Homework 2 Solutions

Transcription

1 Mah 308 Differenial Equaions Fall 2002 & 2. See he las page. Hoework 2 Soluions 3a). Newon s secon law of oion says ha a = F, an we know a =, so we have = F. One par of he force is graviy, g. However, we are using he convenion ha up is posiive, an graviy creaes an acceleraion ownwar, so he graviaional force is really g. In par a), we are ol ha here is a fricional force F v) ha epens on v. We know ha his force shoul ac in he irecion opposie o he irecion of oion, bu we are ol ha F v) < 0 when v > 0, so ha is accoune for in F. So he oal force is F = g + F v). Thus Newon s secon law gives us = g + F v), or = g + F v). This is a firs orer ifferenial equaion for v, he velociy of he objec. If he objec is release fro heigh y 0 a = 0, he velociy a ha insan is zero. Therefore he iniial coniion is v0) = 0. 3b). In his case, F v) = 0, an he equaion is We can inegrae his o obain an he iniial coniion v0) = 0 gives us C = 0, so As increases, he velociy ecreases linearly wih. = g. v = g + C, v) = g. 3c). If F v) = v, he equaion for v is = g v. This equaion for v is separable, so we can following he usual seps o solve i: Separae assuing v g/ ): g + /)v =.

2 Inegrae an solve for v). I ll o his one in eail once ore in he following, expx) e x ): ln g + /)v = + k, ln g + /)v = + k 2 where k 2 = /)k ), g + /)v = exp C ) + k 2 = k 3 exp C ) where k 3 = e k2 ), Noe ha k 3 us be a posiive consan, because i coes fro exponeniaing k 2. Le s ge ri of he absolue value. We have wo cases: g + /)v > 0, or g + /)v < 0. If g + /)v > 0, hen g + /)v = g + /)v, an he las equaion becoes g + /)v = k 3 exp C ) If g + /)v < 0, hen g + /)v = g + /)v), so we have g + /)v = k 3 exp C ) The only ifference beween hese wo cases is he inus sign. Now, k 3 is a posiive consan, so we can cobine hese wo cases ino one forula g + /)v = k 4 exp C ) where k 4 is an arbirary nonzero consan. An hen we observe ha if k 4 = 0, we have g + /)v = 0, or v = g/, which is he equilibriu soluion. Thus we can say ha k 4 is an arbirary consan incluing he possibiliy ha k 4 = 0). Finally, we solve for v: v) = g + k exp C ), where k is an arbirary consan. k is jus /)k 4, an since k 4 is arbirary, so is k.) We wan he soluion where v0) = 0, so v0) = g + k = 0, hence k = g. Thus he soluion o he iniial value proble in his case is v) = g + g exp C ) = g exp C )). As increases, exp ) approaches zero, so he velociy approaches g. This is he erinal velociy.) 3). We now assue ha F v) = v v. Before proceeing, le s use a lile inuiion o ake life easier. If we rop he objec, we expec i o fall; he velociy will be zero iniially, an hen i will becoe negaive. We expec i o say negaive for > 0. Therefore, for he soluion ha we are consiering, we have v = v, an F v) = v 2. The ifferenial equaion is hen = g + C 2. 2

3 This equaion is separable, an we solve i he usual way. Tha is, we wrie g + C 2 =, an inegrae. While i is no necessary, I ll firs ake he coefficien of v 2 one: g + v 2 =, For convenience, I ll efine he paraeer a = g. We can o his because, g, an are all posiive. Noe ha his is no he sae a as he acceleraion in Newon s Law.) We hen have v 2 a 2 =, ) To inegrae he expression on he lef, we will use he parial fracion expansion v 2 a 2 = 2av + a) + 2av a). Then v 2 a 2 = 2av + a) + 2av a) = ln v + a + ln v a 2a 2a = ln v a ln v + a ) 2a = 2a ln v a v + a. Thus, he resul of inegraing boh sies of ) is 2a ln v a v + a = + K. This gives us or ) v a v + a = exp 2aC + 2aK ) v a v + a = K 2aC 3 exp. ) 2aC = K 2 exp, where, as usual, we absorb he ± ino he consan K 3. Now solve his for v: ) 2aC v a = K 3 exp v + a), )) ) )) 2aC K 3 exp 2aC v = a + K 3 exp 2aC a = a + K 3 exp v = a + K 3 exp 2a )) K 3 exp 2a ) 3

4 Recall ha a = g, so we have ) g + K 3 exp 2 v = ) K 3 exp 2 Now we fin he soluion for which v0) = 0: ) g + K3 v0) = = 0 = K 3 =. K 3 So he soluion o he iniial value proble is ) g exp 2 v) = ) + exp 2 As increases, he exponenials in his forula becoe large uch larger han ), an he expression in parenheses approaches. Thus, as increases, v) approaches erinal velociy. g. Again we see ha here is a 3e). In c), we have he equaion = g /)v, so he er /)v us have he sae unis as, which is an acceleraion, wih unis /sec2. v has unis of /sec, an has unis of kg, so us have unis of kg/sec for /)v o have unis of /sec 2. In ), he er /)v 2 us have unis of acceleraion, so in his case, he unis of us be kg/. 4. The logisic equaion is = k ). N The equilibria are ) = 0 an ) = N. Now assue 0 an N. We separae o obain: p ) = k. N To inegrae he lef sie, we use he parial fracion expansion: p ) = N N + ) p = ln N + ln = ln N. I haven inclue he consan of inegraion; his will be inclue in he consan on he righ sie.) So afer inegraing we have ln N = k + C. Now solve for : N = ek+c = C 2 e k, N = C 3e k 4

5 Noe ha C 2 is a posiive consan, while C 3 is an arbirary nonzero consan. As in previous probles, he ±C 2 ha arises fro eliinaing he absolue value is absorbe ino C 3. Wih a lile algebra, we fin solve for : ) = C 3Ne k. 2) + C 3 ek We observe ha seing C 3 = 0 gives us = 0, which is one of he equilibriu soluions, so in fac we can allow C 3 o be an arbirary consan, incluing C 3 = 0. However, here is no value of C 3 ha will give us he oher equilibriu soluion = N, so he general soluion is given by 2) or ) = N. A ifferen bu equivalen) for of he soluion is where C 4 is an arbirary consan. = N + C 4 e k or ) = 0, 5a). The aoun of rug in he bloosrea a ie in hours) is a) illigras. There are wo processes causing he aoun o change:. The rug clears a a rae ha is proporional o he aoun presen. We are ol ha he proporionaliy consan is 0.2 per hour, so he rae of change of a) fro he clearing of he rug is 0.2a). 2. The rug is being ae inravenously. The concenraion of he rug in he soluion is.4 g/l, an he flow rae is 0. l/hour, so he rae of change of a is.4 g/l)0. l/hour) = 0.4 g/hour. Thus he ne rae of change of a) is 0.2a This gives us he ifferenial equaion a = 0.2a We are ol ha here is iniially no rug in he bloosrea, so he iniial coniion is a0) = 0. The ifferenial equaion an he iniial coniion are he iniial value proble. 5b). The ifferenial equaion is separable; in fac, we solve a ore general equaion like his in class. We fin he general soluion o he ifferenial equaion o be a) = Ce 0.2. We use he iniial coniion o fin ha C = 0.7, so he soluion o he iniial value proble is 5c). a) = e 0.2. Afer one hour, he aoun of rug in he bloosrea is a) = e 0.2 = g. Afer one ay, a24) = g. As, e 0.2 0, so li a) =

6 .4/2. The following able an plo show he resul of applying Euler s Meho k k y k f k, y k ) y /5. See he soluion in he ex.).4/6. k k y k f k, y k ) y /0. Euler s Meho is no working very well in eiher case. For exaple, we know he equaion has an equilibriu a w = 3. If w0) > 3, we expec he soluion o approach his equilibriu as. Bu in Exercise 5, he soluion jups fro w = 4 o w = in one sep, an in Exercise 6, he nuerical soluion appears o be oscillaing aroun w = 3. Neiher of hese behaviors is reasonable for his equaion. Any soluion ha crosse w = 3 woul have o have a slope of zero here; neiher of hese approxiaions shows his. Looking ahea o Secion.5, we coul quoe he Uniqueness Theore, an poin ou ha i woul be ipossible for a soluion o cross w = 3.) In boh cases, he sep size is oo big. We nee o use a saller sep size o ge reasonable resuls. 6

7 .4/. Here is a plo of he righ-han sie of he ifferenial equaion in Exercise 6 versus w: w vs. w 4 2 w There are wo equilibriu soluions, w) = an w) = 3. For w < or w > 3, w) is ecreasing, while for < w < 3, w) is increasing. Any soluion w) for which < w0) < 3 will increase onoonically owar 3. If a soluion crosse w = 3 which, as we will see in Secion.5, is no possible for his equaion), i woul have o have a slope of zero here. The nuerical soluion copue in Exercise 6 oscillaes aroun he value w = 3, which our qualiaive analysis shows is ipossible behavior for a soluion o he ifferenial equaion. 7