Exam 1 Solutions. 1 Question 1. February 10, Part (A) 1.2 Part (B) To find equilibrium solutions, set P (t) = C = dp

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1 Exam Soluions Februar 0, 05 Quesion. Par (A) To find equilibrium soluions, se P () = C = = 0. This implies: = P ( P ) P = P P P = P P = P ( + P ) = 0 The equilibrium soluion are hus P () = 0 and P () =.. Par (B) We expand and use he mehod of parial fracions. = P ( P ) P = P P P = P P = P ( + P ) Separaing variables we ge: P ( + P ) = where P ( + P ) = A P + B = = A( + P ) + BP + P Se P = 0 o have A + 0 = = A =. Se P = o have = B = B =. Thus: P ( + P ) = P P + = Inegraing we ge: log P log P + = + C Facoring and simplifing ields: = log P P + = + C = P () P () + = Ce P () î Ce ó = Ce = P () = Ce Ce = C e C

2 .3 Par (C) We plug in he iniial condiion and solve for C: P () = C e C = P (0) = C C = = C = = P () = e

3 APPM 360: Secion exam 7:00pm 8:30pm, Februar, 05. ON THE FRONT OF YOUR BLUEBOOK wrie: () our name, () our suden ID number, (3) reciaion secion (4) our insrucor s name, and (5) a grading able for FIVE quesions. Tex books, class noes, and elecronic devices are NOT permied. A one-page crib shee is allowed. Turn in a signed exam paper wih our BLUEBOOK. Problem : (30 poins) Consider he differenial equaion = P ( P ) P (a) (5 poins) Find all equilibrium soluions. (b) (0 poins) Find he general soluion P (). (c) (5 poins) Solve he IVP where P (0) =. Soluion: Par (A) We expand and use he mehod of parial fracions. = P ( P ) P = P P P = P P = P ( + P ) Separaing variables we ge: P ( + P ) = where P ( + P ) = A P + B = = A( + P ) + BP + P Se P = 0 o have A + 0 = = A =. Se P = o have = B = B =. Thus: P ( + P ) = P P + = Inegraing we ge: log P log P + = + C Facoring and simplifing ields: = log P P + = + C = P () P () + = Ce P () [ Ce ] = Ce = P () = Ce Ce = C e C Par (B) () In sandard form, he ODE is: = P () P () = f(p, ). The funcion f(p, ) is coninuous a and in a recangle around (, P ) = (0, ). Also, he parial derivaive P f(p, ) = P () is also coninuous a and in a recangle around (, P ) = (0, ). Hence, Picard s heorem guaranees he exisence of a unique soluion o he IVP. Par (C) We plug in he iniial condiion and solve for C: P () = C e = P (0) = C C C = = C = = P () = e Problem : (30 poins) Answer he following quesions. (a) (5 poins) Solve he IVP + ( ) =, () = e using he mehod of inegraing facors.

4 (b) (5 poins) If he iniial condiion is changed o ( 0 ) = 4 for which values of 0 does Picard s heorem guaranee a unique soluion? (c) (0 poins) Consider he ODE d + (sin ) = f() Using he wo sage Euler-Lagrange mehod (i.e., mehod of variaion of parameers) show ha he paricular soluion has he form where v() saisfies p () = v()e cos dv = f()e cos Soluion: Inegraing facors (a) Idenif normalized form of he ODE as ( ) + = and see ha p() = ( ). Find he inegraing facor ( ) µ() = exp p() = exp = e Mulipling he ODE b inegraing facor gives d ( e () ) = e Inegrae for general soluion () = ( Ce ) Solve IVP o find paricular soluion (b) Given Idenif rae funcion e = Ce C =. p () = ( e ). = ( ) f(, ) = ( ) ( ), f (, ) = Boh funcions are well-defined provide 0. Thus a uniques soluion exis 0. (c) Subsiue ansaz ino he ODE o find ( ) cos dv e sin ecos v + sin ( v()e cos ) = f() Then cos dv e = f() and answer follows immediael. Problem 3: (30 poins) Answer he following quesions TRUE or FALSE. You DO NOT need o jusif our answer. (a) All isoclines of an auonomous firs-order diffeq mus be horizonal lines.

5 (b) For he IVP =, (0) = he absolue value of he approximaion error in using Euler s mehod wih sep size h = afer one sep is /. (c) All firs-order auonomous diffeqs mus have a leas one equilibrium soluion. (d) Picard s Theorem guaranees ha he IVP = /3, (0) = 0 has he unique equilibrium soluion () = 0. (e) All equilibrium poins of d = (4 ) are sable. (f) The diffeq (4) = sec () + ln(e ) is linear. Soluion: (a) True. If = f(), hen solving = c can onl ield consan -values as soluions. (b) True. Noe he IVP has soluion () = / + = () = 3/. Wih 0 = 0, 0 = and f(, ) =, one applicaion of Euler s mehod gives = 0 + hf( 0, 0 ) = + (0) = so ha () = /. (c) False. = has none. (d) False. Here = f(, ) wih f(, ) = /3 = f (, ) = (/3) /3 which is no coninuous on an recangle conaining (0, 0). In fac, () = (/3) 3/ is anoher soluion. (e) False. = is unsable. (f) True. (4) indicaes a fourh derivaive, and noe ha ln(e ) =.

6 Problem 4: (30 poins) Alice is a graduae suden in chemisr. In her lab she finds a ank ha conains 0 gallons of chemical soluion, which she labels as soluion A. The chemical concenraion of soluion A is measured o be pounds per gallon. She also finds plen of chemical soluion wih chemical concenraion 4 pounds per gallon, which she labels as soluion B. To achieve he chemical concenraion ha she desires, she injecs soluion B ino soluion A a a flow rae of 3 gallons per hour and drains he well-mixed mixure ou a a flow rae of 3 gallons per hour. (a) (8 poins) Se up he iniial value problem for he amoun of he chemical in he mixure as a funcion of ime. (b) ( poins) Solve he iniial value problem. (c) (6 poins) Suppose ha insead of draining he mixure a 3 gallons per hour, Alice evaporaes he mixure a 3 gallons per hour. In his process, waer escapes from he mixure while he chemical remains. In his case, wha is he amoun of he chemical in he mixure as a funcion of ime? (d) (4 poins) To reach a chemical concenraion of 3 pounds per gallon, is i more ime efficien o evaporae he mixure or o drain he mixure? You do no need o jusif our answer. Soluion: (a) Le he chemical amoun as a funcion of ime be x(). Then he differenial equaion is The iniial condiion is x () = (rae in) (rae ou) = 4(lb/gal) 3(gal/hr) x (lb/gal) 3(gal/hr) 0 = 3x() 0 (lb/hr). () x(0) = (lb/gal) 0(gal) = 0(lb). (b) To solve he differenial equaion, we rewrie i in sandard form () x () + 3 x() =. 0 The homogeneous soluion is (3) x h () x h() = 0 x h () = ce 3 0. A paricular soluion is, b inspecion (4) x p () = 40. Therefore he general soluion is (5) x() = x h () + x p () = ce Using he iniial condiion, we have (6) c + 40 = 0 c = 0. Hence he soluion o he iniial value problem is (7) x() = 0e (c) In his case he concenraion of he ougoing flow is 0, so he iniial value problem becomes (8) x () =, x(0) = 0. The soluion is hen (9) x() = 0 +. (d) I is more ime efficien o evaporae he mixure. This is because he rae a which he chemical amoun increases is alwas greaer for evaporaion han drainage.

7 3 (A) 3 (B) (C) 3 (D) Problem 5: (30 poins) Consider he following firs order ordinar differenial equaions: () = sin(4), () = ( + )( ), (3) = ( ), (4) = ( ) +, (5) ( ) = +, (6) =, (a) (4 poins) Which of he above differenial equaions are linear? Which of he above differenial equaions are auonomous? (b) (6 poins) Mach direcion fields(a)-(d) wih is corresponding differenial equaions.(noe ha here are more equaions han direcion fields, so wo equaions have no corresponding direcion fields.) (c) (0 poins) Where appropriae, idenif he equilibria shown in EACH GRAPH and give heir sabili.

8 Soluion: (a) Linear: None. Auonomous: (), (3), (4), (5). (b)-(c) Graph (A): (3); equilibria: = 0: sable, = : unsable. Graph (B): (); equilibria: = : sable, = : unsable. Graph (C): (); equilibria: None. Graph (D): (5); equilibria: = : unsable.