Math 333 Problem Set #2 Solution 14 February 2003

Transcription

1 Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial condiion says y 2 4y = x3 3 + e3x 3 + C. so C = 10, and we have 3 3 = C, y 2 4y = x3 + e 3x To solve his for y, we complee he square. We ge so ha y 2 4y + 4 = x3 + e 3x x3 + e (y 2) = ± 3x The iniial condiion says ha we have o ake he negaive square roo, so he soluion is x3 + e y = 2 3x

2 A2. Consider he iniial value problem y + (co )y = 2 sin ; y(π/2) = y 0. (a) Use he mehod of inegraing facors o solve he iniial value problem. (Your soluion will depend on y 0.) Soluion: The inegraing facor here is e R co d, which urns ou o be simply sin. Muliplying hrough by sin, we ge Inegraing boh sides, we ge We solve for y, geing (sin )y + (cos )y = 2 sin 2. sin 2 (sin )y = + C 2 = cos sin + C. The iniial condiion says ha y = + C sin cos. y 0 = π 2 + C so ha C = y 0 π, and our soluion is 2 y = + y 0 π 2 sin cos. (b) Use he compuer o draw a direcion field for his differenial equaion wih 0 π and 3 y 3. Soluion: Here is he picure.

3 3 y() (c) As you can ell from he direcion field, for some values of y 0 (ha is, y(π/2)), soluions o his IVP go o + as approaches π, and for oher values of y 0, he soluions go o as approaches π. There s a criical value y c, such ha soluions wih y 0 < y c go o and soluions wih y 0 > y c go o +. Deermine he value of y c. Soluion: The soluions go o ± as π because he denominaor sin goes o zero here (from above). A soluion will go o + if he numeraor, + y 0 π 2, is posiive as π. This happens when π + y 0 π > 0 2 π 2 + y 0 > 0, ha is, when y 0 > π 2. Similarly, when y 0 < π 2, he expression + y 0 π 2 is negaive when π, so he soluion runs off o.

4 B1. (B & D, Secion 2.3, problem 14) When an organism dies, he amoun of carbon-14 i conains begins o decay a a rae proporional o he amoun presen. Afer abou 5730 years, only half of he organism s original carbon-14 is lef. Le Q() denoe he amoun of carbon-14 in some organism s remains. (a) Assuming ha Q saisfies he differenial equaion Q = rq, deermine he decay consan r for carbon-14. Give he value of r in is exac form (using logarihms) and hen give a decimal approximaion. Soluion: The soluion o he differenial equaion is The given condiion says ha Q() = Q 0 e r. We solve his for r, geing Q 0 2 = Q 0 e 5730r. r = ln (b) Find an expression for Q() a any ime, if Q(0) = Q 0. Soluion: We ve already done his. We have Q() = Q 0 e (ln(2)/5730). (c) Suppose ha cerain remains are discovered in which he curren residual amoun of carbon-14 is 20% of he original amoun. Deermine he age of hese remains. Soluion: The condiion given says ha Q 0 5 = Q 0 e (ln(2)/5730)

5 We cancel he Q 0 s and ake logarihms o ge ( ) ln 2 ln 5 = ln 5 = ln 2 13, years. B2. (B & D, Secion 2.3, problem 3) A ank originally conains 100 gal of fresh waer. Then waer conaining 1/2 lb of sal per gallon is poured ino he ank a a rae of 2 gal/min, and he mixure is allowed o leave a he same rae. Afer 10 min he process is sopped, and fresh waer is poured ino he ank a a rae of 2 gal/min, wih he mixure again leaving a he same rae. Find he amoun of sal in he ank a he end of an addiional 10 min. Soluion: Le Q() denoe he amoun of sal in he ank (in pounds) a ime (in minues). For he firs en minues of he experimen, Q() saisfies he differenial equaion Q () = Q() 100 wih he iniial condiion Q(0) = 0. inegraing facors. We wrie We can solve his differenial equaion using Q () + Q() = 1 and muliply hrough by e o ge and inegrae o ge e Q () + e Q() d ( ) e Q() d = e = e e Q() = e + C.

6 The iniial condiion implies ha C =. We divide hrough by e o ge Q() = e = ( 1 e Afer en minues, he amoun of sal in he ank is ). Q(10) = (1 e 1 5 ). For he nex en minues, he funcion Q() saisfies he differenial equaion Q () = 2 Q() 100 = Q(). This one we can solve by inspecion; he soluion is Q() = Q 0 e, where Q 0 is he amoun of sal in he ank a he beginning of he second en minues. (We re-se he clock afer he firs en minues.) We ge his amoun from above; i s (1 e 1 5 ). Thus afer anoher en minues, he amoun of sal in he ank is Q(10) = (1 e 1 5 )e 1 5 = (e 1 5 e 2 5 ) 7.42 pounds. Alernaively, we could sar he second differenial equaion running on he same clock as he firs, in which case we have Q() = Ce wih he iniial condiion Q(10) = (1 e 1 5 ). Using his iniial condiion o deermine C, we ge so ha C = (e 1 5 1). We have (1 e 1 5 ) = Ce 1 5 Q() = (e 1 5 1)e

7 where is he ime since he very beginning of he experimen. The answer o he quesion should now be Q(20) = (e 1 5 1)e 2 5 = (e 1 5 e 2 5 ) 7.42 pounds. C1. (Based on B & D, Secion 2.3, problem 10) Consider a morgage in which an iniial amoun B 0 is borrowed, ineres is charged a a rae r (compounded coninuously), and paymens are made a a rae k, assumed coninuous. Le B() denoe he amoun of money sill owed a ime. (a) Consruc a model (using appropriae unis) and solve he iniial value problem. Soluion: Taking r as a yearly percenage rae, k in dollars per year, B() in dollars, and in years, we have he iniial value problem B () = rb() k; B(0) = B 0. The differenial equaion is no difficul o solve. We wrie B () rb() = k and muliply hrough by he inegraing facor e r o ge so ha d d (e r B) = ke r e r B = k r e r + C 1 B = k r + C 1e r. The iniial condiion says ha B 0 = k r + C 1,

8 so we ge C 1 = B 0 (k/r), and he soluion o he IVP is B() = k ( r + B 0 k ) e r. r (b) (From problem 10) A home buyer can afford o spend no more han $800/monh on morgage paymens. Suppose ha he ineres rae is 9% and ha he erm of a morgage is 20 years. Wha is he maximum amoun he buyer can afford o borrow, and how much ineres will be paid on his amoun hrough he erm of he morgage? Soluion: The condiion here is ha B(20) = 0. Tha is, We solve his for B 0, geing 0 = k r + ( B 0 k r ) e 20r. B 0 = k r (1 e 20r ). The given values for k and r in he problem are k = 9600 dollars/year and r = Wih hese values, we ge B 0 = (1 e 1.8 ) $89, Over he erm of he morgage, he borrower pays $192,000. Since he principal is only $89,034.78, he borrower ends up paying $102, in ineres. C2. A ank conains 100 liers of fresh waer. A sal soluion flows ino he ank a he rae of 10 liers per minue, and he mixure in he ank flows ou a he same rae. The concenraion of sal in he incoming soluion is given by + 10 sin() grams per lier a ime (in minues). (a) Wrie an iniial value problem for Q(), he quaniy of sal in he ank a ime. Soluion: We have Q () = (rae in) (rae ou) = 10 ( + 10 sin()) 10 Q() 100 = sin Q().

9 The iniial condiion is Q(0) = 0. (b) Solve he IVP. You may use a compuer o do some of he nasy inegraion, bu don rely on dsolve() or anyhing like ha; ha would ake all he fun ou of i. Soluion: We wrie Q () + 1 Q() = sin 10 and use he inegraing facor e 10 o ge d d e 10 Q() = 0e e 10 sin. Using he compuer, we find ha he inegral of e 10 sin is e 10 (sin 10 cos ). Thus when we inegrae he equaion above, we ge Solving for Q(), we ge e 10 Q() = 00e e 10 (sin 10 cos ) + C. Q() = The iniial condiion says ha so C = Here, hen, is he complee soluion: (sin 10 cos ) + Ce 0 = ( 10) + C, 101 Q() = (sin 10 cos ) ( ) e 10.

10 (c) Use he compuer o draw a graph showing he concenraion of sal in he ank along wih he concenraion of sal in he incoming soluion. Take o a leas 60 minues. How would you describe he long-erm behavior of he concenraion of sal in he ank? Soluion: The concenraion of sal in he ank is Q()/100, so we plo ha funcion along wih he funcion + 10 sin(). Here is he picure Here are some hings o noice abou he picure: The concenraion of sal in he ank sars a zero and rises o a seady-sae oscillaion near grams per lier. The rae a which he concenraion rises varies direcly wih he concenraion of he incoming soluion. Afer abou = 40, he exponenial erm Ce /10 becomes negligible. The seady-sae soluion is a sinusoid wih he same period as he inpu funcion. The seady-sae soluion oscillaes around grams per lier, he same value around which he inpu concenraion oscillaes. The ampliude of he ank-concenraion oscillaion is smaller han he ampliude of he oscillaion of he inpu soluion. This makes sense; he volume

11 of waer in he ank cushions he effec of he inpu. The concenraion of sal in he ank increases whenever he sal concenraion in he incoming soluion is greaer han he concenraion in he ank, and decreases whenever he sal concenraion in he incoming soluion is less han he concenraion in he ank. The oscillaion of he sal concenraion in he ank lags almos 90 behind he concenraion sal in he incoming soluion.