Math 308 Week 2 Solutions

Transcription

1 Mah 308 Week Soluions Here are soluions o he even-numbered suggesed problems. The answers o he oddnumbered problems are in he back of your exbook, and he soluions are in he Soluion Manual, which you can purchase from he campus booksore. Malab 3 8. Perform he following asks on he differenial equaion + )y + 4y =. a) Use dfield7 o plo a few soluions wih differen iniial poins. Sar wih he display window bounded by 0 0 and 5 y 5, and modify i o sui he problem. b) Make a conjecure abou he limiing behavior of he soluions as. c) Find he general analyic soluion o his equaion. d) Verify he conjecure you made in par b), or if you no longer believe i, make a new conjecure and verify ha. Answer: a) The following is a plo of he direcion field wih he specified bounds 0 0 and 5 y 5) and some soluions: 4 y)

2 The following is a plo wih display window 5 5, 5 y 5: 4 y) b) The soluions appear o be limiing o a value slighly above y = 0. The exac value, as we shall see in pars c) and d) is y = /4.) c) This is a separable differenial equaion: + )y + 4y = + )y = 4y) dy d = 4y + 4 ln 4y = ln + ) + C 4y = e ln+ )+C 4y = ±e C e ln/+ ) ) y = 4 C 3 + ) We have o be a lile careful in his calculaion. When we divided by 4y during he iniial separaion, we los he soluion y = /4. Dividing by +, however, did no lose any soluions, since + is always greaer han 0). The los soluion y = /4 comes back when we replace ±e C wih C 3. Thus, he general soluion is y) = 4 C + )

3 d) From he soluion, we see ha as, y) A differenial equaion of he form dx/d = fx), whose righ-side does no explicily depend on he independen variable, is called an auonomous differenial equaion. For example, he logisic model in Example 5 was auonomous. For he auonomous differenial equaion x = xx ), < x < perform each of he following asks. Noe ha he firs hree asks are o be performed wihou he aid of echnology. a) Se he righ-hand side of he differenial equaion equal o zero and solve for he equilibrium poins. b) Plo he graph of he righ-hand side of each auonomous differenial equaion versus x. Draw he phase line below he graph and indicae where x is increasing or decreasing. c) Use he informaion in pars a) and b) o draw sample soluions in he x plane. Be sure o include he equilibrium soluions. d) Check your resuls wih dfield7. soluions. Again, be sure o include he equilibrium e) If x 0 is an equilibrium poin, i.e., if fx 0 ) = 0, hen x) = x 0 is an equilibrium soluion. I can be shown ha if f x 0 ) < 0, hen every soluion curve ha has an iniial value near x 0 converges o x 0 as. In his case x 0 is called a sable equilibrium poin. If f x 0 ) > 0 hen every soluion curve ha has an iniial value near x 0 diverges away from x 0 as, and x 0 is called an unsable equilibrium poin. If f x 0 ) = 0, no conclusion can be drawn abou he behavior of soluion curves. In his case he equilibrium poin may fail o be eiher sable or unsable. Apply his es o each of he equilibrium poins. Answer: a) The equilibrium poins are x = 0 and x =. 3

4 b) Here is he graph of fx) = xx ) along wih he phase line for he differenial equaion. We see ha f is posiive for x < 0 and for x >, and ha f is negaive for 0 < x <. Thus, he soluions x) will be increasing for x < 0 and for x >, and ha x) will be decreasing for 0 < x <. B œ! source B œ # sink If we draw he phase line verically, as we did in class, we ge: B œ # sink B œ! source 4

5 c) Based on he phase line, we expec soluions o look somehing like he following: B œ # B œ! The equilibrium soluions are x = 0 and x =. Soluions beween x = 0 and x = end o 0 as and end o as. Soluions wih x > increase owards we can ell from he phase line if hese soluions have verical asympoes or no); as, hese soluions approach. Soluions wih x < 0 approach 0 as. d) Here is a plo of he direcion field and some soluions: 4 3 x) This maches our predicion. 5

6 e) In class, we used he erminology sinks and sources insead of sable and unsable. Sable soluions are sinks, and unsable soluions are sources. We see from he previous pars of his problem ha x = 0 is a sable soluion or sink), and x = is an unsable soluion or source). We can check if i saisfies he condiions described in e). Recall ha fx) = xx ). Thus, f x) = x. Thus, f 0) =, and f ) =. We see ha f 0) < 0, and hus x = 0 is a sable equilibrium. We see ha f ) > 0, and hus x = is an unsable equilibrium. This gives a differen mehod o deermine wheher an equilibrium soluion is a sink or source. Insead of deermining he sign of fx) beween each of he equilibrium poins, you can compue f x) a each of he equilibrium poins. However, if f x) = 0 a an equilibrium poin, he equilibrium could sill poenially be a sink or a source. Then, you will have o look a he sign of fx) below and above he equilibrium poin. This mehod works for he same reason ha he previous mehod worked. By finding f x), you are deermining wheher fx) is currenly increasing or decreasing a he given equilibrium. If fx) is increasing, hen he sign of fx) is changing from negaive o posiive remember ha f is 0 a he equilibrium poin), and he equilibrium is a source. If fx) is decreasing hen he sign of fx) is changing from posiive o negaive, and fx) is a sink. The logisic equaion P = rp P/K) was discussed in class. Usually he parameer s r and K are consans, and we found in class ha for any soluion P ) which has a posiive iniial value we have P ) K as. For his reason, K is called he carrying capaciy of he sysem. In he following exercises, you will look a cases where he carrying capaciy is no a consan. In Exercises 3-5, you are o examine he long erm behavior of soluions, especially in comparison o he carrying capaciy. In paricular: a) Use dfield7 o plo several soluions o he equaion. I is up o you o find a display window ha is appropriae o he problem a hand). b) Based on he plo done in par a), describe he long erm behavior of he soluions o he equaion. In paricular, compare his long erm behavior o ha of K. I migh be helpful o plo K on he display window. In Exercise 3 he soluions will all be asympoic o a consan. In Exercises 4 and 5 he soluions will all have he same long erm behavior. Describe ha behavior in comparison o he graph of K. 3. K) = e, r =. In his case K) is monoone increasing, and K) is asympoic o. This migh model a siuaion of a human populaion where, due o echnological improvemen, he availabiliy of resources is increasing wih ime, alhough ulimaely limied. 6

7 Answer: a) Here is a graph of he direcion field and some soluions: P) There is no consan soluion ha he soluions are approaching bu hey do appear o be approaching y =. b) Here is a graph of he funcion K): As, K). The soluions o he differenial equaion also approach as. 7

8 4. K) = +, r =. Again K) is monoone increasing, bu his ime i is unbounded. This migh model a siuaion of a human populaion where, due o echnological improvemen, he availabiliy of resources is seadily increasing wih ime, and herefore he effecs of compeiion are becoming less severe. Answer: a) Here is a graph of he direcion field and some soluions: 8 6 P) b) Here is a graph of he funcion K): I looks like he soluions o he differenial equaion approach he funcion K). Thus, he populaion ends o infiniy as. By he way, in Exercises 3 and 4, he soluions evenually approach K) from below even he ones ha begin above). Whenever y > K), he slope of he 8

9 soluion hrough he poin, y) will be negaive, so he soluion will decrease. Since K) is an increasing funcion in boh 3 and 4) and since he soluion evenually approaches K), he soluion will evenually need o have a posiive slope. Wha happens is ha he soluion passes hrough he funcion K). A he poin where he soluion crosses K), i has slope 0 and is a local minima. Afer ha poin, i has posiive slope and is increasing owards K). 5. K) = cosπ), r =. This is perhaps he mos ineresing case. Here he carrying capaciy is periodic in ime wih period, which should be considered o be one year. This models a populaion of insecs or small animals ha are affeced by he seasons. You will noice ha he long erm behavior as reflecs he behavior of K. The soluion does no end o a consan, bu neverheless all soluions have he same long erm behavior for large values of. In paricular, you should ake noice of he locaion of he maxima and minima of K and of P and how hey are relaed. You can use he zoom in opion o ge a beer picure of his. Answer: a) Here is a graph of he direcion field and some soluions: P)

10 b) Here is a graph of K): If you look closely a hese graphs, you will see ha he maxima and minima of K) are no he same as he maxima and minima of P ) noe ha he minima for K) occur when is an ineger, and ha is no he case for P )). I looks like he maxima and minima for P ) occur abou halfway beween he maxima and minima for K) wih each maxima for P ) occurring afer a maxima for K), and each minima for P ) occurring afer a minima for P ). Here is a graph showing he direcion field, one soluion, and he graph of K). The graph of K) is he curve wih he larger ampliude. 3.5 P)

11 If you hink abou wha his means for he populaion where K) is periodic due o seasons, his should make sense. The populaion does no begin decreasing immediaely afer he emperaures begin decreasing in he fall. Insead, he populaion coninues increasing, jus no as quickly as i was increasing before. A shor while laer, he populaion sars decreasing. NSS.4. In Example we approximaed he ranscendenal number e by using Euler s mehod o solve he iniial value problem y = y, y0) = Show ha he Euler approximaion y n obained by using he sep size /n is given by he formula y n = + n) n, n =,... Recall from calculus ha lim + n = e n n) and hence Euler s mehod converges heoreically) o he correc value. Answer: There s a possible confusion in his problem wih n meaning wo differen hings he nh sep in Euler s mehod and he sep size /n. To simplify hings, we will use k o denoe he curren sep in Euler s mehod. We wan o show ha he formula for he y value afer n seps of sep-size /n is y n = + n) n, n =,... Using sep size /n, he x values beween 0 and are 0, /n, /n, 3/n,..., n )/n,. To ge he nex y-value a each sep in Euler s mehod, we use he formula y k+ = y k + hfx k, y k ) In he formula, h = /n he sep size) and fx k, y k ) = y k since we are considering he differenial equaion y = y). Using his informaion, he formula becomes This is equivalen o y k+ = y k + y k n y k+ = y k + ) n This formula looks promising. Using i, Euler s mehod gives us he following able:

12 k x k y k y k+ = y k + ) n 0 0 y = + n /n + n y = + ) n /n 3 3/n + n) y 3 = + ) 3 n + n) 3 y 4 = + ) 4 n. n. n )/n.. + n) n y n = + n ) n NSS. Thus, we ge he formula we were rying o show: y n = + ) n n And, from calculus, we know ha lim n + n) n = e Thus, Euler s mehod converges o he correc value. By he way, he above limi is a relaively imporan limi, and is ofen used as he definiion of e. There s no a sandard place for he limi in he Calculus curriculum, hough, so you may no have seen i before.) 30. As saed in his secion, he separaion of equaion ) on page 40 requires division by py), and his may disguise he fac ha he roos of he equaion py) are acually consan soluions o he differenial equaion. a) To explore his furher, separae he equaion o derive he soluion dy dx = x 3)y + )/3 y = + x /6 x + C) 3

13 b) Show ha y saisfies he original equaion dy/dx = x 3)y + ) /3 c) Show ha here is no choice of he consan C ha will make he soluion in par a) yield he soluion y. Thus, we los he soluion y when we divided by y + ) /3. Answer: a) We separae: Inegraing, we ge Thus: dy y + ) = /3 x 3) dx 3y + ) /3 = x 3x + C y = + x /6 x + C) 3 b) The consan soluion y = is a soluion o he differenial equaion, because he derivaive of he consan funcion is 0, and if we plug y = ino he equaion x 3)y + ) /3 we ge x 3) + ) /3 = 0. c) For every value of C, he soluion y = + x /6 x + C) 3 is a cubic equaion and no a line. Thus, we los he consan soluion y = when we divided by y + ) /3. When you solve a separable differenial equaion, you should be careful abou his. Check during he separaion sep o see if you are dividing by a consan soluion. If you are, hen check your final answer o see if i incorporaes he consan soluion someimes i does). If i does no, hen you should make sure o include he consan soluion as par of your final answer. 38. Free Fall. In Secion., we discussed a model for an objec falling oward Earh. Assuming ha only air resisance and graviy are acing on he objec, we found ha he velociy v mus saisfy he equaion m dv = mg bv d where m is he mass, g is he acceleraion due o graviy, and b > 0 is a consan see Figure.). If m = 00 kg, g = 9.8 m/sec, b = 5 kg/sec, and v0) = 0m/sec, solve for v). Wha is he limiing i.e., erminal) velociy for he objec? Answer: We have he differenial equaion 00 dv = 980 5v d This equaion is separable: 00 dv = d 980 5v 0 ln 980 5v = + C 980 5v = e /0)+C 980 5v = Ae /0 3 v = 96 Ae /0

14 We have o be a lile careful in his calculaion. When we divided by 980 5v during he iniial separaion, we los he soluion v = 96. The los soluion v = 96 comes back when we replace ±e C wih A. The iniial condiion v0) = 0, ells us ha 0 = 96 Ae 0. Solving for A, we ge A = 86. Thus, he soluion is v) = 96 86e /0 From his soluion, we can see ha as, v) 96. velociy is 96 m/sec Thus, he erminal 4