Math 334 Test 1 KEY Spring 2010 Section: 001. Instructor: Scott Glasgow Dates: May 10 and 11.

Transcription

1 1 Mah 334 Tes 1 KEY Spring 21 Secion: 1 Insrucor: Sco Glasgow Daes: Ma 1 and 11. Do NOT wrie on his problem saemen bookle, excep for our indicaion of following he honor code jus below. No credi will be given for work wrien on his bookle. Raher wrie in a blue book. Also wrie our name, course, and secion number on he blue book. Honor Code: Afer I have learned of he conens of his exam b an means, I will no disclose o anone an of hese conens b an means unil afer he exam has closed: Signaure:

2 2 1) Skech he direcion field of he ODE ds 2S 2 d, (1.1) wih special aenion o nullclines, i.e. poins where he direcion field is horizonal. (4 p.s) Soluion: We make a vecor field wih x -axis ime, and -axis S, he slopes of each vecor a each poin S, given b 2S 2. Below is a Mahemaica rendiion generaed b he following command: (Division b he radical is necessar o keep all he vecors roughl he same lengh.) ) True or false, he following equaion is a effecivel an ordinar differenial equaion (ODE): Eiher wa, explain. x ( x, ) x (, ). (1.2) Hin: If ou can solve i b he mehods of his course, i is an ODE, raher han a PDE.

3 3 (3 p.s) Soluion: True: Despie he presence of a parial derivaive, (1.2) is effecivel an ODE. The variable is effecivel a parameer since no (parial) derivaives wih respec o i appear. The general soluion o i, achieved b inegraing facor or separaion, is x (, ) Cexp x, (1.3) where C is an arbirar funcion of an and all parameers, including, bu no he bona-fide variable (of inegraion) x. (3 p.s) 3) True or false, he equaion (1.2) is a nonlinear differenial equaion. Eiher wa, explain. Soluion: False: The equaion is linear because i can be expressed as F, ;, x x, where F xis (affine) linear in is firs 2 argumens. x, ;, 4) True or false, he following differenial equaion is linear: Eiher wa, explain. (3 p.s) 2 ( x, ) x (, ) x (, ) 2. (1.4) x x Soluion: False: The equaion is nonlinear because i can be expressed as,, ;, F,, ;, x is nonlinear in i firs 3 argumens. F x, where xx 3 p.s) x 5) Wha is he order of equaion (1.4)? Explain. xx x Soluion: The order is second because he highes order derivaive presen is second order. 6) Fill in he blank: Le he funcions f (, ) and blank be coninuous in some recangle,,. Then, in some inerval conaining he poin

4 4 h hconained in here is a unique soluion () of he iniial value problem f (, ), ( ). d (1.5) 3 p.s) Soluion: blank= f : f : f /. 7) Fill in he blank: If he funcions p and g are coninuous on an open inerval I : conaining he poin, hen here exiss a unique funcion () ha saisfies he differenial equaion p() g() d (1.6) for each in blank, and also saisfies he iniial condiion ( ), where is an arbirar prescribed iniial value. 3 p.s) Soluion: blank= I : :. (3 p.s) 8) Solve he I.V.P. d f (), ( ). Soluion: Inegraion of boh sides of he equaion gives hence s () () ( ) () s () s ds f () s ds f () s ds f() f( ), s s s (1.7) () f() f( ). (1.8) 9)

5 5 Solve he I.V.P. a b, ( ), d in erms of ( b/ a, a ) wihou using an inegraing facor. (14 p.s) Soluion: Separaion is abou he onl remaining recourse: a b a b / a ad ln b / a a C d b / a a a b/ a Ce b/ ace. (1.9) So hen a a ( ) b/ ace C b/ a e, (1.1) and he soluion o he I.V.P. is a b/ ace b/ a a a b/ a e e b/ a b/ a e e a e b a 1. a a (1.11) 1) Solve he I.V.P. d 1 1, (). 1 (1.12) (18 p.s) Soluion: We firs pu he equaion in he sandard form for calculaing an inegraing facor: Thus an inegraing facor is 1 d 1 1. (1.13) use of which giving 1 d 1 1 ln(1 ) ln(1 ) 1 e e e (1.14) 1

6 6 1 1 d 1 1 c. 1d d (1.15) The iniial daa specifies ha cc 1 (1.16) so ha he soluion o he I.V.P. is 1 1. (1.17) 11) Wha does he exisence and uniqueness heorem for linear equaions (EUTL) sa abou he soluions of (1.12) specificall? (8 p.s) Answer: The heorem assers ha a soluion persiss abou he iniial poin for as long as he coefficien funcions are coninuous. For (1.12) hen, i specificall saes ha a soluion will persis in he inerval,1, since,1 conains he iniial poin, and one of he coefficiens is disconinuous (indeed does no exis) a 1. 12) Explain wh he less experienced suden ma hink ha our formula obained for he soluion of (1.12) conradics he EUTL. (4 p.s) Answer: The less experienced suden ma noe ha he soluion obained acuall solves he IVP (a leas in a limiing sense assuming in paricular ha 1 " " 1 for all 1 ) over he inerval,,1. 13) Explain wh he more experienced suden sees no conradicion beween our formula and he EUTL. (6 p.s) Answer: The more experienced suden simpl noes ha,,1 : he heorem does no claim ha soluions will definiel no persis beond an inerval in which he coefficien funcions are coninuous. I simpl makes no claim here (bu cerainl

7 7 suggess ha one ough o be able o find an example I.V.P. for which he soluion persiss no longer han claimed). 14) Find he general soluion of he DE (12 p.s) 1 cos( x ) dx 1xcos( x ). Soluion: Noe ha he equaion is exac, x 1 cos( x ) 1sin( x ) 1xcos( x ) (1.18) and, so, perform he compaible inegraions ( x, ) 1 cos( x ) x ( x, ) 1xcos( x ) ( x, ) xxsin( x ) f( ) ( x, ) xsin( x ) g( x) f( ), g( x) x. (1.19) So hen we can ake and he O.D.E. is inegraed wih he expression 15) Find he general soluion of he nearl exac DE ( x, ) x xsin( x ), (1.2) x xsin( x ) C. (1.21) (2 p.s) x 4xdx 4x3x. Soluion: Noe ha he equaion is no exac, x 3x 2 2 4x 3 6x 2 4x 3 12x 2 12x 3 4x 3 3x 4, (1.22)

8 8 bu ha he following quoien is a pure funcion of : x x x x 6x8x x x 2. x x x x x x (1.23) Thus we can find an inegraing facor ha is a funcion of onl. As per formulae alrea developed (or ou can re-derive i) he inegraing facor ( ) saisfies he ODE d 2 d 2 2 ln 2ln. (1.24) Use of his facor gives he exac equaion So he following inegraions are compaible 3x 4x dx 4x3x (1.25) ( x, ) 3x 4x x ( x, ) 4x3x x x x f (, ) ( ) x x x g x (, ) ( ) f( ), g( x). (1.26) So hen we can ake ( x, ) x x, (1.27) and he O.D.E. is inegraed wih he expression x x C. (1.28) 16) A ank holds iniiall a soluion wih A (mass) unis of a conaminan and V (volume) unis of (solven/conaminan) soluion. Afer his iniial preparaion, soluion wih he consan conaminaion concenraion of C in (mass) unis per uni volume pours in o he ank a he consan rae of R in (volume) unis per uni ime. The well-mixed soluion pours ou of he ank a he consan rae of R ou (volume) unis per uni ime. Se up he IVP whose soluion gives A A (), he amoun of conaminan (in mass) wihin he ank a ime unis afer he iniial preparaion.

9 9 (1 p.s) Soluion: We have dvin dvou dv Rin, Rou Rin Rou d d d dain dvin daou dvou Cin CinRin ; Cou CouRou d d d d da dain daou CinRin CouRou. d d d (1.29) Bu for he well-sirred mixure A () Cou Cou (). (1.3) V () dv Using V() V, and Rin Rou consan, we ge d V V() V ( R R ). (1.31) Thus in ou A () A () Cou Cou (), V () V( R R ) in ou (1.32) and he required IVP is da() Rou CinRin A( ); A() A. d V ( Rin Rou ). (1.33) 17) The I.V.P. ds rs p, S() S, d wih r and p posiive consans, has a soluion S which depends on ime, he parameers r and p, and he iniial daa S, i.e. we can wrie ha S S(; r, p, S). Given a specific ime T such ha STr ( ;, ps, ), find he iniial daa S. Noe: evidenl we will ge ha

10 1 S S( T; r, p). So find his funcion S( T; r, p) of hree variables. (Recall ha ou have solved his annui or (in reverse) morgage problem as homework, excep ha here he hree variables T, r, and pwhere specified. The problem is acuall much easier o solve leaving hem indeerminae, so be happ.) (16 p.s) Soluion: Find he general soluion o he ODE b separaion or inegraing facor. I choose he laer: he form ds rs p, (1.34) d suggess he inegraing facor e r, giving r r ds r r de S r r p r e re S pe pe e S e C d d r p r S Ce. r The iniial daa hen demands ha (1.35) p p S C C S r r (1.36) and he soluion o he IVP is hen p p S S e r r r. (1.37) The final daa gives rt rt p p rt p pe p(1 e ) S e S. r r r r r (1.38)