Project #1 Math 285 Name:

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1 Projec #1 Mah 85 Name: Solving Orinary Differenial Equaions by Maple: Sep 1: Iniialize he program: wih(deools): wih(pdeools): Sep : Define an ODE: (There are several ways of efining equaions, we sar wih he basic one firs, jus basically give he ODE a name: Firs Orer DE: y 3 Example: y sin x x > wih(deools): > eq1:=iff(y(),)-(1/3)*y()=sin(); Higher Orer ODE: y y Example: m b ky 0 eq1 := y( ) y( ) = sin( ) > eq:=m*iff(y(),$)+b*iff(y(),)+k*y()=0; Example: eq := m y( ) + b 4 3 y y y 35 50y 4 5e y( ) + k y( ) = 0 > eq3:=iff(y(),$4)-10*iff(y(),$3)+35*iff(y(),$)-50*y()+4=5*exp(); 4 eq3 := 4 y( ) y( ) + 35 y( ) - 50 y( ) + 4 = 5 e Solving ODE: > solve(eq1,y()); y( ) = cos( ) sin( ) + e 3 _C1 > solve(eq,y()); - y( ) = _C1 e b - b - 4 k m m - + _C e b + b - 4 k m m Solving IVP Now, wih equaion #1, we wan o solve i wih iniial value say y 0 > solve({eq1,y(0)=pi},y()); y( ) = cos( ) sin( ) + e 3 p

2 For equaion #, we wan o solve wih iniial value y 010, y' 0 We can efine he iniial value as: > IV:=y(0)=10, D(y)(0)=; IV := y( 0) = 10, D( y) ( 0) = an for equaion #, for insance, m =1, b = -, k =0.1, we hen subsiue ino equaion # wih new equaion name: > eqm:=subs(m=1,b=-,k=0.1,eq); eqm := y( ) - y( ) y( ) = 0 Now, we can solve eqm wih he iniial value IV as follows: > solve({eqm,iv},y()); y( ) = e 1 10 ( ) e ( ) 1. Now, use he ouline above o solve he following ODE: y a) 1 e e y 0; y 0 1 y y c) 4 7y 0; y 0 0; y' 0 1 x e) x sin ; x 0 1; x' 0 1 x g) x sin ; y b) y sin x; y 1 x ) 4 4x cos ; x 0 f) y' '' 3y'' 3y' y x 4e x x h) Applicaion: Mixing Problems y''' 3y'' 3y' y e Ex1: A ank iniially conains 40 gal of sugar waer having a concenraion of 3 lb of sugar for each gallon of waer. A ime zero, sugar waer wih a concenraion of 4 lb of sugar per gal begins pouring ino he ank a a rae of gal per minue. Simulaneously, a rain is opene a he boom of he ank so ha he volume of he sugar-waer soluion in he ank remains consan. (a) How much sugar is in he ank afer 15 minues? (b) How long will i ake he sugar conen in he ank o reach 150 lb? 170 lb? (c) Wha will be he evenual sugar conen in he ank? Iniialize: > resar; > wih( DEools ): > wih( plos ): > wih( PDEools ): The mahemaical formulaion of his problem mus express he physical requiremen ha (amoun of sugar in ank) = (rae sugar is ae o ank ) - ( rae sugar is remove from ank ) Le x enoe he amoun of sugar (pouns) in he ank a ime (minues). Then, he raes in an ou are respecively. x e

3 > rae_in := 4 * ; > rae_ou := (x()/40) * ; so ha he governing ODE is > oe := iff( x(), ) = rae_in - rae_ou; The amoun of sugar in he ank iniially, ha is, when 0, gives he iniial coniion > ic := x(0)=40 * 3; The ODE in his IVP is firs-orer an linear. The inegraing facor is > mu() = infacor( oe ); The soluion o he IVP is > sol := solve( { oe, ic }, x(), [linear] ); (a) The amoun of sugar (in pouns) in he ank afer 15 minues is > eval( sol, =15. ); (b) The ank will conain 150 pouns of sugar a a ime (in minues) saisfying > eq150 := eval( sol, x()=150 ); Thus, he esire ime is foun by he calculaions > 150 := solve( eq150, ): > [`150 lbs`] = 150; > `` = evalf(150); Repeaing he same seps for he ime when 170 pouns of sugar are in he ank leas o he equaion > eq170 := eval( sol, x()=170 ); whose soluion is > 170 := solve( eq170, ): > [`170 lbs`] = 170; `` = evalf(170); This complex-value soluion is clearly no physically realisic. A quick inspecion of he soluion, graph wih ile One ank mixing problem > plo( rhs(sol), =0..10, ile="one ank mixing problem" ); shows ha he amoun of sugar in he ank reaches a seay-sae limi ha is well below 170 pouns. Therefore, a no ime is here ever 170 pouns of sugar in he ank. (c) In (b) i was noe ha he amoun of sugar in he ank levels off below 170 pouns. The exac limi can be eermine from he soluion by looking a he limi as, ha is, a > seay_sae := map( Limi, sol, =infiniy ); whose value is > value( seay_sae ); Noe ha x 160 is an equilibrium soluion for his ODE. However, be careful o avoi he common error of concluing ha he limi is 160 pouns because x 160 is an equilibrium soluion.. Use he ouline above o solve he following: a) Iniially, a 100 lier ank conains a sal soluion wih concenraion 0.5 kg/lier. A fresher soluion wih concenraion 0.1 kg/lier flows ino he ank a he rae of 4 lier/min. The conens of he ank are kep well sirre, an he mixure flows ou a he same rae i flows in. i. Fin he amoun of sal in he ank as a funcion of ime. ii. Deermine he concenraion of sal in he ank a any ime. iii. Deermine he seay sae amoun of sal in he ank. iv. Fin he seay sae concenraion of sal in he ank. b) A he sar, 5 lbs of sal are issolve in 0 gal of waer. Sal soluion wih concenraion lb/gal is ae a a rae of 3gal/min, an he well sirre mixure is raine ou a he same rae of flow. How long shoul his process coninue o raise he amoun of sal in he ank o 5 lbs?

4 Ex: Consier he previous problem, excep ha he ouflow from he ank is a a rae of 3 gallons per minue. (a) Fin he formula for he volume of sugar waer in he ank a any ime. When is he ank empy? (b) Fin he IVP for he amoun of sugar in he ank. (c) Fin he IVP for he concenraion of sugar in he waer. () When is he ank empy? Wha is he concenraion of sugar immeiaely before he ank is empy? How much sugar is in he ank a his ime? (e) Plo he amoun of sugar an concenraion of sugar in he ank up o he ime he ank becomes empy. Wha happens o hese soluions a laer imes? Soluion: a) The volume sars a 40 gallons. Every minue gallons of sugar waer are ae o he ank an 3 gallons are remove; he ne change is a loss of 1 gallon per minue. The raes "in" an "ou" are respecively > Vrae_in := ; > Vrae_ou := 3; so he ime-varying volume in he ank is > V := 40 + (Vrae_in-Vrae_ou)*; Noe ha V is he soluion of he IVP > oev := iff( v(), ) = - 3; > icv := v(0) = 40; as confirme via > solve( { oev, icv }, v(), [separable] ); b) The IVP for he amoun of sugar in he ank is similar o he one in he previous example. There is no x ifference in he rae a which sugar eners he ank. The concenraion of sugar exiing he ank is an his is ifferen because V V is no longer consan. The raes "in" an "ou" are now respecively > Srae_in := rae_in; > Srae_ou := (x()/v) * 3; Thus, he governing ODE is > oe := iff( x(), ) = Srae_in - Srae_ou; The iniial coniion is unchange: > ic := ic; c) The IVP for he concenraion of sugar in he ank is obaine from he ODE in (b) an he efiniion of he concenraion, c, which is > conc_eq := c() = x()/'v'; Raher han eriving he ifferenial equaion for c manually, he change comman from he PDEools package will be use o auomae he process. I gives > oec := change( x() = c()*v, oe, [c] ); The iniial coniion for he concenraion is > icc := c(0) = 3; ) The ank is empy when he volume of sugar waer is zero. This occurs afer 40 minues, as obaine by > _empy := solve( V=0, {} ); The concenraion is foun by solving he (linear) IVP foun in (c), which is one via > solc := solve( { oec, icc }, c(), [linear] ); Thus, he concenraion a he insan he ank empies is > subs( _empy, solc ); The amoun of sugar in he ank as he ank empies is obaine by evaluaing he soluion > sol := solve( {oe,ic}, x(), [linear] ); a he ime he ank empies. This gives > subs( _empy, sol ); which is exacly wha one woul expec. (If no, hink abou i!) e) The requese plos are > plo( rhs(sol), =0..40, ile=`amoun of sugar` ); V

5 an > plo( rhs(solc), =0..40, ile=`concenraion of sugar` ); The concenraion remains posiive unil 10, bu afer 40, he volume an amoun of sugar become negaive. Even hough he IVPs have soluions for all ime, for 40 hese resuls are no physically meaningful. Now, follow he ouline above o solve he following: 1. A large ank is parially fille wih 100 gallons of flui in which 10 pouns of sal is issolve. Brine conaining ½ lb of sal per gallon is pumpe ino he ank a a rae of 3 gal/min. The well-mixe soluion is hen pumpe ou a a slower rae of gal/min. i) Fin he IVP for he amoun of sal in he ank a ime. ii) Fin he IVP for he concenraion of sugar in he waer a ime. iii) When will he ank overflow? iv) Wha will be he number of pouns of sal in he ank a he insan i overflows? v) Deermine he number of pouns of sal in he ank as. Does your answer agree wih your vi) inuiion? Plo he graph of x on he inerval 0,500. Problems 4, 5, an 6 on page