2.2.2: Find general soluions for he equaion APPM 2360 Homework Soluions, Due June 10 Soluion: Finding he inegraing facor, dy + 2y = 3e µ) = e 2) = e 2 Muliplying he differenial equaion by he inegraing facor, Inegraing boh sides, e 2 y + 2y ) = 3e 3 d ) ye 2 = 3e 3 ye 2 = e 3 + C y) = e + Ce 2 2.2.8: Find general soluions for he equaion Soluion: Finding he inegraing facor, dy + 1 y = 1 2 µ) = e 1/ = e ln) = Muliplying he differenial equaion by he inegraing facor, Inegraing boh sides, y + y = 1 d ) 1 y = y = ln) + C y) = ln) + C 2.2.14: Find general soluions for he equaion 2 + 9 ) dy + y = 0 Soluion: Firs, have o divide hrough by y coefficien, dy + y 2 + 9 ) = 0
Finding he inegraing facor, µ) = e 2 + 9 ) = e 1/2 ln2 +9) = 2 + 9 ) 1/2 Muliplying he differenial equaion by he inegraing facor, Inegraing boh sides, y 2 + 9 ) 1/2 + y 2 + 9 ) 1/2 = 0 d y 2 + 9 ) ) 1/2 = 0 y 2 + 9 ) 1/2 = C y) = C 2 + 9 ) 1/2 2.2.20: Find he soluion of he following IVP, 1 + e ) dy + e y = 0, y0) = 1 Soluion: Dividing hrough by he y coefficien, Finding he inegraing facor, µ) = e e y + y ) = 0 1 + e e 1 + e ) = e lne +1) = e + 1 Muliplying he differenial equaion by he inegraing facor, Inegraing boh sides, Using he iniial value, e + 1)y + ye = 0 d ye + 1) ) = 0 ye + 1) = C y) = C e + 1 1e 0 + 1) = C C = 2 Hence, he paricular soluion is y) = 2 e + 1 2.2.28: Solve he differenial equaion by he inegraing facor mehod, seps 1 4 y + 3 2 y = 2
Soluion: Sep 1: Find he inegraing facor, µ) = e 3 2 = e 3 = e 3 Sep 2: Muliply he differenial equaion by he inegraing facor, Sep 3: Find he aniderivaive, e 3 y + e 3 3 2 y = 2 e 3 d ye 3) = 2 e 3 ye 3 = 2 e 3 = 1 3 e3 Sep 4: Solve for y, y) = 1 3 2.2.29: Solve he differenial equaion by he inegraing facor mehod, seps 1 4 y + 1 y = 1 2 Soluion: Sep 1: Finding he inegraing facor, µ) = e 1/ = e ln) = Sep 2: Muliplying he differenial equaion by he inegraing facor, Sep 3: Inegraing boh sides, y + y = 1 d ) 1 y = y = ln) + C Sep 4: Solving for y, y) = ln) + C 2.3.2: Doubling Time The ime d required for he soluion y of he growh problem y = ky, k > 0, y0) = y 0 o reach wice is original value is called he doubling ime. Find d in erms of k.
Soluion: Solving he differenial equaion, dy y = Using he iniial value, k lny) = k + C y) = Ce k y 0 = Ce 0 y) = y 0 e k We wan o find he soluion ha is wice he original value, or 2y 0. Solving for d using his informaion, 2y 0 = y 0 e k d k d = ln2) d = ln2) k 2.3.5: Deermining Decay from Half-Life A cerain radioacive subsance has a half-life of 5 hours. Find he ime for a given amoun o decay o one-enh of is original mass. Soluion: We know he soluion o he decay equaion, y) = y 0 e k Given ha he half-life is 5 hours, hen we can solve for y 0, y5) = y 0 e 5k = 1 2 y 0 k = ln2) 5 Finding ha i akes for he amoun o decay o 1/10 of is original mass, y 0 10 = y 0e ln2)/5 ln2)/5 = ln10) = 5 ln10) ln2) 16.6 hours 2.3.15: Sodium Penahol Eliminaion Ed is undergoing surgery for an old fooball injury and mus be anesheized. The aneshesiologis knows Ed will be under when he concenraion of sodium penahol in his blood is a leas 50 milligrams per kilogram of body weigh. Suppose ha Ed weighs 100 kg 220 pounds) and ha sodium penahol is eliminaed from he bloodsream a a rae proporional o he amoun presen. If he half-life of he drug is 10 hours, wha single dose should be given o keep Ed aneheized for hree hours? Soluion: From he soluion o he decay equaion, hen solving for k, y) = y 0 e k y 0 2 = y 0e k 10 k = ln2) 10
Since Ed weighs 100 kg, hen we need 100 50 = 5000 miligrams of sodium penahol for Ed o remain asleep by hree hours. In oher words, Or abou 6.16 grams iniially. 5000 = y 0 e 3ln2)/10 y 0 = 5000e 3ln2)/10 6155.72 milligrams 2.3.29: How o Become a Millionaire Upon graduaion from college, Sergei has no money. However, during each year afer ha, he will deposi d = $1,000 ino an accoun ha pays ineres a a rae of 1% compounded coninuously. a) Find he fuure value A) of Sergei s accoun. Soluion: The accumulaed value is expressed by he following differenial equaion da = 0.01A + 1000, A0) = 0 Solving his yields A) = 1000 e 0.01 1 ) 0.01 b) Find he value for an annual deposi d ha would produce a balance of one million dollars when he reires 40 years laer. Soluion: Because our iniial deposi is 0, we wan o solve he following equaion, 1000000 = a e 0.01 40 1 ) 0.01 for a. Therefore, a = 10000 e 0.01 40 1 ) $4918.25 c) If d = $2,500, wha should be he value of he ineres rae r in order for Sergei s balance o be one million dollars afer 40 years? Soluion: The previous equaion changes o 1000000 = 2500 e r 40 1 ) r Then solving for r, using sofware o achieve an approximaion, we see ha r 9.04%. 2.4.4: Saly Goal A he sar, 5 lb of sal are dissolved in 20 gal of waer. Sal soluion wih concenraion 2 lb/gal is added a a rae of 3 gal/min, and he well-sirred mixure is drained ou a he same rae of flow. How long should his process coninue o raise he amoun of sal in he ank o 25 lb? Soluion: We have he following relaionship, dx = RATE IN RATE OUT = 2 lb/gal ) 3 gal/min ) ) x 20 lb/gal 3 gal/min ) = 6 3 20 x
Solving his differenial equaion, we can use he mehod of inegraing facor, µ = e 3/20 = e 3/20 Muliplying his hrough he original equaion, we obain Using he iniial value, x0) = 5, So d ) xe 3/20 = 6e 3/20 xe 3/20 = 40e 3/20 + C x) = 40 + Ce 3/20 5 = 40 + C = C = 35 x) = 40 35e 3/20 Since we wan o find he ime i akes unil here are 25 lbs of sal in he ank, hen we have o solve he following equaion 25 = 40 35e 3/20 ) = 20 3 ln 3 5.65 mins 7 2.4.6: Saly Overflow A 600-gallon ank is filled wih 300 gallons of pure waer. A spigo is opened and a sal soluion conaining 1 lb of sal per gallon of soluion begins flowing ino he ank a a rae of 3 gal/min. Simulaneously, a drain is opened a he boom of he ank allowing he soluion o leave he ank a a rae of 1 gal/min. Wha will be he sal conen in he ank a he precise momen ha he volume of soluion in he ank reaches he ank s capaciy of 600 gal? Soluion: We have he following relaionship, dx Using he mehod of inegraing facors, = RATE IN RATE OUT = 1 lb/gal ) 3 gal/min ) ) x 300 + 3 1) lb/gal 1 gal/min ) x = 3 300 + 2 µ) = e 1/300+2) = e 1/2 ln 300+2 = 300 + 2) 1/2 Muliplying he original differenial equaion, 300 + 2 ) 1/2x + 300 + 2 ) 1/2 x = 3300 + 2) 1/2 From he iniial value, we know ha x0) = 0, so d 300 + 2) 1/2 x ) = 3300 + 2) 1/2 300 + 2) 1/2 x = 300 + 2) 3/2 + C C300) 1/2 = 300) x) = 300 + 2) + C300 + 2) 1/2 0 = 300 + 2 0) + C300 + 2 0) 1/2 C = 300 3/2
We know ha 2 gal/min are begin added, and we need o fill 300 gals o mee he capaciy, so i will ake 150 minues o fill he ank. Therefore, x150) = 300 + 2 150) 300 3/2 300 + 2 150) 1/2 387.868 lbs 2.4.15: Using he Time Consan A noon, wih he emperaure in your house a 75 o F and he ouside emperaure a 95 o F, your air condiioner breaks down. Suppose ha he ime consan 1/k for your house is 4 hours. a) Wha will he emperaure in your house be a 2:00 pm? Solving, Soluion: Our differenial equaion becomes Using he iniial condiion, T 0) = 75, hen Then, a = 2, dt = 1 4 95 T ) dt 95 T = 4 ln 95 T = 4 + C T ) = 95 Ce /4 75 = 95 C = C = 20 T 2) = 95 20e 2/4 82.97 book s soluion is off, see second par b) When will he emperaure in your house reach 80 o F? Soluion: We need o solve he equaion, for. Therefore, 80 = 95 20e /4 e /4 = 15 20 ) 3 = 4ln 1.15 hours 4 Consider 0.15 60 9 minues, so we expec i o be 80 o a abou 1:09 pm. 2.4.19: The Coffee and Cream Problem John and Maria are having dinner, and each orders a cup of coffee. John cools his coffee wih some cream. They wai 10 minues and hen Maria cools her coffee wih he same amoun of cream. The wo hen began o drink. Who drinks he hoer coffee? Soluion: Define J as he ime ha John pours cream ino his coffee, and le M be he ime Maria pours her cream, so M = J + 10. A J, we know ha John s coffee is 10 o cooler han Maria s coffee. In beween J and M in he en minue inerval), John s coffee is cooling more slowly han Maria s coffee, and Maria s coffee is always hoer han John s. A M, we suspec ha here will be less han a 10 o F difference beween John and Maria s coffee emperaures. So when Maria adds cream o her coffee, i will cause her coffee emperaure o be lower han John s. Hence, John drinks he warmer coffee.