Mah 14 Soluion Tes 1B Fall 017 Problem 1: A ank has a capaci for 500 gallons and conains 0 gallons of waer wih lbs of sal iniiall. A soluion conaining of 8 lbsgal of sal is pumped ino he ank a 10 galsmin. A well sirred mixure flows ou of he ank a he same rae. a) Se up he differenial equaion ha models his siuaion and hen find he soluion ha gives he amoun of sal in he ank a an ime using he inegraing facor. b) Find and explain he limiing value of our soluion. c) Suppose he rae a which he soluion flows ou changes o 8 galsmin. Deermine how long i will ake o fill he ank. Soluion: (18)ps The DE model for amoun of sal a an ime is (0) = lbs of sal 80 wih iniial value The inegraing facor u() = e 80e e d 80e d so e 400e C 400 Ce where C =-398 when (0) = and he soluion is 400 398 e e 1d e o give us ) (400 398 ) 400 0 400 b Limi Limi e lbs of sal of he long erm period. c) The volumn will now be V = 0 + se 500 = 0 + o ge ha = 100 min or 1hr and 40 min. so i will ake one hours and 40 min o fill he ank.
Problem : Consider he iniial value problem an( ), (0) 3 ( 1) Find he larges open recangular region on which he soluion for his iniial value problem is guaraneed and is unique. Soluion: (14ps) Since his is a non-linear DE we will consider i o be muli-variable equaion f(,) = an( ) Therefore f (, ), for (0) = 3 ( 1) f an( ) and he parial derivaive is ( 1) Noice ha for boh he funcion and he parial derivaive wih he iniial value (0) = -3, he region mus be and 0 Problem 3: Given he differenial equaion p g ( ) ( ) wih he paricular soluion 5 e, find p() and g(). Show all work and explain reasoning. Soluion: (1ps) Consider he given soluion and work backward in he process o solve a 1 s order linear DE o find he original DE for his soluion. e e 5 5 5 e 1 5 5 e e 1 5 5 5 e 10e 0e 0 10 0 5 e Now noice ha p() = 10 and g() = 0 for he DE +P() = g()
Problem 4: Given he following direcion field, find he possible auonomous differenial equaion associaed wih i and clearl jusif our answer b indicaing exacl wh or wh no each DE is valid or no valid. 1) ( 3)( 1) ) ( 3)( 1) 3) ( 3)( 1) Soluion: (14ps) Since his is an auonomous DE, all isoclines are horizonal. We are ineresed in he equilibrium soluions which are he isoclines wih zero slopes a ever poin. The equilibrium soluions in he Direcion field occur a = -1 and a = -3. When we invesigae he DE in 1) we see ha when = 0 he equilibrium soluions occur a = -1 and = 3. Therefore his is no he correc DE for his direcion field. When we invesigae he DE in ) we see ha he equilibrium are correc when = 0. Bu on furher invesigaion we noice he isocline a = 0 will be = (0 + 3)(0 +1) > 0 bu = 0 clearl has negaive slopes and does no fi his DE. When we invesigae he DE in 3) we see ha he equilibrium are correc when = 0. And on furher invesigaion we noice he isocline a = 0 will be = (-0-3)(0 +1) < 0. B esing a few oher isoclines such as = o ge = (--3)(+1) = - 15 < 0, he direcion field seems o be a correc fi. So 3) is he correc answer.
Problem 5: A bod of weigh 18 lbs falls downward wih no iniial veloci. The resisance consan due o air is given o be k = 8. a) Se up iniial value differenial equaion ha models his scenario bu do no solve. b) Is veloci posiive or negaive in he scenario? Wh? c) Is resisance posiive or negaive in he scenario? Wh? d) Suppose ha air resisance is now given o be R = S0 where S is speed. Express R in erms of veloci. Soluion: (15ps) Using he equaion mv mg kv, we have he following; Since weigh is given o be 18 lbs and we know ha mg = weigh, we have ha m(3) = 18 and mass m = 4. So 4v 18 8 v wih iniial value v(0) = 0 To pu in he proper form for a 1 s order linear DE, we have 4v 8v 18 b) Veloci of a dropped falling objec is consider negaive. c) The resisance force is consider o be opposie in sign o he veloci and is herefore posiive. d)-the resisance in he above formula is defined o be R = kv bu is now express as R = S0 where S is speed and R = v 0 where v is negaive so R = -v0
Problem 6: Suppose a viral disease is spreading hrough a large populaion a a rae of change modeled b he differenial equaion k where he iniial number of people infeced was equal o 100. In a week a oal of 500 people were infeced. Use he mehod separaion of variables o solve for he number people infeced a an ime. Soluion: (15ps) Given k, wih iniial value = 0, = 100 and = 1 (in weeks), = 500 we have ha 1 d kd 1 1 k C, and C = for = 0 and = 100 100 and k = 4500 for = 1, = 500 So 1 1 4 5 k C (4 500) ( 1100) 500 people a ime. Problem 7: Given he iniial value problem 0, (), 1) Use Euler s Mehod o esimae (4) wih sep value h = 1. ) Do ou hink his esimae would be close? Explain wh or wh no. Soluion: (1ps) I) Since () = - we have he slope a he poin (, -) will be = -3 and he equaion of he angen line is = -3 + 4. We will now evaluae he nex poin = 3 o ge ha (3) = -5. II) Our new poin is now ( 3, -5) wih new slope = 8-5. The equaion of he angen line will be = (-85) + -(15). Now evaluae he nex poin, = 4, o ge ha (4) = -335.