pr final

7182019 Pr Final

httpslidepdfcomreaderfullpr-final-5691a1c806a46 18

1048573

1048573

1803 Twopracticefinalexams

OperatorFormulas

bull

Exponential

Response

Formula

x p

=

Aertp(r)

solves

p(D)x

=

Aert provided

p(r)

=

0

bull

Resonant

Response

Formula

If

p(r)

=

0

then

x p

=

Atertp1048573(r)

solves

p(D)x

=

Aert

provided

p1048573

(r)

=

0

bullExponentialShiftLaw p(D)(ertu)=ert p(D+rI )u

Propertiesof theLaplacetransform

infin

0 Definition L[f (t)]=F (s)= f (t)eminusstdtforResgtgt00minus

1 Linearity [af (t)+bg(t)]=aF (s)+bG(s)L

2

Inverse

transform

F

(s)

essentially

determines

f

(t)

3 s-shiftrule [eatf (t)]=F (sminusa)L

4 t-shiftrule L[f a(t)]=eminusasF (s) f a(t)=

f (tminusa) if tgta

0 if tlta

5 s-derivativerule [tf (t)]=minusF 1048573(s)L

6 t-derivativerule [f 1048573(t)]=sF (s)minusf (0+)L

[f 10485731048573(t)]=s2F (s)minussf (0+)minusf 1048573(0+)L

if weignoresingularitiesinderivativesatt=0

t

L7 Convolution

rule

[f

(t)

lowast

g(t)]

=

F

(s)G(s)

f

(t)

lowast

g(t)

=

f

(τ

)g(t

minus

τ

)d

τ

0

1

8 Weight

function

[w(t)]

=

W

(s)

=

w(t)

the

unit

impulse

response

L p(s)

Formulas

for

the

Laplace

transform

1 1 n

L[1]= L[e

at] =

sminusa

[tn] =

sn+1s

L

s ωL[cos(ωt)]=

s2 +ω2

L[sin(ωt)]=

e

s2 +ω2

minusas

[ua(t)]

=

L[δ a(t)]

=

eminusasL s

whereu(t)istheunitstepfunctionu(t)=1fortgt0u(t)=0fortlt0

7182019 Pr Final


1048573

1048573

Fourierseries

a0f (t)= +a1

cos(t)+b1

sin(t)+a2

cos(2t)+b2

sin(2t)+middot middot middot2

π

π1 1

am

=

f

(t)

cos(mt)

dt

bm

=

f

(t)

sin(mt)

dt

π π

π

minusπ

π

minusπ

cos(mt)cos(nt)dt= sin(mt)sin(nt)dt=0 for m=n

minusπ π

minusππ

2cos

(mt)

dt

=

sin2(mt)

dt

=

π

for

m

gt

0

minusπ

minusπ

If

sq(t)

is

the

odd

function

of

period

2π

which

has

value

1

between

0

and

π

then

4

sin(3t)

sin(5t)sq(t)= sin(t)+ +

π 3 5 +middot middot middot

Variation

of

parameters

The solution to u = Au+q(t) is given by u=Φ(t) Φ(t)minus1q(t)dt where Φ(t) is any

fundamental

matrix

for

A

(In

fact

this

true

even

if

the

coefficient

matrix

A

=

A(t)

is

nonconstant The1times 1casewasstudiedearlyon)

Defectivematrix formula

If

A

is

a

defective

2

times 2

matrix

with

eigenvalue

λ1

and

nonzero

eigenvectorv1

then

you

can

solveforw in(Aminus λ1I )w=v1

andu=eλ1t(tv1

+w)isasolutiontou =Au

ExamI

1 (a) A vial of pure Kryptonite undergoes radioactive decay but Lex Luthor keeps it

continually

resupplied

at

the

rate

of

q (t)

grams

per

hour

Set

up

the

ODE

describing

the

numberx(t)of gramsof Kryptoniteinthevial Youranswerwillinvolveq (t)andanasyet

undetermined

decay

rate

(b)

Lex

actually

puts

in

the

Kryptonite

in

01

gram

doses

once

an

hour

on

the

half

hour

startingatt= 12 Writedownanexpression forq (t)thatmodelsthis (Assume itstarts

at

t

=

0

and

goes

on

forever)

2 (a)Sketchthegraphof afunctiong(x)suchthatthephaselineof theautonomousODE

x =

g(x)

looks

like

this

w

w

w

-1 0 1

(b)Use the Euler methodwith 3 stepsto estimatethe valueat x= 03 of the solution to

y1048573 =

x

+

y

with

y(0)

=

1

3

(a)

Find

the

general

solution

to

the

ODE

tx +

x

=

cos

t

7182019 Pr Final


1048573

1114109

917501

1114109

917501

1114109

917501

1114109 917501

(b)Supposethatx1(t)andx2(t)aresolutionsof afirstorderlinearODEandthatx1

=x2Write

down

a

nonzero

solution

to

the

associated

homogeneous

linear

ODE

(in

terms

of

x1

andx2) Thenwritedownthegeneralsolutiontotheoriginalequation(intermsof x1

and

x2)

2

4 (a)Express(1+i)21 intheforma+biwithaandbreal (Itmaybeusefultoknowthat

10

=

1024)

(b)

Write

each

of

the

three

cube

roots

of

8i

in

the

form

a

+

bi

with

a

and

b

real

uml5 (a)Findaparticularrealsolutiontox+5x=4eminustcos(2t)

(b)Findthegeneralrealsolutiontox+2x+2x=2t2+2uml

6 (a)

A

certain

periodic

function

f (t)

has

Fourier

series

given

by

cos(3t) cos(5t)f (t)

=

cos(t)

+

+

9

25 +

middot middot middot

What

is

a

periodic

solution

of

the

equation

x

+

ω2x

=

f (t)

(if

one

exists)

n

(b)

What

is

the

Fourier

series

of

the

function g(t)

which

is

periodic

of

period

2

and

such

thatg(t)=2for0lttlt1andg(t)=0forminus1lttlt0

7 (a)

For

what

values

of

c

and

k

does

the

LTI

operator

L

=

D2+

cD

+

kI

have

unit

impulse

responsegivenbyw(t)=eminus2tsin(t)fortgt0

e(b)ForthissameoperatorLwritedowntheconvolutionintegralforthesolutiontoLx=

minus2t with

rest

initial

conditions

and

evaluate

it

4

(c)

Find

the

inverse

Laplace

transform

of

F

(s)

=

(s2

+

2s

+

5)(s

+

1)

(d)

Sketch

the

pole

diagram

of

F

(s)

8 This problem concerns a 2times2 real matrix A whose eigenvalues are 2 andminus1 with

1 2

correspondingeigenvectors and 2 1

(a)

Write

down

a

fundamental

matrix

for

the

vector

equationu =

Au

(b)

Compute

eAt

0

(c)

Find

a

solution

tou =

Au

+

1

a

4

9 LetA= andconsiderthesystemu =Au Foreachof thefollowingconditionsa 3

determineallvaluesof awhicharesuchthatthesystemsatisfiesthecondition

(a)

Asymptotically

stable

(b)

Defective

node

(c)Spiral(includingcenters)

(d)Node

(e)

Saddle

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

(f )Thereissomeconstantsolutionotherthan0

10 Two

ant

species

compete

with

each

other

for

the

same

food

Each

population

depresses

thegrowthrateof theotheraswellasitsown(logistically) Herersquosamodelof this

x =

(6

minus

2x

minus

y)x

y = (6

minus

x

minus

2y)y

(a)

Find

where

the

vector

field

is

horizontal

where

it

is

vertical

and

locate

the

critical

points

(b)Thereisonecriticalpointintheupperrightquadrantwithpositivevaluesof xandyFind

the

Jacobian

and

evaluate

it

at

that

critical

point

(c)

Sketch

the

phase

portrait

of

the

linearization

at

this

critical

point

Plot

any

eigendirec-tionscarefullyandnamethetype(nodesaddlespiralstableunstable)

(d) Identify the types of the other critical points and sketch a phase portrait of this au-tonomous

system

in

the

upper

right

quadrant

(e)Canatrajectoryleavetheupperrightquadrant (YesorNo)

ExamII

1 Salt

water

enters

a

twenty

gallon

tank

at

a

rate

of

five

gallons

per

minute

and

leaves

it

atthesameratethroughaholeinthebottom Arotorkeepsthesolutionwellmixed Write

x(t)

for

the

number

of

pounds

of

salt

in

the

tank

at

time

t

and

suppose

that

at

t

=

0

the

tankisfullof freshwater Supposethattheconcentrationof saltinthewaterbeingadded

is

q (t)

pounds

per

gallon

(a)

Write

down

a

differential

equation

that

controls

x(t)

(Measure

time

in

minutes)

(b)

Salt

is

added

to

the

tank

in

four

sudden

discrete

packets

of

half

a

pound

each

once

a

minutestartingatt=0 Whatisq (t)

dy2 (a)UsetheEulermethodwithstepsize12toestimatey(2)if =xyminus1andy(1)=2

dx

dy(b)Sketchtheisoclinesforslopes0andplusmn1atleastfortheODE =x

2minusyandusethis

dx

to

sketch

the

graph

of

the

solution

with

y(minus1)

=

1

between

x

=

minus2

and

x

=

2

or

more

(Thinkaboutwhatthesignof theslopefieldisabovetheisoclineforslopezero)

3

(a)

Find

the

general

solution

of

the

ODE

y1048573

minus

xy

=

2ex22

sin(x)

1 1

(b) Suppose et and eminust minus1

are both solutions to the homogeneous linear system

1

0

u =

Au

What

is

A

Find

a

solution

of u minus

Au

=

4

4 (a)

Express

cos(πt)

minus

sin(πt)

in

the

form

A

cos(ωt

minus

φ)

1+i(b)Express intheforma+biabreal

1

minus

i

(c)Express1+iasreiθ withrandθrealandrge0

uml5

(a)

Find

a

particular

solution

of

x

+

2

x

+

x

=

4et

+

5

7182019 Pr Final


1114109

917501

(b)Findthegeneralrealsolutionof x +4x +13x =0uml

uml(c)Findtheamplitudeof thesinusoidalsolutionof x +2x +x =2sin(3t)

6 What

is

the

Fourier

series

for

f (x)

=

1

+

cos(x minus

π4)

7 (a)

What

is

the

unit

impulse

response

for

the

operator

D2 +

2D +

3I

(12)eminustsin(2t) fort gt 0

(b)

The

unit

impulse

response

of

a

certain

system

is

given

by

w(t)

=

0

for

t lt 0

Write down the integral expressing the system response to the signal eminust (with rest initialconditions)

and

evaluate

it 1114109

917501

8 (a)

Compute

eAt where

A =

1

0

1

2

1114109

917501

1114109

917501

(b)

Solveu

=

Au

+

2et

etwith

initial

conditionu(0)

=

0

0

1114109

917501

costt(c)

Write

down

a

real

matrixA for

whichu =

Au

has

as

a

solutionu

=e

2

sint

1 1

9 As in 8(a) let A = Sketch the phase portrait for u = Au Mark the

0 2

eigenlines (the straight solutions) with their corresponding eigenvalues and at least four

other

trajectories

and

name

the

type

of

phase

portrait

you

have

(saddle

spiral

node

stableunstable)

10 Writex forthepopulationof bugs(insomeconvenientunits)andy forthepopulation

of

birds

Birds

eat

bugs

and

the

two

together

satisfy

the

nonlinear

autonomous

system

x = (2minusx minusy)x

y = (xminus

1)y

(so

that

in

the

absence

of

birds

the

bug

population

grows

logistically

and

in

the

absence

of

bugsthebirdsdieoutexponentially)

(a)Findallthecriticalpointsof thissystem

(b)Findthelinearizationatthecriticalpointwithpositivex andy coordinatesandsketch

the

trajectories

near

that

critical

point

(c)Nowmalathionisintroducedinanattempttoreducethebugpopulation Thisreducestherateof reproductionof bothspeciessothenewsystemisgivenby

x = (2

minus

x minus

y minus

a)xy = (x minus1minusb)y

for certain positive constants a b What happens to the critical point studied in (b) Is

this

measure

successful

in

reducing

the

bug

population

7182019 Pr Final


1048573

SolutionstoExamI

1 (a)x +kx=q (t)wherek isthedecayrate

(b)

q (t)

=

(01)(δ (t

minus 5)

+

δ (t

minus 15)

+

δ (t

minus 25)

+

middot middot middot)

2 The

g(x)

is

positive

for

x

lt

minus1

and

0

lt

x

lt

1

negative

for

minus1

lt

x

lt

0

and

x

gt

1

and

zeroatx=

minus10and1

k

0

xk

0

yk

1

Ak

=

xk

+

yk

1

hAk

1

(b)

h

=

1 1

2

1

2

11

122

12

142

12

142

3 3 1362

d

sint+c

3 (a)

The

equation

is

tx

=

cos

t

so

tx

=

cos

t

dt

=

sin

t

+

c

and

x

=

dt t

(b)Thedifferencex2

minus x1

mustbeasolutiontothehomogeneousequationwhosegeneralsolutionisthusc(x2

minusx1) Thegeneralsolutionof theoriginalequationisthenx1

+c(x2

minusx1)

4

(a)

|1

+

i|

=

radic 2

and

arg(1

+

i) =

π4

Therefore

|(1

+

i)21|

= 2212 =

1024radic 2

and arg((1+i)21) =21π4 or 5π4 The complex number with these polar coordinates is

minus1024(1

+

i)

(b)

|8i =

8

and

arg(8i) =

π2

so

the

cube

roots

have

magnitude

2

and

arguments

π6|5π6and9π6=3π2 Thesenumbersare(

radic 3+i)(minusradic 3+i)andminus2i

5 (a)Thisistherealpartof z +5z =4e(minus1+2i)t Since p(s)=s2 +5 p(minus1+2i)=(minus1+

4e(minus1+2i)t

2i)2+5=1minus4iminus4+5=2minus4isotheExponentialResponseFormulagivesz p

=2minus 4i

2(1+2i) 2

x p

=Rez p

=Re eminust(cos(2t)+isin(2t))= eminust(cos(2t)minus 2sin(2t))5 5

2] x p

= at2 + bt + c

2]

x p

= 2at

+

b

(b)

Undertermined

coefficients

1] x p

= 2a

2t2 +2 = 2at2 + (2b+4a)t + 2c+2b+2a

soa=1b=minus2andc=2 x p

=t2minus 2t+2 Thecharacteristicpolynomials2 +2s+2=

(s

+

1)2+

1

has

roots

minus1

plusmn i

so

basic

homogeneous

solutions

are

eminustcos

t

and

eminustsin

t

The

generalsolutionisthusx p

=t2minus 2t+2+eminust(acost+bsint)

cos(t)

cos(3t)

cos(5t)6 (a)x p

= + + ω2

n

minus9) 25(ω2

n

minus1 9(ω2

n

minus25)

+middot middot middot

4

sin(3πt)

sin(5πt)(b)g(t)=1+sq(πt)=1+ sin(πt)+ +

π

3

5 +middot middot middot

7 (a)

The

unit

impulse

response

is

a

homogeneous

solution

and

to

get

eminus2tsin(t)

the

roots

of thecharacteristicpolynomialmustbeminus2plusmn i Thesumof therootsisminus4andtheproduct

is5sothecharacteristicpolynomialiss2 +4s+5 c=4k=5t

(b)

x

=

eminus2tsin(t)

lowast eminus2t =

eminus2tlowast eminus2tsin(t)

=

eminus2(tminusτ )eminus2τ sin(τ )

dτ

0

=

et

minus2t sin(τ )

dτ

=

eminus2t(minus cos(t)

+

1)

0

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s

minus cos(2t)+1andf (t)=eminust(minus cos(2t)+1)

(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2

2e2tminus 2eminustAte =Φ(t)Φ(0)minus1 = 1 minuse2t +

4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1

(c)Lookforaconstantsolution u=minusAminus1

1

TofindA noticethatA =2

0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3

4 minus1 4 3 2 minus1 3 1minus11114109 917501

minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2

whena2 +10a +9=0 ie whena =minus1ora =minus9 Forminus9lt a lt minus1theyarenon-realThe real parts of the eigenvalues are negative when trA lt 0 and detA gt 0 ie a +3lt 0

andminusa gt 0

ora lt minus3

(a)a lt minus3 (b)a =minus1minus9 (c)minus9lt a lt minus1 (d)a lt minus9andminus1lt a lt 0 (e)0lt a(f )

a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where

y =0 y =0ory =3minus (x2) Therearecriticalpointsat(0 0) (0 3) (3 0) (2 2)

x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4

(c)WithA =J (2 2) pA(λ) =λ2 +8λ +12sotheeigenvaluesareminus2andminus6 Anonzero

1 1

eigenvector forminus2 is minus1

and forminus6 is This gives a stable node in which the

1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0

(d)J (0 0)= givesanunstablestar0 6

3 0

J (0 3)=

minus3

minus6

haseigenvalues3and

minus6andgivesasaddle Anonzeroeigenvector

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1

forλ=minus6is andfor3is

minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)

+middot middot middot)

k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown

3 (a)y=ex22(C minus 2cosx)

0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15

6 1+cos(π4)cosx+sin(π4)sinx

7 (a)

(1sqrt2)eminustsin(radic

2t)

(b)

w(t)

lowast eminust =

0

t(12)eminususin(2u)eminus(tminusu)du

=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1

uastablespiral(counterclockwise)1 0

(c)Thenewcriticalpointis(1+b1minus a+b) Itisstillstable Clearlytherearemorebugs

7182019 Pr Final


1048573

1048573

Fourierseries

a0f (t)= +a1

cos(t)+b1

sin(t)+a2

cos(2t)+b2

sin(2t)+middot middot middot2

π

π1 1

am

=

f

(t)

cos(mt)

dt

bm

=

f

(t)

sin(mt)

dt

π π

π

minusπ

π

minusπ

cos(mt)cos(nt)dt= sin(mt)sin(nt)dt=0 for m=n

minusπ π

minusππ

2cos

(mt)

dt

=

sin2(mt)

dt

=

π

for

m

gt

0

minusπ

minusπ

If

sq(t)

is

the

odd

function

of

period

2π

which

has

value

1

between

0

and

π

then

4

sin(3t)

sin(5t)sq(t)= sin(t)+ +

π 3 5 +middot middot middot

Variation

of

parameters

The solution to u = Au+q(t) is given by u=Φ(t) Φ(t)minus1q(t)dt where Φ(t) is any

fundamental

matrix

for

A

(In

fact

this

true

even

if

the

coefficient

matrix

A

=

A(t)

is

nonconstant The1times 1casewasstudiedearlyon)

Defectivematrix formula

If

A

is

a

defective

2

times 2

matrix

with

eigenvalue

λ1

and

nonzero

eigenvectorv1

then

you

can

solveforw in(Aminus λ1I )w=v1

andu=eλ1t(tv1

+w)isasolutiontou =Au

ExamI

1 (a) A vial of pure Kryptonite undergoes radioactive decay but Lex Luthor keeps it

continually

resupplied

at

the

rate

of

q (t)

grams

per

hour

Set

up

the

ODE

describing

the

numberx(t)of gramsof Kryptoniteinthevial Youranswerwillinvolveq (t)andanasyet

undetermined

decay

rate

(b)

Lex

actually

puts

in

the

Kryptonite

in

01

gram

doses

once

an

hour

on

the

half

hour

startingatt= 12 Writedownanexpression forq (t)thatmodelsthis (Assume itstarts

at

t

=

0

and

goes

on

forever)

2 (a)Sketchthegraphof afunctiong(x)suchthatthephaselineof theautonomousODE

x =

g(x)

looks

like

this

w

w

w

-1 0 1

(b)Use the Euler methodwith 3 stepsto estimatethe valueat x= 03 of the solution to

y1048573 =

x

+

y

with

y(0)

=

1

3

(a)

Find

the

general

solution

to

the

ODE

tx +

x

=

cos

t

7182019 Pr Final


1048573

1114109

917501

1114109

917501

1114109

917501

1114109 917501


=x2Write

down

a

nonzero

solution

to

the

associated

homogeneous

linear

ODE

(in

terms

of

x1


and

x2)

2


10

=

1024)

(b)

Write

each

of

the

three

cube

roots

of

8i

in

the

form

a

+

bi

with

a

and

b

real



6 (a)

A

certain

periodic

function

f (t)

has

Fourier

series

given

by


=

cos(t)

+

+

9

25 +


What

is

a

periodic

solution

of

the

equation

x

+

ω2x

=

f (t)

(if

one

exists)

n

(b)

What

is

the

Fourier

series

of

the

function g(t)

which

is

periodic

of

period

2

and

such


7 (a)

For

what

values

of

c

and

k

does

the

LTI

operator

L

=

D2+

cD

+

kI

have

unit

impulse



minus2t with

rest

initial

conditions

and

evaluate

it

4

(c)

Find

the

inverse

Laplace

transform

of

F

(s)

=

(s2

+

2s

+

5)(s

+

1)

(d)

Sketch

the

pole

diagram

of

F

(s)


1 2


(a)

Write

down

a

fundamental

matrix

for

the

vector

equationu =

Au

(b)

Compute

eAt

0

(c)

Find

a

solution

tou =

Au

+

1

a

4



(a)

Asymptotically

stable

(b)

Defective

node


(d)Node

(e)

Saddle

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501


10 Two

ant

species

compete

with

each

other

for

the

same

food

Each

population

depresses


x =

(6

minus

2x

minus

y)x

y = (6

minus

x

minus

2y)y

(a)

Find

where

the

vector

field

is

horizontal

where

it

is

vertical

and

locate

the

critical

points


the

Jacobian

and

evaluate

it

at

that

critical

point

(c)

Sketch

the

phase

portrait

of

the

linearization

at

this

critical

point

Plot

any



system

in

the

upper

right

quadrant


ExamII

1 Salt

water

enters

a

twenty

gallon

tank

at

a

rate

of

five

gallons

per

minute

and

leaves

it


x(t)

for

the

number

of

pounds

of

salt

in

the

tank

at

time

t

and

suppose

that

at

t

=

0

the


is

q (t)

pounds

per

gallon

(a)

Write

down

a

differential

equation

that

controls

x(t)

(Measure

time

in

minutes)

(b)

Salt

is

added

to

the

tank

in

four

sudden

discrete

packets

of

half

a

pound

each

once

a



dx


2minusyandusethis

dx

to

sketch

the

graph

of

the

solution

with

y(minus1)

=

1

between

x

=

minus2

and

x

=

2

or

more


3

(a)

Find

the

general

solution

of

the

ODE

y1048573

minus

xy

=

2ex22

sin(x)

1 1



1

0

u =

Au

What

is

A

Find

a

solution

of u minus

Au

=

4

4 (a)

Express

cos(πt)

minus

sin(πt)

in

the

form

A

cos(ωt

minus

φ)


1

minus

i


uml5

(a)

Find

a

particular

solution

of

x

+

2

x

+

x

=

4et

+

5

7182019 Pr Final


1114109

917501



6 What

is

the

Fourier

series

for

f (x)

=

1

+

cos(x minus

π4)

7 (a)

What

is

the

unit

impulse

response

for

the

operator

D2 +

2D +

3I


(b)

The

unit

impulse

response

of

a

certain

system

is

given

by

w(t)

=

0

for

t lt 0


and

evaluate

it 1114109

917501

8 (a)

Compute

eAt where

A =

1

0

1

2

1114109

917501

1114109

917501

(b)

Solveu

=

Au

+

2et

etwith

initial

conditionu(0)

=

0

0

1114109

917501

costt(c)

Write

down

a

real

matrixA for

whichu =

Au

has

as

a

solutionu

=e

2

sint

1 1


0 2


other

trajectories

and

name

the

type

of

phase

portrait

you

have

(saddle

spiral

node

stableunstable)


of

birds

Birds

eat

bugs

and

the

two

together

satisfy

the

nonlinear

autonomous

system


y = (xminus

1)y

(so

that

in

the

absence

of

birds

the

bug

population

grows

logistically

and

in

the

absence

of




the

trajectories

near

that

critical

point


x = (2

minus

x minus

y minus



this

measure

successful

in

reducing

the

bug

population

7182019 Pr Final


1048573

SolutionstoExamI


(b)

q (t)

=

(01)(δ (t

minus 5)

+

δ (t

minus 15)

+

δ (t

minus 25)

+


2 The

g(x)

is

positive

for

x

lt

minus1

and

0

lt

x

lt

1

negative

for

minus1

lt

x

lt

0

and

x

gt

1

and

zeroatx=

minus10and1

k

0

xk

0

yk

1

Ak

=

xk

+

yk

1

hAk

1

(b)

h

=

1 1

2

1

2

11

122

12

142

12

142

3 3 1362

d

sint+c

3 (a)

The

equation

is

tx

=

cos

t

so

tx

=

cos

t

dt

=

sin

t

+

c

and

x

=

dt t

(b)Thedifferencex2

minus x1



+c(x2

minusx1)

4

(a)

|1

+

i|

=

radic 2

and

arg(1

+

i) =

π4

Therefore

|(1

+

i)21|

= 2212 =

1024radic 2


minus1024(1

+

i)

(b)

|8i =

8

and

arg(8i) =

π2

so

the

cube

roots

have

magnitude

2

and

arguments




4e(minus1+2i)t


=2minus 4i

2(1+2i) 2

x p

=Rez p


2] x p

= at2 + bt + c

2]

x p

= 2at

+

b

(b)

Undertermined

coefficients

1] x p

= 2a

2t2 +2 = 2at2 + (2b+4a)t + 2c+2b+2a



(s

+

1)2+

1

has

roots

minus1

plusmn i

so

basic

homogeneous

solutions

are

eminustcos

t

and

eminustsin

t

The



cos(t)

cos(3t)

cos(5t)6 (a)x p

= + + ω2

n

minus9) 25(ω2

n

minus1 9(ω2

n

minus25)


4

sin(3πt)


π

3


7 (a)

The

unit

impulse

response

is

a

homogeneous

solution

and

to

get

eminus2tsin(t)

the

roots



(b)

x

=

eminus2tsin(t)

lowast eminus2t =


=


dτ

0

=

et

minus2t sin(τ )

dτ

=


+

1)

0

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s


(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2


4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1


1


0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3


minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2


andminusa gt 0

ora lt minus3


a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where


x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4


1 1



1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0


3 0

J (0 3)=

minus3

minus6

haseigenvalues3and


7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



7182019 Pr Final


1048573

1114109

917501

1114109

917501

1114109

917501

1114109 917501


=x2Write

down

a

nonzero

solution

to

the

associated

homogeneous

linear

ODE

(in

terms

of

x1


and

x2)

2


10

=

1024)

(b)

Write

each

of

the

three

cube

roots

of

8i

in

the

form

a

+

bi

with

a

and

b

real



6 (a)

A

certain

periodic

function

f (t)

has

Fourier

series

given

by


=

cos(t)

+

+

9

25 +


What

is

a

periodic

solution

of

the

equation

x

+

ω2x

=

f (t)

(if

one

exists)

n

(b)

What

is

the

Fourier

series

of

the

function g(t)

which

is

periodic

of

period

2

and

such


7 (a)

For

what

values

of

c

and

k

does

the

LTI

operator

L

=

D2+

cD

+

kI

have

unit

impulse



minus2t with

rest

initial

conditions

and

evaluate

it

4

(c)

Find

the

inverse

Laplace

transform

of

F

(s)

=

(s2

+

2s

+

5)(s

+

1)

(d)

Sketch

the

pole

diagram

of

F

(s)


1 2


(a)

Write

down

a

fundamental

matrix

for

the

vector

equationu =

Au

(b)

Compute

eAt

0

(c)

Find

a

solution

tou =

Au

+

1

a

4



(a)

Asymptotically

stable

(b)

Defective

node


(d)Node

(e)

Saddle

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501


10 Two

ant

species

compete

with

each

other

for

the

same

food

Each

population

depresses


x =

(6

minus

2x

minus

y)x

y = (6

minus

x

minus

2y)y

(a)

Find

where

the

vector

field

is

horizontal

where

it

is

vertical

and

locate

the

critical

points


the

Jacobian

and

evaluate

it

at

that

critical

point

(c)

Sketch

the

phase

portrait

of

the

linearization

at

this

critical

point

Plot

any



system

in

the

upper

right

quadrant


ExamII

1 Salt

water

enters

a

twenty

gallon

tank

at

a

rate

of

five

gallons

per

minute

and

leaves

it


x(t)

for

the

number

of

pounds

of

salt

in

the

tank

at

time

t

and

suppose

that

at

t

=

0

the


is

q (t)

pounds

per

gallon

(a)

Write

down

a

differential

equation

that

controls

x(t)

(Measure

time

in

minutes)

(b)

Salt

is

added

to

the

tank

in

four

sudden

discrete

packets

of

half

a

pound

each

once

a



dx


2minusyandusethis

dx

to

sketch

the

graph

of

the

solution

with

y(minus1)

=

1

between

x

=

minus2

and

x

=

2

or

more


3

(a)

Find

the

general

solution

of

the

ODE

y1048573

minus

xy

=

2ex22

sin(x)

1 1



1

0

u =

Au

What

is

A

Find

a

solution

of u minus

Au

=

4

4 (a)

Express

cos(πt)

minus

sin(πt)

in

the

form

A

cos(ωt

minus

φ)


1

minus

i


uml5

(a)

Find

a

particular

solution

of

x

+

2

x

+

x

=

4et

+

5

7182019 Pr Final


1114109

917501



6 What

is

the

Fourier

series

for

f (x)

=

1

+

cos(x minus

π4)

7 (a)

What

is

the

unit

impulse

response

for

the

operator

D2 +

2D +

3I


(b)

The

unit

impulse

response

of

a

certain

system

is

given

by

w(t)

=

0

for

t lt 0


and

evaluate

it 1114109

917501

8 (a)

Compute

eAt where

A =

1

0

1

2

1114109

917501

1114109

917501

(b)

Solveu

=

Au

+

2et

etwith

initial

conditionu(0)

=

0

0

1114109

917501

costt(c)

Write

down

a

real

matrixA for

whichu =

Au

has

as

a

solutionu

=e

2

sint

1 1


0 2


other

trajectories

and

name

the

type

of

phase

portrait

you

have

(saddle

spiral

node

stableunstable)


of

birds

Birds

eat

bugs

and

the

two

together

satisfy

the

nonlinear

autonomous

system


y = (xminus

1)y

(so

that

in

the

absence

of

birds

the

bug

population

grows

logistically

and

in

the

absence

of




the

trajectories

near

that

critical

point


x = (2

minus

x minus

y minus



this

measure

successful

in

reducing

the

bug

population

7182019 Pr Final


1048573

SolutionstoExamI


(b)

q (t)

=

(01)(δ (t

minus 5)

+

δ (t

minus 15)

+

δ (t

minus 25)

+


2 The

g(x)

is

positive

for

x

lt

minus1

and

0

lt

x

lt

1

negative

for

minus1

lt

x

lt

0

and

x

gt

1

and

zeroatx=

minus10and1

k

0

xk

0

yk

1

Ak

=

xk

+

yk

1

hAk

1

(b)

h

=

1 1

2

1

2

11

122

12

142

12

142

3 3 1362

d

sint+c

3 (a)

The

equation

is

tx

=

cos

t

so

tx

=

cos

t

dt

=

sin

t

+

c

and

x

=

dt t

(b)Thedifferencex2

minus x1



+c(x2

minusx1)

4

(a)

|1

+

i|

=

radic 2

and

arg(1

+

i) =

π4

Therefore

|(1

+

i)21|

= 2212 =

1024radic 2


minus1024(1

+

i)

(b)

|8i =

8

and

arg(8i) =

π2

so

the

cube

roots

have

magnitude

2

and

arguments




4e(minus1+2i)t


=2minus 4i

2(1+2i) 2

x p

=Rez p


2] x p

= at2 + bt + c

2]

x p

= 2at

+

b

(b)

Undertermined

coefficients

1] x p

= 2a

2t2 +2 = 2at2 + (2b+4a)t + 2c+2b+2a



(s

+

1)2+

1

has

roots

minus1

plusmn i

so

basic

homogeneous

solutions

are

eminustcos

t

and

eminustsin

t

The



cos(t)

cos(3t)

cos(5t)6 (a)x p

= + + ω2

n

minus9) 25(ω2

n

minus1 9(ω2

n

minus25)


4

sin(3πt)


π

3


7 (a)

The

unit

impulse

response

is

a

homogeneous

solution

and

to

get

eminus2tsin(t)

the

roots



(b)

x

=

eminus2tsin(t)

lowast eminus2t =


=


dτ

0

=

et

minus2t sin(τ )

dτ

=


+

1)

0

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s


(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2


4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1


1


0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3


minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2


andminusa gt 0

ora lt minus3


a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where


x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4


1 1



1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0


3 0

J (0 3)=

minus3

minus6

haseigenvalues3and


7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501


10 Two

ant

species

compete

with

each

other

for

the

same

food

Each

population

depresses


x =

(6

minus

2x

minus

y)x

y = (6

minus

x

minus

2y)y

(a)

Find

where

the

vector

field

is

horizontal

where

it

is

vertical

and

locate

the

critical

points


the

Jacobian

and

evaluate

it

at

that

critical

point

(c)

Sketch

the

phase

portrait

of

the

linearization

at

this

critical

point

Plot

any



system

in

the

upper

right

quadrant


ExamII

1 Salt

water

enters

a

twenty

gallon

tank

at

a

rate

of

five

gallons

per

minute

and

leaves

it


x(t)

for

the

number

of

pounds

of

salt

in

the

tank

at

time

t

and

suppose

that

at

t

=

0

the


is

q (t)

pounds

per

gallon

(a)

Write

down

a

differential

equation

that

controls

x(t)

(Measure

time

in

minutes)

(b)

Salt

is

added

to

the

tank

in

four

sudden

discrete

packets

of

half

a

pound

each

once

a



dx


2minusyandusethis

dx

to

sketch

the

graph

of

the

solution

with

y(minus1)

=

1

between

x

=

minus2

and

x

=

2

or

more


3

(a)

Find

the

general

solution

of

the

ODE

y1048573

minus

xy

=

2ex22

sin(x)

1 1



1

0

u =

Au

What

is

A

Find

a

solution

of u minus

Au

=

4

4 (a)

Express

cos(πt)

minus

sin(πt)

in

the

form

A

cos(ωt

minus

φ)


1

minus

i


uml5

(a)

Find

a

particular

solution

of

x

+

2

x

+

x

=

4et

+

5

7182019 Pr Final


1114109

917501



6 What

is

the

Fourier

series

for

f (x)

=

1

+

cos(x minus

π4)

7 (a)

What

is

the

unit

impulse

response

for

the

operator

D2 +

2D +

3I


(b)

The

unit

impulse

response

of

a

certain

system

is

given

by

w(t)

=

0

for

t lt 0


and

evaluate

it 1114109

917501

8 (a)

Compute

eAt where

A =

1

0

1

2

1114109

917501

1114109

917501

(b)

Solveu

=

Au

+

2et

etwith

initial

conditionu(0)

=

0

0

1114109

917501

costt(c)

Write

down

a

real

matrixA for

whichu =

Au

has

as

a

solutionu

=e

2

sint

1 1


0 2


other

trajectories

and

name

the

type

of

phase

portrait

you

have

(saddle

spiral

node

stableunstable)


of

birds

Birds

eat

bugs

and

the

two

together

satisfy

the

nonlinear

autonomous

system


y = (xminus

1)y

(so

that

in

the

absence

of

birds

the

bug

population

grows

logistically

and

in

the

absence

of




the

trajectories

near

that

critical

point


x = (2

minus

x minus

y minus



this

measure

successful

in

reducing

the

bug

population

7182019 Pr Final


1048573

SolutionstoExamI


(b)

q (t)

=

(01)(δ (t

minus 5)

+

δ (t

minus 15)

+

δ (t

minus 25)

+


2 The

g(x)

is

positive

for

x

lt

minus1

and

0

lt

x

lt

1

negative

for

minus1

lt

x

lt

0

and

x

gt

1

and

zeroatx=

minus10and1

k

0

xk

0

yk

1

Ak

=

xk

+

yk

1

hAk

1

(b)

h

=

1 1

2

1

2

11

122

12

142

12

142

3 3 1362

d

sint+c

3 (a)

The

equation

is

tx

=

cos

t

so

tx

=

cos

t

dt

=

sin

t

+

c

and

x

=

dt t

(b)Thedifferencex2

minus x1



+c(x2

minusx1)

4

(a)

|1

+

i|

=

radic 2

and

arg(1

+

i) =

π4

Therefore

|(1

+

i)21|

= 2212 =

1024radic 2


minus1024(1

+

i)

(b)

|8i =

8

and

arg(8i) =

π2

so

the

cube

roots

have

magnitude

2

and

arguments




4e(minus1+2i)t


=2minus 4i

2(1+2i) 2

x p

=Rez p


2] x p

= at2 + bt + c

2]

x p

= 2at

+

b

(b)

Undertermined

coefficients

1] x p

= 2a

2t2 +2 = 2at2 + (2b+4a)t + 2c+2b+2a



(s

+

1)2+

1

has

roots

minus1

plusmn i

so

basic

homogeneous

solutions

are

eminustcos

t

and

eminustsin

t

The



cos(t)

cos(3t)

cos(5t)6 (a)x p

= + + ω2

n

minus9) 25(ω2

n

minus1 9(ω2

n

minus25)


4

sin(3πt)


π

3


7 (a)

The

unit

impulse

response

is

a

homogeneous

solution

and

to

get

eminus2tsin(t)

the

roots



(b)

x

=

eminus2tsin(t)

lowast eminus2t =


=


dτ

0

=

et

minus2t sin(τ )

dτ

=


+

1)

0

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s


(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2


4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1


1


0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3


minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2


andminusa gt 0

ora lt minus3


a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where


x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4


1 1



1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0


3 0

J (0 3)=

minus3

minus6

haseigenvalues3and


7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



7182019 Pr Final


1114109

917501



6 What

is

the

Fourier

series

for

f (x)

=

1

+

cos(x minus

π4)

7 (a)

What

is

the

unit

impulse

response

for

the

operator

D2 +

2D +

3I


(b)

The

unit

impulse

response

of

a

certain

system

is

given

by

w(t)

=

0

for

t lt 0


and

evaluate

it 1114109

917501

8 (a)

Compute

eAt where

A =

1

0

1

2

1114109

917501

1114109

917501

(b)

Solveu

=

Au

+

2et

etwith

initial

conditionu(0)

=

0

0

1114109

917501

costt(c)

Write

down

a

real

matrixA for

whichu =

Au

has

as

a

solutionu

=e

2

sint

1 1


0 2


other

trajectories

and

name

the

type

of

phase

portrait

you

have

(saddle

spiral

node

stableunstable)


of

birds

Birds

eat

bugs

and

the

two

together

satisfy

the

nonlinear

autonomous

system


y = (xminus

1)y

(so

that

in

the

absence

of

birds

the

bug

population

grows

logistically

and

in

the

absence

of




the

trajectories

near

that

critical

point


x = (2

minus

x minus

y minus



this

measure

successful

in

reducing

the

bug

population

7182019 Pr Final


1048573

SolutionstoExamI


(b)

q (t)

=

(01)(δ (t

minus 5)

+

δ (t

minus 15)

+

δ (t

minus 25)

+


2 The

g(x)

is

positive

for

x

lt

minus1

and

0

lt

x

lt

1

negative

for

minus1

lt

x

lt

0

and

x

gt

1

and

zeroatx=

minus10and1

k

0

xk

0

yk

1

Ak

=

xk

+

yk

1

hAk

1

(b)

h

=

1 1

2

1

2

11

122

12

142

12

142

3 3 1362

d

sint+c

3 (a)

The

equation

is

tx

=

cos

t

so

tx

=

cos

t

dt

=

sin

t

+

c

and

x

=

dt t

(b)Thedifferencex2

minus x1



+c(x2

minusx1)

4

(a)

|1

+

i|

=

radic 2

and

arg(1

+

i) =

π4

Therefore

|(1

+

i)21|

= 2212 =

1024radic 2


minus1024(1

+

i)

(b)

|8i =

8

and

arg(8i) =

π2

so

the

cube

roots

have

magnitude

2

and

arguments




4e(minus1+2i)t


=2minus 4i

2(1+2i) 2

x p

=Rez p


2] x p

= at2 + bt + c

2]

x p

= 2at

+

b

(b)

Undertermined

coefficients

1] x p

= 2a

2t2 +2 = 2at2 + (2b+4a)t + 2c+2b+2a



(s

+

1)2+

1

has

roots

minus1

plusmn i

so

basic

homogeneous

solutions

are

eminustcos

t

and

eminustsin

t

The



cos(t)

cos(3t)

cos(5t)6 (a)x p

= + + ω2

n

minus9) 25(ω2

n

minus1 9(ω2

n

minus25)


4

sin(3πt)


π

3


7 (a)

The

unit

impulse

response

is

a

homogeneous

solution

and

to

get

eminus2tsin(t)

the

roots



(b)

x

=

eminus2tsin(t)

lowast eminus2t =


=


dτ

0

=

et

minus2t sin(τ )

dτ

=


+

1)

0

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s


(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2


4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1


1


0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3


minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2


andminusa gt 0

ora lt minus3


a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where


x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4


1 1



1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0


3 0

J (0 3)=

minus3

minus6

haseigenvalues3and


7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



7182019 Pr Final


1048573

SolutionstoExamI


(b)

q (t)

=

(01)(δ (t

minus 5)

+

δ (t

minus 15)

+

δ (t

minus 25)

+


2 The

g(x)

is

positive

for

x

lt

minus1

and

0

lt

x

lt

1

negative

for

minus1

lt

x

lt

0

and

x

gt

1

and

zeroatx=

minus10and1

k

0

xk

0

yk

1

Ak

=

xk

+

yk

1

hAk

1

(b)

h

=

1 1

2

1

2

11

122

12

142

12

142

3 3 1362

d

sint+c

3 (a)

The

equation

is

tx

=

cos

t

so

tx

=

cos

t

dt

=

sin

t

+

c

and

x

=

dt t

(b)Thedifferencex2

minus x1



+c(x2

minusx1)

4

(a)

|1

+

i|

=

radic 2

and

arg(1

+

i) =

π4

Therefore

|(1

+

i)21|

= 2212 =

1024radic 2


minus1024(1

+

i)

(b)

|8i =

8

and

arg(8i) =

π2

so

the

cube

roots

have

magnitude

2

and

arguments




4e(minus1+2i)t


=2minus 4i

2(1+2i) 2

x p

=Rez p


2] x p

= at2 + bt + c

2]

x p

= 2at

+

b

(b)

Undertermined

coefficients

1] x p

= 2a

2t2 +2 = 2at2 + (2b+4a)t + 2c+2b+2a



(s

+

1)2+

1

has

roots

minus1

plusmn i

so

basic

homogeneous

solutions

are

eminustcos

t

and

eminustsin

t

The



cos(t)

cos(3t)

cos(5t)6 (a)x p

= + + ω2

n

minus9) 25(ω2

n

minus1 9(ω2

n

minus25)


4

sin(3πt)


π

3


7 (a)

The

unit

impulse

response

is

a

homogeneous

solution

and

to

get

eminus2tsin(t)

the

roots



(b)

x

=

eminus2tsin(t)

lowast eminus2t =


=


dτ

0

=

et

minus2t sin(τ )

dτ

=


+

1)

0

7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s


(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2


4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1


1


0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3


minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2


andminusa gt 0

ora lt minus3


a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where


x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4


1 1



1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0


3 0

J (0 3)=

minus3

minus6

haseigenvalues3and


7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

4

a(s +

1)

+

b c

(c)

F (s)

=

=

+

Multiply

through

by

the

first

de-(s2 +

2s +

5)(s +

1)

(s +

1)2 +

4 s +

1

4

nominator

and

set

s =

minus1

+2i(minus1

+

2i)

+

1

=

a(2i)

+

b

or

minus2i =

2ai+

b so

a =

minus1 b =

0

4

Multiply

through

by

the

second

denominator

and

set

s

=

minus1

1

minus 2

+

5

=

c

or

c

=

1

1

s

1

Thus

F (s)

=

minus(s +

1)

+

Write

G(s)

=

minus +

so

F (s) =

G(s +

1)

g(t) =(s +1)2 +4 s +1 s2 +4 s


(d)

There

are

poles

at

s =

minus1

plusmn 2i and

at

s =

minus1

2te 2eminust8 (a)

Φ(t)

=

2e2t eminust

(b)

Φ(0)

=

1 2

Φ(0)minus1 = 1

1 minus2

= 1

minus1 2

2 1

1

9175013 2

minus1

1114109

minus3

minus2


4eminust

4e2t

minuseminust

3

minus2e2t +2eminust

1 1


1


0 2 2

2 2 1 2

andA

=

or

putting

these

equations

side

by

side

in

a

matrix A

=

1

minus

1 1114109

2 11114109 917501

917501 1114109 917501 1114109 917501

1114109 917501

2 minus2 2

minus2 1 2 2

2 1

Thus A =

minus1=

minus2

and Aminus1 =

minus3


minus2 minus1

1

The

constant

solution

is

thus

minus

1

(a +

3)2 +

4a

9

pA(λ)

=

λ2

minus (a +

3)λ minus a

The

roots

are

(a +

3)

plusmn

They

are

repeated

2


andminusa gt 0

ora lt minus3


a =

0

10 (a)

The

vector

field

is

vertical

where

x =

0

x =

0

or

y =

6

minus 2x

It

is

horizontal

where


x

(b)J (x y)=

6

minus 4x minus y

minus soJ (22)=

minus4

minus2

minusy

6

minus x minus 4y

minus2

minus4


1 1



1

1

non-ray

trajectories

become

tangent

to

the

eigenline

through

minus1

as

t rarrinfin

6 0


3 0

J (0 3)=

minus3

minus6

haseigenvalues3and


7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



7182019 Pr Final


1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109

917501

1114109 917501

3 0

for

λ

=

3

is minus1

for

minus6

is

1

J

(3

0)

=

minus6 minus3

has

eigenvalues

minus6

and

3

and

gives

a

saddle

A

nonzero

eigenvector

0 3

1


minus1

0 3

(e)No

SolutionstoExamII

1 (a)

x +

(14)x

=

5q (t)

(b)q (t)

=

(12)(δ (t)

+δ (tminus 1)

+δ (tminus 2)

+δ (tminus 3)


k

0

2 (a)

h

=

5

1

2

xk

yk

Ak

=xk

+yk

hAk

1

2

1

5

15 25 275 1375

2

3875

(b)

Not

shown


0 1

(b)A= u=

minus4

1 0 0

4 (a)ω=πt A=

radic 2 φ=minusπ4

radic 2e

(b)i

iπ4 (c)

1

+

i

=

5 (a)

et +

5

(b)

eminus2t(a

cos(3t)

+

b

sin(3t))

(c)15


7 (a)


2t)

(b)

w(t)

lowast eminust =

0


=

(14)eminust(1

minus cos(2t))

t 2t

8 (a)

e minuset +e

0

e2t

2t

(b)

(t

minus1)et +e

minuset +

e2t

(c)

1 minus12

2 1

9 unstablenode

10 (a)(00)(20)(11)

(b)u =

minus1

minus1



pr final

Documents