u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x
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1 . 1 Mah 211 Homework #3 February 2, y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih u(x) = e a(x)dx 2/x = e dx = e 2ln x = x 2 = x 2. Muliply boh sides of our equaion by he inegraing facor and noe ha he lef-hand side of he resuling equaion is he derivaive of a produc. ( x 2 y + 2 ) x y = cos x Inegrae and solve for x. x 2 y + 2xy = cos x (x 2 y) = cos x x 2 y = sin x + C y = sin x + C x (1 + x)y + y = cos x Answer: Divide boh sides by 1 + x and solve for y. y = x y + cos x 1 + x Compare his resul wih y = a(x)y + f(x) and noe ha a(x) = 1/(1 + x). Consequenly, an inegraing facor is found wih u(x) = e a(x)dx = e 1/(1+x)dx = e ln 1+x = 1 + x. If 1 + x>0, hen 1 + x =1 + x. If1+ x<0, hen 1 + x = (1 + x). In eiher case, if we muliply boh sides of our equaion by eiher inegraing facor, we arrive a (1 + x)y + y = cos x. Check ha he lef-hand side of his resul is he derivaive of a produc, inegrae, and solve for y. ((1 + x)y) = cos x (1 + x)y = sin x + C y = sin x + C 1 + x
2 2 Chaper y = y + 2xe 2x, y(0) = 3 Answer: Compare y = y + 2xe 2x wih y = a(x)y + f(x)and noe ha a(x) = 1. Consequenly, an inegraing facor is found wih u(x) = e a(x)dx 1 = e dx = e x. Muliply boh sides of our equaion by he inegraing facor and noe ha he lef-hand side of he resuling equaion is he derivaive of a produc. e x y e x y = 2xe x ( e x y ) = 2xe x Inegraion by pars yields 2xe x dx = 2xe x 2e x = 2xe x 2e x + C. Consequenly, e x y = 2xe x 2e x + C y = 2xe 2x 2e 2x + ce x The iniial condiion provides 3 = y(0) = 2(0)e 2(0) 2e 2(0) + Ce 0 = 2 + C. Consequenly, C = 5 and y = 2xe 2x 2e 2x + 5e x (1 + 2 )y + 4y = (1 + 2 ) 2, y(1) = 0 Answer: Solve for y. y = y Compare his wih y = a()y + f() and noe ha a() = 4/(1 + 2 ). Consequenly, an inegraing facor is found wih u() = e a()dx 4/(1+ = e 2 )d = e 2ln 1+ 2 = (1 + 2 ) 2. Muliply boh sides of our equaion by he inegraing facor and noe ha he lef-hand side of he resuling equaion is he derivaive of a produc. (1 + 2 ) 2 y + 4(1 + 2 ) = ( (1 + 2 ) 2 y ) 1 = (1 + 2 ) 2 y = an 1 + C The iniial condiion y(1) = 0gives ( ) 2 (0) = an C. Consequenly, C = π/4 and y = an 1 π 4 (1 + 2 ) 2.
3 xy + 2y = sin x, y(π/2) = 0 Answer: Solve xy + 2y = sin x for y. y = 2 x y + sin x x Compare his wih y = a(x)y + f(x) and noe ha a(x) = 2/x and f(x) = (sin x)/x. I is imporan o noe ha neiher a nor f is coninuous a x = 0, a fac ha will heavily influence our inerval of exisence. An inegraing facor is found wih u(x) = e a(x)dx 2/x = e dx = e 2ln x = x 2 = x 2. Muliply boh sides of our equaion by he inegraing facor and noe ha he lef-hand side of he resuling equaion is he derivaive of a produc. x 2 y + 2xy = x sin x ( x 2 y ) = x sin x Inegraion by pars yields x sin xdx= xcos x + cos xdx= x cos x + sin x + C. Consequenly, x 2 y = x cos x + sin x + C, y = 1 x cos x + 1 x 2 sin x + C x 2. The iniial condiion provides 0 = y(π/2) = 4 π 2 + 4C π 2. Consequenly, C = 1 and y = (1/x) cos x + (1/x 2 ) sin x 1/x 2. We canno exend any inerval o include x = 0, as our soluion is undefined here. The iniial condiion y(π/2) = 0 forces he soluion hrough a poin wih x = π/2, a fac which causes us o selec (0, + ) as he inerval of exisence. The soluion curve is shown in Figure. Noe how i drops o negaive infiniy as x approaches zero from he righ.
4 4 Chaper. 5 (π/2,0) y x Soluion of xy + 2y = sin x, y(π/2) = A ank conains 100 gal of pure waer. A ime zero, sugar-waer soluion conaining 0.2 lb of sugar per gal eners he ank a a rae of 3 gal per minue. Simulaneously, a drain is opened a he boom of he ank allowing sugar-soluion o leave he ank a 3 gal per minue. Assume ha he soluion in he ank is kep perfecly mixed a all imes. Answer: Le S() denoe he amoun of sugar in he ank, measured in pounds. The rae in is 3g/m 0.2lb/g = 0.6lb/m. The rae ou is 3g/m S/100lb/g = 3S/100lb/m. Hence ds = rae in rae ou d = 0.6 2S/100 This linear equaion can be solved using he inegraing facor u() = e 3/100 o ge he general soluion S() = 20 + Ce 3/100. Since S(0) = 0, he consan C = 20 and he soluion is S() = 20(1 e 3/100. (a) S(20) = 10(1 e 0.6 ) (b) S() = 15 when e 3/100 = 1 15/20 = 1/4. Taking logarihms his ranslaes o = (100 ln 4)/ (c) As S() A ank iniially conains 100 gal of waer in which is dissolved 2 lb of sal. Sal-waer soluion conaining 1 lb of sal for every 4 gal of soluion eners he ank a a rae of 5 gal per minue. Soluion leaves he ank a he same rae, allowing for a consan soluion volume in he ank. (a) Use an analyic mehod o deermine he evenual sal conen in he ank. Answer: Le x() represen he number of pounds of sal in he ank a ime. The rae a which he sal in he ank is changing wih respec o ime is equal o he rae a which sal eners he ank minus he rae a which sal leaves he ank, i.e., dx = rae in rae ou. d
5 . 5 In order ha he unis mach in his equaion, dx/d, he rae In, and he rae Ou mus each be measured in pounds per minue (lb/min). Soluion eners he ank a 5 gal/min, bu he concenraion of his soluion is 1/4 lb/gal. Consequenly, rae in = 5 gal/min 1 4 lb/gal = 5 4 lb/min. Soluion leaves he ank a 5 gal/min, bu a wha concenraion? Assuming perfec mixing, he concenraion of sal in he soluion is found by dividing he amoun of sal by he volume of soluion, c() = x()/100. Consequenly, rae ou = 5 gal/min x() 100 lb/gal = 1 20 x()lb/min. As here are 2 lb of sal presen in he soluion iniially, x(0) = 2 and dx d = x, x(0) = Muliply by he inegraing facor, e (1/20), and inegrae. ( e (1/20) x ) 5 = 4 e(1/20) e (1/20) x = 25e (1/20) + C x = 25 + Ce (1/20) The iniial condiion x(0) = 2 gives C = 23 and x() = 25 23e (1/20). Thus, he concenraion a ime is given by c() = x() e (1/20) =, 100 and he evenual concenraion can be found by aking he limi as e (1/20) lim = lb/gal Noe ha his answer is quie reasonable as he concenraion of soluion enering he ank is also 1/4 lb/gal. (b) Use a numerical solver o deermine he evenual sal conen in he ank and compare your approximaion wih he analyical soluion found in par (a). Answer: We found i convenien o manipulae our original differenial equaion before using our solver. The key idea is simple: we wan o skech he concenraion c(), no he sal conen x(). However, c() = x() 100 or x() = 100c().
6 6 Chaper. Consequenly, x () = 100c (). Subsiuing hese ino our balance equaion gives x = x, 100c = (100c), c = c, wih c(0) = x(0)/100 = 2/100 =. The numerical soluion of his ODE is presened in Figure. Noe how he concenraion approaches 0.25 lb/gal. 0.3 c = 1/80 1/20 c 0.2 c Soluion of c = 1/80 (1/20)c, c(0) = A ank iniially conains 100 gal of a sal-waer soluion conaining 0.05 lb of sal for each gallon of waer. A ime zero, pure waer is poured ino he ank a a rae of 2 gal per minue. Simulaneously, a drain is opened a he boom of he ank ha allows sal-waer soluion o leave he ank a a rae of 3 gal per minue. Wha will be he sal conen in he ank when precisely 50 gal of sal soluion remain? Answer: The volume in he ank is decreasing a 1 g/m, so he volume is V() = 100. There is no sugar coming in, and he rae ou is 3g/m S()/V (). Hence he differenial equaion is ds d = 3S 100. This equaion is linear and homogenous. I can be solved by separaing variables. The general soluion is S() + A(100 ) 3. Since S(0) = = 5, we see ha A = , and he soluion is S() = (100 ) 3. When V()= 100 = 50g, S() = = 0.625lb.
7 Suppose ha a soluion conaining a drug eners an organ a he rae a cm 3 /s, wih drug concenraion κ g/cm 3. Soluion leaves he organ a a slower rae of b cm 3 /s. Furher, he faser rae of infusion causes he organ s volume o increase wih ime according o V()= V 0 + r, wih V 0 is iniial volume. If here is no iniial quaniy of he drug in he organ, show ha he concenraion of he drug in he organ is given by [ c() = aκ ( ) ] (b+r)/r V0 1. b + r V 0 + r Answer: Le x() represen he amoun of drug in he organ a ime. The rae a which he drug eners he organ is rae In = a cm 3 /s κ g/cm 3 = aκ g/s. The rae a which he drug leaves he organ equals he rae a which fluid leaves he organ, muliplied by he concenraion of he drug in he fluid a ha ime. Hence, Consequenly, rae ou = b cm 3 /s The inegraing facor is u() = e dx d b/(v 0 +r)d x() b V 0 + r g/cm3 = V 0 + r x()g/s. = aκ b V 0 + r x. Muliply by he inegraing facor and inegrae. ( (V0 + r) b/r x ) = aκ(v0 + r) b/r = e (b/r) ln(v 0+r) = (V 0 + r) b/r. (V 0 + r) b/r aκ x = r(b/r + 1) (V 0 + r) b/r+1 + L x = aκ b + r (V 0 + r) + L(V 0 + r) b/r No drug in he sysem iniially gives x(0) = 0 and L = aκv b/r+1 0 /(b + r). Consequenly, x = aκ b/r+1 b + r (V aκv0 0 + r) (V 0 + r) b/r, b + r x = aκ [ b + r (V 0 + r) 1 V b/r+1 0 (V 0 + r) b/r 1], x = aκ b + r (V 0 + r) [ 1 ( V0 ) ] b/r+1. V 0 + r The concenraion is found by dividing x() by V()= V + r. Consequenly, [ c() = aκ ( ) ] (b+r)/r V0 1. b + r V 0 + r
8 8 Chaper Is he iniial value problem y = 4 + y 2, y(0) = 1 guaraneed a unique soluion by he hypoheses of Theorem 7.12? Jusify your answer. Answer: The righ hand side of he equaion is f(,y)= 4+y 2. f is coninuous in he whole plane. Is parial derivaive f/ y = 2y is also coninous on he whole plane. Hence he hypoheses are saisfied and he heorem guaranees a unique soluion Is he iniial value problem y = y, y(4) = 0 guaraneed a unique soluion by he hypoheses of Theorem 7.12? Jusify your answer. Answer: The righ hand side of he equaion is f(,y) = y. f is defined only where y 0, and i is coninuous here. Our iniial condiion is a y 0 = 0, and 0 = 4. There is no recangle conaining ( 0,y 0 ) where f(,y) is defined and coninuous. Consequenly he hypoheses of he heorem are no saisfied y = 2y, y(0) = 2 (i) Find he general soluion of he differenial equaion. Skech several members of he family of soluions porrayed by he general soluion. Answer: The equaion is linear. The general soluion is y() = + 2C y (ii) Show ha here is no soluion saisfying he given iniial condiion. Explain why his lack of soluion does no conradic he exisence heorem. Answer: Since he general soluion is y() = + 2C 2, every soluion saisfies y(0) = 0. There is no soluion wih y(0) = 2. If we pu he equaion ino normal form dy = 2y, d we see ha he righ hand side f(,y) = 2y fails o be coninuous a = 0. Consequenly he hypoheses of he exisence heorem are no saisfied.
9 Suppose ha y is a soluion o he iniial value problem y = (y 2 1)e y and y(1) = 0. Show ha 1 <y()<1 for all for which y is defined. Answer: Noice ha he righ hand side of he equaion is f(,y)= (y 2 1)e y.f is coninuous on he whole plane. Is parial derivaive f/ y = 2ye y + (y 2 1)e y is also coninuous on he whole plane. Thus he hypoheses of he uniqueness heorem are saisfied. By direc subsiuion we discover ha y 1 () = 1 and y 2 () = 1 are boh soluions o he differenial equaion. If y is a soluion and saisfies y(1) = 0, hen y 1 (1) <y(1) <y 2 (1). By he uniqueness heorem we mus have y 1 ()<y()< y 2 () for all for which y is defined. Hence 1 <y()<1 for all for which y is defined. M2.8. Answer: We derive he model in erms of P(), he amoun of polluan in he lake, measured in km 3. The rae in is p km 3 /h. he rae ou is (r i + p) P/V. Hence he model is P = p (r i + p)p. V To pu his in erms of he concenraion we noice ha c = P/V. Since he volume is consan he derivaive is c = P /V. Hence c = 1 V P = p V (r i + p)p V 2 = p V (r i + p)c. V Hence we have c + (r i + p)c V = p V. M2.9. Answer: (a) From problem 8, one obains he differenial equaion c = p/v ((p + r i )/V )c. Subsiuing p = 2, r = 50, and V = 100 and simplifying gives c = 1/50 (26/50)c. So, wrie c = 1/50 - (26/50)*c in he dialog for dfield5, le run from 0 o 5 and c from - o 0.05, and obain Figure 1.
10 10 Chaper. c = (13/25) c + (1/50) c Figure 1. (b) Using he zoom box echnique, one can obain a picure much like ha given in Figure 2, showing ha when = , hen c =, which is 2%. c = (13/25) c + (1/50) c Figure 2. (c) The limiing value of c can be found eiher by explicily solving he differenial equaion or by looking a long-erm behaviour of he differenial equaion in dfield5. If one solves he differenial equaion, one should obain c() = 1/26 1/26e 13/25.
11 . 11 Taking he limi lim c(), he exponenial erm goes o zero, leaving 1/26. Hence he limiing value of c() is 1/26, or abou 3.85%. Using dfield5 and seing he parameer o go from 0 o 50, for insance, one can see ha he value of c approaches , as expeced. To find he approximae value of when c = ha is, 3.5% use he zoom box echnique o obain a picure like Figure 3. From his, one can see ha = c = (13/25) c + (1/50) c Figure 3. M2.10. Answer: This problem can be answered analyically. Since here is no inpu of poluan, we now have p = 0 and he model equaion changes o c = c = c 2. The soluion o his linear, homogenous equaion is c() = c 0 e /2. For c 0 =, we ge c() = when = 2ln(7/4) We ge he same answer (approximaely) using dfield5. See Figure 4.
12 12 Chaper. c = c/ c Figure 4. M2.11. Answer: (a) In he dfield5 seup dialog, ener c = 1/50 - ((2 + r)/100)*c and in one parameer box, ener r = *cos(2*pi*( - 1/3)). In Figure 5 presened below, he iniial condiions given were = 0,c = 0; = 0,c = 0.01; = 0,c = ; = 0,c = 0.03; = 0,c = As one can see, each soluion ends owards a single oscillaory paern.
13 c = 2/100 ((2 + r)/100) c r = cos(2 pi ( (1/3))). 13 c Figure 5. (b) In Figure 6 below, he varies from 10 o 11, represening he passage of ime over a single year. As one can see, he concenraion reaches a maximum a = 10.07, which is shorly before he beginning of February. c = 2/100 ((2 + r)/100) c r = cos(2 pi ( (1/3))) c Figure 6.
14 14 Chaper. M2.25. Answer: In Figure 7 below, he circle indicaes he poin = 4,x = 0, and he four curves are given by iniial condiions = 0,x = 0.728; = 0,x = ; = 0,x = (he one which his he arge); and = 0,x = x = x x Figure 7. M2.27. Answer: In Figure 8 below, he circle indicaes he poin = 5,x = 0, and he four curves are given by he iniial condiions = 0,x = 3.2; = 0,x = 3.232; = 0,x = (he one which hi he arge); and = 0,x = 3.3.
15 . 15 x = x sin(x) x Figure 8.
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