Reminders a Problem Set 10 is postedproblem Set 11 will be due on

the last day of class

Exam 2 will take place thisThursday in

the following mannerAt 6pm on

Thursday you will

receive the examvia email you

must submit theocean by 6pm on

FridayIf you foresee this

presents

Scheduling problems emailme

immediately

todayWronski au 3.2 3 6Mechanical and electrical vibrations

3.7

Undetermined coefficients 3.25and

Vaiak of parameter3 6

Wronsk ian

Consider the equakinatt c bet it CH U O 7

Suppose you havetwo solutm u Lt and

zit and you want to solve the

initial value problemfor with

Uco Uo ie co ie

Uo Io Sone given number

we look for aSouther of the fon

Ult C UH Cz Udt

Ci Cz thumhento be determined

The numbers Ci Czare given by the

system of algebraic equatuC U Cos Cz UzCo7 Uo

e ie Ca c Cz UID lio

There is one condition onthe coefficients

Udo Uz lo ieK cider that

guarantees this always has oneand

exactly are soluhr 9,4

CaioThe condition is

der I o

The Wronskian of two solutions

u and uz 13the functor of

t given by

triumitt dett Ya

U A Walt il A Udt

Exera.se SeeProklensetWe have seen Thor

for a Z d

system I AX

and two fundamentalsolders 71,72

that theirwronshian f der five 5k

solves the equatorW tr A W

when for A 94 91 the

tr A au 1922

bell

E For each equation bellae

let's find a set of fundamentalHf

solutions and compete their Wronskian

1 is 144 0Characteristicpolynomial d 4Roots are 112 4

0 X I 2 I

Sit and eZit are solution their

real and imaging pantsare trees also

sobbing u wee getµ it cos

Zt Uz Ct sink 1

Their Wronski an i's

D u Uz t Ulitz El Uzcos Zt 2Costa f zsin

sink

zcoscztf zsinczc72z.to

Hi cuz it 2

2 it Zvi Yu Ocharacteristic polynomial 12

211 4

Roots use quadraticformula D 2IF2

l I I V127I I jI iT t arealpart imaginaryport

Then we haveU Lt e cos Dt

Uch e thin Pt

Wronski anEr Its etoes t B e

tsin Dt

insect Etsi Dt Bet us taftersomecancellations

u iz U 42 e cos 32e si e

Z

Z tBe 2T

True uz it Be

3 it 5 it 4 u O TAK2 154 4 0

moons51 25102

Note boaEhueigseamaml.es

5235 4utt

Uz It eU It e utu H e

t ie Ct Ye

wit et fue at f Et e

ut

5TMtle51

3 e

u ii 3 at 44 0

d2 3 4 0

moon 3 0296 31722

Zz I IA

Tal parTimagm past

wilt e Ztcos htt

Uz It e Etsi AtTutti exercise

A autbn

The charactersher equatorfor

This equateis

a Rtbdt C 0

The roots are

D EaVb2

za

Xz Fa Visz a

Note that the sign of5 Uac

determines if the roots are real

or not

If b 0 anda c 30 Then

the roots are purely imaginary

Ii aand solutions are

ascfat in fatIf b 0 and a c

0 then if

b is smell we willget tf Ya Ceo

and the roots well de

de Ia iVua

29

112 Ia ib2Za

If we unite ft FaW

Uac 52Ta

the solutes beam

u A enters cwt uz Henthicwt

functors oh

so there are functus oscillating butalso deaagins exponentiallylastly if bSO and sufficiently lazethem b2 Uac

O and the roots willbe real no oscillation

Damping tone andrestorative

restorativeforce forcemoi K x k

o

Hooke's Law

m I e KK 25gdamp.rs frichn

Iq9mneeThe results quite has thefirm Moitra Kx 0

ahu Mgr KH

In summaryNo damping b o

perfectly oscillatingsolutes

Some damp'sb 0 small oscillatingexponential

decay

More daeemi b 0 laze wemore

oscillate purelyexponentialbehave

One moreterm

m KK JI fitExternefForce

This leads to consider

inhomogeneous equationsacit bit cu fees

InhomogeneousequationTwo methods

Dyariatu of parameter2 Undetermined coefficients

Vor of parameters Two waysof thinking about

it

The books's Sechin 3.6

Redneck to 2 d systemsUw of pants in Chapter 7

I'll discuswhen dealing

with

a t bit cue fees

one approach is toconsider

the corresponding2 d system

iI

ex Ea ba xt Isto the system we

can applyaide of parameters get

a solute xH7 of which

the first componentwill be

the solution tothe 2nd order

We get the followingif

UK weltsare a fondurutd

ergshen ofsolutes to

a iitbie tea D

Then a particular solutesto

aiitb.ci Ccn fris green by

Upa

f sina.I.fisds um iI d was

The guard solute isRen

Singhacts C UCH 1 Cz

Uit t 4pctI