ass 4 answr - Assignment 4 Solutions Total=12 points....

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Unformatted text preview: Assignment 4 Solutions Total=12 points. Assignment 4 Solutions 1. (a) Find the equilibrium solutions of the following differential equations. You should find three 1. (a) Find the equilibrium solutions of the yol= win− 6yffere11y −e6.uations. You should find f lo y 3 g di 2 + ntial q three Solution: (2 points) We find the− 6y 2 + 11y ) = .y 3 − 6y 2 +11y − 6 = (y − 1)(y − 2)(y − 3) y = y 3 roots of f (y − 6 to be 1, 2, 3. (b) Dr(b) the phase lphaseiagram. aw Draw the ine d line diagram. SolutioSolution: d thpoints)of f (y ) f is 3a polynomial 6 =easily )(y − 2)(ythe ) to b e 1, 2, 3. f is n: We fin (2 e roots Since = y − 6y 2 + 11y − we (y − 1 identify − 3 intervals where Since f is a p olynomial we easily identify us e ingraph sthe phase ilinecdiagram.decreasing. 1 increasing/decreasing. This allows th to terval where f s in reasing/ See Figure This allows us to graph the phase line diagram: 1 2 3 Of course, we an also use the derivative criterion todiagramab out stability. Figure 1: Phase-line decide (c) Graph the equilibrium solutions and some solution curves y (0) = 0, y (0) = 0.5, y (0) = 1.5, y (0) = 2.5, y (0) = 4. Of course, we can also use the derivative criterion to decide about stability. You should clearly indicate the b ehaviour of the solution curves as t → ∞. Here are (c) raphs: some gGraph the equilibrium solutions and some solution curves Solution: (2 points)You should clearly indicate the behaviour of the solution curves as t → ∞. See Figure 2. 2. Suppose that size N if a populations satisfies = 3 following differential equation: y the dN y =N 2 52 = − 2N. dt 1 + N2 (a) Find all equilibrium points. y=1 (b) Use the derivative criterion to decide if the equilibria are stable or unstable. −5 −4 −3 −2 −1 (c) Draw the phase line diagram. 0 1 2 2 3 4 5N Solution: (a) (2 points) f (N ) = 1+N 2 − 2N = 0 ⇐⇒ 5N 2 = 2N (1 + N 2 ) ⇐⇒ N (2N 2 − 1 5N + 2) = 0 ⇐⇒ 2N (N − 2 )(N − 2) = 0. Hence the equilibria 0, 1/2 et 2. 2. Supp ose that size N if a p opulations satisfies the following differential equation: dN 5N 2 1 − 2N . = dt 1 + N2 1 You should clearly indicate the b ehaviour of the solution curves as t → ∞. Here are some graphs: y=3 y=2 y=1 (a) Find all equilibrium p oints. (b) Use the derivati−5 cr−erion3to −2 cide if the equilibria a3 e st4 ble or unstable. ve it4 − de −1 r a 01 2 (c) Draw the phase line diagram. 2 Solution: (a) f (N ) = 15N 2 − 2N = 0 ⇐⇒ 5N 2 = 2N (1 + N 2 ) ⇐⇒ N (2N 2 − 5N + 2) = 0 +N ⇐⇒ e 2Nat si− 1N(N a p opul0. ions cattsfieequieibollow,ing 2 iete2ential equation: (N ze 2 ) if − 2) = at Hen se ihe s th l f ria 0 1/ d ff r . 2. Supp os th Figure 2: Some solution curves 1 (b) The derivative of f is f (N ) = (1+0N2 )2 − 2 , and we evaluate f at the equilibria : f (0) = 2 dN N 5N −2 < 0, f (1/2) = 1.2 > 0 and f (2) = −1.2 − 20. . Hence, 0 and 2 are stable and 1/2 is 2<N = dt 1+N 10N (b)b(2 points) The derivative of f is f (N ) = (1+N 2 )2 − 2, and we evaluate f at the equilibria unsta le. (c) : f (0) = −2 < 0, f (1/2) = 1.2 > 0 and f (2) = −1.2 < 0. Hence, 0 and 2 are stable and 1/2 1 is unstable. (c) (2 points) See Figure 3. 0 1/2 2 Figure 3: Phase-line diagram 3. The Zombies have invaded Campus again! The campus p opulation is ab out 36, 000. The 3.ombiesZombiesecting humans at a rateagain! hTheprop ortional to the product ,of t. The Z The are inf have invaded Campus whic is campus population is about 36 000 he ZoZombies opulation and the p ortion owhich ans on campus to here products of e mbie p are infecting humans at a rate f hum is proportional (w the α > 0 i th the proZombie lipopulation . and the portion of humans on campus (where α > 0 is the p ortiona ty constant) At proportionality the Biology department has develop ed a method to turn Zombies back into the same time, constant). At the same time, the drug thdepartment nti developed a . Zomb to turn ansform d in into humans by distributing aBiology rough the vehaslation systemmethodies are trZombiesebackto humans a distributing is pro through the ventilation system. Zombie are here β > 0 s humans at by rate which a drug p ortional the p opulation of Zombies s (wtransformed iinto thehumanstatnalirate nstant). is proportional the population of Zombies (where β > 0 is prop or io ty co which the proportionality constant). (a) Denote by z (t) the numb er of Zombies at time t. Find the differential equation which (a) t) satisfieby ccordthe numbernof rZombies iat n. Not. that the total p opulatequation swhich z ( Denote s a z (t) ing to the i fo mation g ve time te Find the differential ion is of ize z t 6 000 + z according to s t e numb er o invad rs. T =(3) ,satisfies (0) and z (0) ithehinformationfgiven. eNote that the total population is of size T = 36, 000 + z (0) and z (0) is the number of invaders. (b) Find the equilibria of the differential equation in (a). These equilibria will dep end on α (b) d β . the equilibria of the differential equation in (a). These equilibria will depend on α an Find and β . (c) Assume that α = 1/2 and β = 2/3. Without solving the differential equation, what is (c) eAssumerthat om= e1/i2 and lβ ng run (when t →solving thet differential and β = 1/2? is th numb e of z α bi s n the o = 2/3. Without ∞)? Wha if α = 2/3 equation, what the number of zombies in the long run (when t → ∞)? What if α = 2/3 and β = 1/2? (a) The Zombie p opulation is given by z (t) and the p ortion of humans by (36000 + z (0) − (a) The Zombie population is given by z (t) and the portion of humans by (36000 + z (0) − z (t))/36000 + z (0)). Write Ptot = 36, 000 + z (0). So, z (t))/36000 + z (0)). Write Ptot = 36, 000 + z (0). So, (Ptot − z (t)) )) z (t) (t) αz (t) (t) (Ptot − z (t− β z (t).(t). z = = αz − βz PtoP t tot (b) Solve αz (Ptot − z )/Ptot − β z = 0 ⇐⇒ z (α − αz /Ptot − β ) = 0 and hence there are 2 β two equilibrium p oints 0 and Ptot (1 − α ). (c) If α = 1/2 and β = 2/3, then the equilbrium t p oints are 0 et −Ptot /3. Calculating the derivative of 1/2z (t) (PtoP−z (t)) − 2/3z (t) gives that to t 0 is a stable equilibrium. Hence the Zombie p opulation approached zero and will eventually disapp ear. If α = 2/3 and β = 1/2, then the equilbrium p oints are 0 et Ptot /4. Calculating the derivative (b) Solve αz (Ptot − z )/Ptot − βz = 0 ⇐⇒ z (α − αz/Ptot − β ) = 0 and hence there are β two equilibrium points 0 and Ptot (1 − α ). (c) If α = 1/2 and β = 2/3, then the equilbrium points are 0 et −Ptot /3. Calculating the derivative of 1/2z (t) (Ptot −z (t)) − 2/3z (t) gives that Ptot 0 is a stable equilibrium. Hence the Zombie population approached zero and will eventually disappear. If α = 2/3 and β = 1/2, then the equilbrium points are 0 et Ptot /4. Calculating the derivative P o − )) z o 2/3z t)((P tPt−oz (t)) − 1/2z t) give tha Pto /4 i a stable equilibrium Hence the Zombie off 2/3z ((t)(PttotPtzt(tt)) − 1/2z ((t) givess thatt Ptott/4 iss a stable equilibrium.. Hence the Zombie tot ot − ( op o2/3z (ton willotmake up 2a (q) aritersoffhthe Ptotm4 us populationqiniltheiulm. g run;;cthatteisZ900bi+ f pulati) P − 1 p z t u g ve at Ca / p i a p ab e n i br on ru n t a om 0 e population witll make u/ a quarter ot the Campuss p ostulaltioe un ithe long Hen ehth is 9000 + p (0))/ation will make up a quarter of the Campus p opulation in the long run; that is 9000 + z (0 / 4 z opul4.. z (0)/4. 4 Find a differentia equation whose phase line diagram i the following and graph the solution 4.. Find a differentiall equation whose phase line diagram iss the following and graph the solution 4. Fiurve foriffy(0)n=a3:: quation whose phase line diagram is the following and graph the solution c nd o re = curvea fd r ye(0) ti l3 e curve for y (0) = 3: -1 -1 1 1 2 2 Solution : We know that the equilibrium p oints are given by -1, 1 et 2. We try the following diffSoeutiion e:quaeion ow =h(y +h1)(y uilibrium 2)ointutasincevthe byer-1,a1ive critWie nrrevhalsfollowing er l nt al : Wt kn y t at t e eq − 1)(y − p . B s are gi en by -iv t et 2. er o t y te e fothat ithe Solution We know that the equilibrium points re given d 1, 1 et 2. We try the llow ng dirrerwntiin loequfirsonry p oi(y + t1)(y e wroy g d2recBun,sancorrhct derutaoin e critdrb enyreve−l(yt+at)(the ff o e s a equation y = ny + 1)(ty − 1)( n − 2). tio t since te e derivative oul e io reveals that ty − a differential ur atit t = ( t in o h − 1)(y − i ). But i ce the soliv it v wcriterion = a s h 1 he a1)(y − iin .ouhefisotuttriy n ourt enpo sthngwrong ghrecti3) ,isa correct solution would b e y = −(y + 1)(y − rrows 2) our first try p cin viinto the wthrou di (0, on a correct solution would be y = −(y + 1)(y − n T r rs l o point t a si e rong direction, arrows 1)(y − 2).The solution curve passing through (0,, 3) iis given in Figure 4. 1)(y − 2).The solution curve passing through (0 3) s y=2 y=2 y=1 y=1 −5 −4 −3 −2 −1 0 −5 −4 −3 −2 −1 0 y = −1 y = −1 1 1 2 2 Figure 4: Solution curve 3 3 3 4 4 ...
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This note was uploaded on 09/09/2011 for the course MAT 1332 taught by Professor Munteanu during the Winter '07 term at University of Ottawa.

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