><p>( (a b a b a b a b a bºa b a b a b! !</p><p>> >ww w w</p><p>œ!</p><p>œ></p><p>w</p><p>> ? . œ > ? ? .</p><p>œ >? ! ? > ? ! Þ</p><p>0 0 0 0 0 0 00</p><p>0</p><p>Hence</p><p>? > ? > > ? ! ? ! œ =38 #> Þww wa b a b a b a ba b a b, ? >. Substituting the given for the function ,<i>initial conditions</i></p><p>? > ? > œ =38 #> Þwwa b a bHence the solution of the IVP is equivalent to solving the integral equation in Part .a b+a b a b c da b- = œ >. Taking the Laplace transform of the integral equation, with ,F _ 9</p><p>F Fa b a b= † = œ Þ' #= = %# #</p><p>Note that the was applied. Solving for the transform ,<i>convolution theorem </i>Fa b=Fa b a ba b= œ Þ# ' = %</p><p>#</p><p># #</p><p>Using partial fractions, we can write</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 332</p><p>#= # % '</p><p>= ' = % $ = % = 'œ Þ</p><p>#</p><p># # # #a ba b ” •Therefore the solution of the is<i>integral equation</i></p><p>9a b> œ =38 #> =38 >% #$ $</p><p>.</p><p>a b a b c da b. Y = œ ? >. Taking the Laplace transform of the ODE, with ,_= Y = Y = œ Þ</p><p>#</p><p>= %#</p><p>#a b a b</p><p>Solving for the transform of the solution,</p><p>Y = œ Þ#</p><p>= ' = %a b a ba b# #</p><p>Using partial fractions, we can write</p><p># ' # #</p><p>= ' = % $ = ' = %œ Þa ba b ” •# # # #</p><p>It follows that the solution of the IVP is</p><p>? > œ =38 > =38 #> Þ# '</p><p>$ $a b</p><p>We find that ? > œ =38 > =38 #> Þww # %$ $a b22 . First note thata b+</p><p>( a bÈ È a b!, 0 C '</p><p>, C.C œ ‡ 0 , Þ</p><p>C</p><p>Take the Laplace transformation of both sides of the equation. Using the <i>convolutiontheorem</i>, with ,J = œ 0 Ca b c da b_</p><p>X ' '</p><p>= Cœ J = † Þ</p><p>#1</p><p>! È a b – —È_It was shown in Prob. , Section , that#( - 'Þ'a b</p><p>_1– —È Ê'C =œ Þ</p><p>Hence</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 333</p><p>X '</p><p>= =œ J = †</p><p>#1</p><p>! È a b Ê1 ,with</p><p>J = œ † Þ#1 X</p><p>=a b Ê È1 !</p><p>Taking the inverse transform, we obtain</p><p>0 C œ ÞX #1</p><p>Ca b Ë!1</p><p>a b a b a b, 3 3@. Combining equations and ,#1 X .B</p><p>C .Cœ ' Þ!</p><p>#</p><p>#</p><p>#</p><p>1Œ </p><p>Solving for the derivative ,.BÎ.C</p><p>.B # C</p><p>.C CœË ! ,</p><p>in which ! 1œ 1X Î Þ!# #</p><p>a b a b- C œ # =38 Î# Þ. Consider the Using the chain rule,<i>change of variable </i>! )#.C .</p><p>.B .Bœ # =38 Î# -9= Î# †! ) )</p><p>)a b a band</p><p>.B ' .B</p><p>.C # =38 Î# -9= Î# .œ †! ) ) )a b a b .</p><p>It follows that</p><p>.B -9= Î#</p><p>. =38 Î#œ # =38 Î# -9= Î#</p><p>œ # -9= Î#</p><p>œ -9=</p><p>) )! ) )</p><p>)</p><p>! )</p><p>! ! )</p><p>a b a bË a ba ba b</p><p>#</p><p>#</p><p>#</p><p>.</p><p>Direct integration results in</p><p>B œ =38 Ga b) ! ) ! ) .Since the curve passes through the , we require Hence ,<i>origin </i>C ! œ B ! œ ! Þ G œ !a b a band . We also haveB œ =38a b) ! ) ! )</p></div></div><div><div><p><b>—————————————————————————— ——</b>CHAPTER 6. </p><p><b>________________________________________________________________________</b> page 334</p><p>C œ # =38 Î#</p><p>œ -9= Þ</p><p>a b a b) ! )! ! )</p><p>#</p></div></div></div></div></div></div></div></div><div><div><div><div></div></div></div></div></main><div><div><div><div></div></div></div></div><div><div><div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div></div></div></div><div><div><div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div></div></div></div></aside></div></div></div></div></div></div></div></div></body>
Apr 16, 2018 Livro Boyce e Diprima, Equacoes Diferenciais. EMBED (for wordpress.com hosted blogs and archive.org item tags).
Written from the perspective of the applied mathematician, the latest edition of this bestselling book focuses on the theory and practical applications of Differential Equations to engineering and the sciences.
Emphasis is placed on the methods of solution, analysis, and approximation. Use of technology, illustrations, and problem sets help readers develop an intuitive understanding of the material. Historical footnotes trace the development of the discipline and identify outstanding individual contributions. This book builds the foundation for anyone who needs to learn differential equations and then progress to more advanced studies.