Lecture ee : Klose modeling and Review midterms#

lastteine Prey - Predator modelold = IR - 1.2 RFat§ = - ft 0.9 RF

today A vaziaeeti eoceepeteeig species .

Ki follows logistic population model with growth ratek=2 and capacity N - 3

y : follows logistic population model with growth rate£-3 and capacity N=3

Suppose that the chance for x and yto meet is 30cg ,

and I deduce y bills x i € deduce d kills y .

date - 2x ( i - E) - say£7 - Sy ( i-÷) - sexy

• kloreeuodeEuigsCleeeeeieel reaction .

A and B substances ⇒ C

( low concentration)

Molecules of A and B react if they come closer together

, act)= aceeooceet of A cat tune t.

- but = aueoceei of B at tiene t .

- rate of reaction is proportional to actbut with ratio K.

daat

=tab

¥f = Kab

• if we add A and B at Cousteau rate,

then thee eg becomes

da_ = Kab t d rate for ARE in{ dolby = Keb tp rate for B .

Beaektoeocupewgspea#- Equilibrium pouts : 2K ( l - I) - xy -o x (2-x-g) so{ syce - ÷) - say. - o { y ( s- y - 24=0⇒ 2=0 or 2- K- y =D

Cf-o or 3- g-2x =D

So we have four ceases :

④ x -o and y - o ⇒ eqaieebaeu point Ceo)

② Seo and 3-y - 2x -o ⇒ eqceieebzeeu poceit Co, 3)

③ 2-x-y - o and y - o ⇒ eqceieebaeu point ( 2. o )

④ 2-x- y - o and3-y -2x - o ⇒ eqaieebaeu posit Cc

, i)

Phaaepereraie Hlf ) - Rus - ( LII:3! )"":#f

)x

Ceo) (2,0)

Nfo' ) - ( o' ) NC :) - (j' ) " (9) =

'" l -

- C's:) MI's) -- Eyal

hint plug ai points close to quibble've points .this will help yea drawing thee all please portrait .

Reviewforeuidterc#. population model

* See-up differential equations . . bank interest problems- euixaig problems

°

separation of wearables

* Analytic methods . uitegrol factors• linear differential equations

° slope field t please lues t existence E eeepveceessthroes

* Qualitative methods . bifurcation

→ Euler's method } see notes.

Examples

① linear equation old-t

de- 3g -- e

2e

solution move terms defy = 3g te

3T

hooey eg .

old = 3g ⇒ solution yn- e

alt

guess yp = Cest for a particular solution

2C eft = 3C @It

t e

"2C = SC ti C = - l

So the general solution is y # =test -est KEIR

meeee if dY_ = 3g t cost ⇒ guess gpct) - C. cost tczseietdt

3.t 3T

if d£ = Sy t e ⇒ guess ypG7= Cte

② Integration factor . doff = Ey t te

Mt : move terms

dyat

- Ey - tee

integration factor is

nets = exp ( f -Edt) -- exp ( - sleet) = t- 2

Multiply : E - oldat- ⇐ g - et

qf-( t-g) = et y

EE et to g-tacet to)

③ Interest problem ,

Initial deposit $1,000 i interest rate 2% .

Additional year

deposit $ I,ooo eoutinuesby .

When does the money reach $ lo,

ooo ?

Solution let yes = money in the bank at year t .

date = 2% y t (ooo

÷y+e•- dy - ott ⇒ I lie to -02g that = Etc ,

0.02

lie to -02g that = O -02£ to . @2.C ,

(oozy teapot =eat to. 02C,

D. cozy teooo =±e°°2" I'"t go geese Ceo

-02£- tooo

c- 2

= : Cz0.02T

yCE) = loadGEF

"- soooo ⇒ yet) = C e - 50000

-=: C

y = tea I,eoo=yCoJ= C -50,000 e = 51,000

y = 1000 (51 Edt- 50)

we have to fund t* oude thecae y Ctx) = toooo

yet) = load (51 @°"- 50)

(oooo = y Ctx) = ( ooo (518024-50)0.

C-*

1000=1# (51¥+*- 50) 10=51e - 50

sie -*⇐*= Go 0.02¥ = be 6051

+* = 50 lee @51

④ Drew please eerie for d£= yseeigEqceieibzeeie points .

yseiey - o ⇒ y - o and sing -o ⇒ Yeo, It, ten, - . -

yr=

5¥ Feet -Eso-to ✓✓✓ y= -E off - Eso-

ye - 3£ LIE = f- ZI) suite) = -II co- 25

y - ZE off = - 35-20

OI : been a solution with central value yco) = - i even become

pest've when t> o ?

Noe because of the existence and uniqueness theorem .

yn# Since fcyleyseiey is coaxiueuo and different.

try,,⇒.# ←we know there exists a euieue ooaet.ae to

- thee IVP-a

dy-

q= fly ) y 6) = - l.

- 25

therefore thee red curve aan'T intersect the

line YCEIEO, which is another solution .