Math 308 Week 10 Solutions

Transcription

1 Mah 8 Week Soluions Here are soluions o he even-numbered suggesed problems. The answers o he oddnumbered problems are in he back of our ebook, and he soluions are in he Soluion Manual, which ou can purchase from he campus booksore. NSS = g(); () =, () = Answer: Firs, we ake he Laplace ransform of boh sides, we ge: L{ } + 9L{} = G where L{g} = G. We use he proper of Laplace ransforms ha L{f } = sl{f} f(). s L{} s() () + 9L{} = G This becomes: (s + 9)L{} s = G We solve for L{}: L{} = G s s s + 9 Thus: { } { } G s = L + L s + 9 s + 9 { } s From he able of Laplace ransforms, we can see ha L = s + 9 cos(). We can use convoluion o ake he inverse Laplace ransform of he oher par: { } { } G L = L {G} L s + 9 s + 9 ( ) = g sin() Thus, we ge = = g( v) sin(v) dv g( v) sin(v) dv + cos() 4. + = g(); () =, () = Answer: Firs, we ake he Laplace ransform of boh sides, we ge: L{ } + L{} = G

2 where L{g} = G. We use he proper of Laplace ransforms ha L{f } = sl{f} f(). s L{} s() () + L{} = G This becomes: (s + )L{} = G We solve for L{}: L{} = G s + + s + Thus: { } { } G = L + L s + s + { } From he able of Laplace ransforms, we can see ha L = sin. We can s + use convoluion o ake he inverse Laplace ransform of he oher par: { } { } G L = L {G} L s + s + = g sin Thus, we ge = = g( v) sin(v) dv g( v) sin(v) dv + sin 6. Use he convoluion heorem o find he inverse laplace ransform of he given funcion: f(s) = (s + )(s + ) Answer: B he convoluion heorem: { } { } L = L (s + )(s + ) s + From he able of Laplace ransforms, we see he following: { } L = e s + { } L = e s + { } L s +

3 Thus: Thus, he answer is: { } L (s + )(s + ) = e e = = e +v e v dv e e v dv = [ e e v] = e e + e = e + e L {f} = e + e 8. Use he convoluion heorem o find he inverse laplace ransform of he given funcion: f() = (s + 4) Answer: B he convoluion heorem: { } { } L = L (s + 4) s + 4 { } L s + 4 From he able of Laplace ransforms, we see he following: { } L = s + 4 sin() Thus: { } L (s + 4) = = 4 ( ) sin() ( ) sin() sin( v) sin(v) dv = 8 cos() + 6 sin() Malab was used o evaluae he inegral. Thus, he answer is: L {f} = 8 cos() + 6 sin()

4 . Use he convoluion heorem o find he inverse laplace ransform of he given funcion: f() = s (s + ) Answer: B he convoluion heorem: { } { } L s = L s (s + ) { } L s + From he able of Laplace ransforms, we see he following: { } L = s Thus: { } L s (s + ) { } L = sin() s + = = ( ) sin ( v) sin v dv = cos + Malab was used o evaluae he inegral. Thus, he answer is: L {f} = cos + NSS.4. Verif ha he pair (), () is a soluion o he given ssem. Skech he rajecor of he given soluion in he phase plane. d d =, () = +, d d = ; () = + + Answer: We can verif ha i is a soluion b aking he derivaives and plugging ino he ssem of differenial equaions: () = + () = + + d d = d d = + 6 +

5 We have alread verified ha d/d =, we jus need o show ha d/d =. We can do ha b compuing : = ( + ) = This is equal o d/d, so we have verified ha () and () are a soluion o he ssem of differenial equaions. We can skech his soluion using he following Malab command: >> ezplo(³+³, ³^+*^+*³,[,]) The inerval [, ] was chosen so ha he resuling graph would have a smaller range. 7 = +, = Find he criical poin se for he following ssem: d d d d = = + +

6 Answer: The criical poins are he poins where d/d and d/d are boh zero. = + + = The soluions o hese wo equaions are = and = 6. Thus, here is one criical poin a he poin ( 6, ) 6. Find he criical poin se for he following ssem: d d d d = + = ( )( ) Answer: The criical poins are he poins where d/d and d/d are boh zero. + = ( )( ) = The firs equaion has soluions = and =. The second equaion has soluions = and =. Thus, he poin (, ) is a criical poin, and an poin of he form (, ) is a criical poin. Thus, he criical poins are (, ) and all poins on he horizonal line = 8. Deermine he inegral curves for he given ssem b solving he phase plane equaion. d d d d = = Answer: We have ha Thus: Thus: d d = ( ) d = ( ) d ( ) d ( ) d = This differenial equaion is eac, because M N = =

7 Thus: f(, ) = ( ) d = + g() Thus: So, g () =. Thus: g() = f = + g () d = + C Thus: f(, ) = + C Thus, he inegral curves for he ssem are: + C = 6. Find all criical poins for he given ssem. Then use a sofware package o skech he direcion field in he phase plane and form his describe he sabili of he criical poins. d d d d = + = 4 Answer: The criical poins occur when + = and 4 =. If we solve hese wo equaions (eiher b hand or wih Malab), we ge = and = as he onl soluion. Thus, he poin (, ) is he onl criical poin. We can use Malab o plo he phase plane and some soluion curves:

8 = + = 4 From his we see ha he criical poin (, ) is sable, because an poin nearb (, ) approaches he origin. 8. Find all criical poins for he given ssem. Then use a sofware package o skech he direcion field in he phase plane and form his describe he sabili of he criical poins. d d d d = (7 ) = ( ) Answer:The criical poins are he poins where d/d and d/d are boh zero. (7 ) = ( ) =

9 From he firs equaion we ge ha eiher = or + = 7. From he second equaion, we ge ha = or + =. Thus, if =, hen he second equaion gives us ha = or =. If + = 7, hen he second equaion gives us ha eiher = and hus =, or ha + =, and hus = and = (from solving + = 7 and + = ). Thus, he criical poins are (, ), (7, ), (, ), (, ) (I also works o jus solve he wo equaions using Malab.) We can plo he phase plane and some soluions: = (7 ) = ( ) From he phase plane, we can deermine he sabili of he criical poins: (a) The criical poin (, ) is unsable. (b) The criical poin (7, ) is sable. (c) The criical poin (, ) is sable. (d) The criical poin (, ) is unsable.

10 Malab 7 The planar auonomous ssem having he form = a + b = c + d where a, b, c, and d are arbirar real numbers, is called a linear ssem of firs order differenial equaions. Eercises - 6 each conain a soluion of some linear ssem. Use MATLAB o creae a plo of versus, versus, and a plo of versus in he phase plane. Use he subplo command, as in subplo(), plo(,), ais igh subplo(), plo(,), ais igh subplo(), plo(,), ais equal o produce a plo conaining all hree plos. Use he suggesed ime inerval.. = e + e = 4e e [., ] Answer: The imporan hing for hese problems (Eercises -6) is o be able o plo he graphs. The subplo command is nice for displaing he resuling graphs, bu is no so imporan (in paricular, here will no be an quesions on he es requiring he subplo command). If ou are curious abou he subplo and ais commands, here is some informaion (oherwise, feel free o skip his paragraph). The command subplo(mnp) divides he figure window ino an m b n grid. Then, i places he plo in spo p in ha m b n grid. So commands given in he insrucions, divide he figure window ino a b grid, hen place he plo of versus in he firs square, and he plo of versus in he second square. Then, he hird subplo command divides he figure ino a b grid, puing he figure in he second recangle. The ais igh command means ha ais includes all of he values. The ais equal command means ha he verical ick marks are he same disance apar as he horizonal ick marks. I works o use ezplo insead of plo. To plo versus, he command is >> ezplo(³-*ep(-*)+*ep(-*)³,[-., ]) To plo versus he command is >> ezplo(³4*ep(-*)-*ep(-*)³,[-., ]) To plo he parameric equaion ( versus ), he command is >> ezplo(³-*ep(-*)+*ep(-*)³, ³4*ep(-*)-*ep(-*)³, [-., ])

11 We can ener all of his along wih he subplo command: >> subplo() >> ezplo(³-*ep(-*)+*ep(-*)³,[-., ]) >> ais igh >> subplo() >> ezplo(³4*ep(-*)-*ep(-*)³,[-., ]) >> ais igh >> subplo() >> ezplo(³-*ep(-*)+*ep(-*)³, ³4*ep(-*)-*ep(-*)³,[-., ]) >> ais equal Here are he resuling graphs: 8 ep( )+ ep( ) 4 ep( ) ep( ) = ep( )+ ep( ), = 4 ep( ) ep( ) = 4e + e = e + e [,.] Answer: To plo versus, he command is >> ezplo(³-4*ep(*)+*ep(*)³,[-,.]) To plo versus he command is

12 >> ezplo(³-*ep(*)+*ep(*)³,[-,.]) To plo he parameric equaion ( versus ), he command is >> ezplo(³-4*ep(*)+*ep(*)³, ³-*ep(*)+*ep(*)³, [-,.]) Here are he resuling plos: 4 ep( )+ ep( ) ep( )+ ep( ) = 4 ep( )+ ep( ), = ep( )+ ep( ). = e e = e + e [.,.] Answer: To plo versus, he command is >> ezplo(³-*ep(-*)-ep(*)³,[-.,.]) To plo versus he command is >> ezplo(³ep(-*)+ep(*)³,[-.,.]) To plo he parameric equaion ( versus ), he command is >> ezplo(³-*ep(-*)-ep(*)³, ³ep(-*)+ep(*)³, [-.,.])

13 Here are he resuling plos: ep( ) ep( ) ep( )+ep( ) = ep( ) ep( ), = ep( )+ep( ) 4 4. = cos() sin() = sin() + cos() [ π, 4π] Answer: To plo versus, he command is >> ezplo(³cos(*)-*sin(*)³,[-*pi, 4*pi]) To plo versus he command is >> ezplo(³*sin(*)+cos(*)³,[-*pi, 4*pi]) To plo he parameric equaion ( versus ), he command is >> ezplo(³cos(*)-*sin(*)³, ³*sin(*)+cos(*)³, [-*pi, 4*pi]) Here are he resuling plos:

14 cos( ) sin( ) sin( )+cos( ) = cos( ) sin( ), = sin( )+cos( ) = e (cos + sin) = e ( 8 sin + cos ) [ π, π/] Answer: To plo versus, he command is >> ezplo(³ep()*(cos()+*sin())³,[-pi, pi/]) To plo versus he command is >> ezplo(³ep()*(-8*sin()+cos())³,[-pi, pi/]) To plo he parameric equaion ( versus ), he command is >> ezplo(³ep()*(cos()+*sin())³, ³ep()*(-8*sin()+cos())³, [-pi, pi/]) Here are he resuling plos:

15 ep() (cos()+ sin()) ep() ( 8 sin()+cos()) = ep() (cos()+ sin()), = ep() ( 8 sin()+cos()) = e (cos + sin ) = e (cos sin ) [ π, π] Answer: To plo versus, he command is >> ezplo(³ep(-)*(cos()+sin())³,[-pi, pi]) To plo versus he command is >> ezplo(³ep(-)*(cos()-sin())³,[-pi, pi]) To plo he parameric equaion ( versus ), he command is >> ezplo(³ep(-)*(cos()+sin())³, ³ep(-)*(cos()-sin())³, [-pi, pi]) Here are he resuling plos:

16 ep( ) (cos()+sin()) ep( ) (cos() sin()) = ep( ) (cos()+ sin()), = ep( ) (cos() sin()) 6 4 For Eercises 7-, selec Galler linear ssem in he PPLANE6 Seup window. Adjus he parameers o mach he indicaed linear ssem. Accep all oher defaul seings in he PPLANE6 Seup window. Selec Soluions Keboard inpu in he PPLANE6 Displa window o sar a soluion rajecor wih he given iniial condiion. Finall, selec Graph Boh and click our soluion rajecor o obain plos of and versus. For Eercise n compare our oupu o Eercise n 6 above. 7. = 4 = () =, () = Answer: The phase plane for his ssem wih he soluion hrough (, ) marked is:

17 = A + B = C + D A = 4 C = B = D = The plo of and versus for he marked soluion is: = A + B = C + D A = 4 C = B = D = and..... If ou look back a he soluion o Eercise, ou will see ha he soluion indicaed in he phase plane looks like he parameric plo in Eercise, and he plos of and versus look like he plos for and versus in Eercise (he are in fac he same, as he equaions in Eercise are a soluion o he ssem in Eercise 7 wih he given iniial condiions.)

18 8. = + = + 4 () =, () = Answer: The phase plane for his ssem wih he soluion hrough (, ) marked is: = A + B = C + D A = C = B = D = The plo of and versus for he marked soluion is: = A + B = C + D A = C = B = D = 4 and..... If ou look back a he soluion o Eercise, ou will see ha he soluion indicaed in he phase plane looks like he parameric plo in Eercise, and he plos of and versus look like he plos for and versus in Eercise (he are in fac he same, as he equaions in Eercise are a soluion o he ssem in Eercise 8 wih he given iniial condiions.)

19 9. = 7 = + 8 () =, () = Answer: The phase plane for his ssem wih he soluion hrough (, ) marked is: = A + B = C + D A = 7 C = B = D = The plo of and versus for he marked soluion is: = A + B = C + D A = 7 C = B = D = 8 and.. If ou look back a he soluion o Eercise, ou will see ha he soluion indicaed in he phase plane looks like he parameric plo in Eercise, and he plos of and versus look like he plos for and versus in Eercise (he are in fac he same, as he equaions in Eercise are a soluion o he ssem in Eercise 9 wih he given iniial condiions.)

20 . = 4 = + () =, () = Answer: The phase plane for his ssem wih he soluion hrough (, ) marked is: = A + B = C + D A = C = B = 4 D = The plo of and versus for he marked soluion is: = A + B = C + D A = C = B = 4 D = and If ou look back a he soluion o Eercise 4, ou will see ha he soluion indicaed in he phase plane looks like he parameric plo in Eercise 4, and he plos of and versus look like he plos for and versus in Eercise 4 (he are in fac he same, as he equaions in Eercise 4 are a soluion o he ssem in Eercise wih he given iniial condiions.)

21 . = 4 + = () =, () = Answer: The phase plane for his ssem wih he soluion hrough (, ) marked is: = A + B = C + D A = 4 C = B = D = The plo of and versus for he marked soluion is: and = A + B = C + D A = 4 C = B = D = 6 4 If ou look back a he soluion o Eercise, ou will see ha he soluion indicaed in he phase plane looks like he parameric plo in Eercise, and he plos of and versus look like he plos for and versus in Eercise (he are in fac he same, as he equaions in Eercise are a soluion o he ssem in Eercise wih he given iniial condiions.)

22 . = + = () =, () = Answer: The phase plane for his ssem wih he soluion hrough (, ) marked is: = A + B = C + D A = C = B = D = The plo of and versus for he marked soluion is: = A + B = C + D A = C = B = D = and 4 6 If ou look back a he soluion o Eercise 6, ou will see ha he soluion indicaed in he phase plane looks like he parameric plo in Eercise 6, and he plos of and versus look like he plos for and versus in Eercise 6 (he are in fac he same, as he equaions in Eercise 6 are a soluion o he ssem in Eercise wih he given iniial condiions.)

23 . In he predaor-pre ssem L P = L + LP = P LP L represens a lad bug populaion and P represens a pes ha lad bugs like o ea. Ener he ssem in he PPLANE6 Seup window, se he displa window so ha L and P, hen selec Arrows for he direcion field. (a) Use he Keboard inpu window o sar a soluion rajecor wih iniial condiion (.,.). Noe ha he lad bug-pes populaion is periodic. As he lad bug populaion grows, heir food suppl dwindles and he lad bug populaion begins o deca. Of course, his gives he pes populaion ime o flourish, and he resuling increase in he food suppl means ha he lad bug populaion begins o grow. Pre soon, he populaions come full ccle o heir iniial saring posiion. (b) Suppose ha he pes populaion is harmful o a farmer s crop and he decides o use a poison spra o reduce he pes populaion. Of course, his will also kill he lad bugs. Quesion: Is his a wise idea? Adjus he ssem as follows: L P = L + LP HL = P LP HP Noe ha his model assumes ha he growh rae of each populaion is reduced b a fied percenage of he populaion. Ener his ssem in he PPLANE6 Seup window, bu keep he original displa window seings. Creae and se a parameer H =.. Sar a soluion rajecor wih iniial condiions (.,.). (c) Repea par (b) wih H =.4,.6, and.8. Is his an effecive wa o conrol he pess? Wh? Describe wha happens o each populaion for each value of he parameer H.

24 Answer: (a) Here is he phase plane showing he soluion hrough he poin (.,.): L = L + L P P = P L P. P L We see ha he lad bug populaion (he horizonal ais) ranges from approimael. o.7 and he pes populaion (he verical ais) ranges from approimael. o.7 (presumabl hese numbers are in housands or millions).

25 (b) Here is he phase plane wih H =. and showing he soluion hrough he poin (.,.): L = L + L P H L P = P L P H P H =.. P L In his siuaion, he pes and lad bug populaions are again periodic. Here he lad bug populaion ranges from approimael.47 o., and he pes populaion ranges from approimael.7 o.7 (hese figures are ver approimae). We see ha he pes populaion has no decreased. In fac, a some imes he pes populaion has increased.

26 (c) Here is he phase plane wih H =.4: L = L + L P H L P = P L P H P H =.4. P L The pes and lad bug populaions are periodic. The lad bug populaion ranges from approimael. o.9, and he pes populaion ranges from approimael o.9. The pes populaion has acuall increased. The lad bug populaion has decreased.

27 Here is he phase plane wih H =.6: L = L + L P H L P = P L P H P H =.6. P L The pes and lad bug populaions are periodic. The lad bug populaion ranges from approimael.4 o.9, and he pes populaion ranges from approimael o.4. The pes populaion has acuall increased. The lad bug populaion has decreased.

28 Here is he phase plane wih H =.8: L = L + L P H L P = P L P H P H =.8. P L Malab The pes and lad bug populaions are periodic. The lad bug populaion ranges from approimael. o.9, and he pes populaion ranges from approimael.9 o.. We see ha he poison spra did no effecivel reduce he pes populaion, because i also killed lad bugs. Fewer lad bugs allowed he pes populaion o increase, since he lad bugs ea he pess.. Use he echnique of Eamples and o verif ha is a soluion o he ssem = e (cos sin ) = e ( sin cos ) = = + Furhermore, use he subs command o verif ha he soluion saisfies he iniial condiions () = and () =. Answer: Firs, we ell Malab ha = e (cos sin ) and = e ( sin cos ).

29 >> sms >> = ep(-)*(cos(*) - sin(*)) >> = ep(-)*(*sin(*) - cos(*)) Ne, we check ha and saisf he firs equaion equivalen o + + =. =, which is >> diff() + * + * >> simple(ans) Malab reurns, so and do saisf he firs equaion. Now, we check ha and saisf he second equaion = +, which is equivalen o =. >> diff() - * - * >> simple(ans) Malab reurns, so he given and are a soluion o he ssem of differenial equaions. Now, we check ha and saisf he iniial condiions () = and () =. To check ha () =, we ener: >> subs(,, ) Malab reurns, so he equaion does saisf () =. Noe ha he did no need o be in quoes because we had defined i o be a smbolic variable previousl (sms ). To check ha () =, we ener: >> subs(,, ) Malab reurns, so he equaion does saisf () =.. Use dsolve o solve he iniial value problem v = v = v + cos wih () = and v() =, over he ime inerval [, ]. Answer: We ask Malab o solve he ssem: >> sol = dsolve(³d=v³, ³Dv=-*v-*+cos(*)³, ³()=³, ³v()=-³, ³³) >> sol. >> sol.v We ge he following: = e sin + e cos + sin() cos() v = 4 e sin 7 e cos + cos() + sin()