This preview shows page 1. Sign up to view the full content.
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: Phaseline 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: Phaseline 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 byer1,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 ...
View
Full
Document
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.
 Winter '07
 MUNTEANU
 Calculus, Equations

Click to edit the document details