Section 2.2 Homogeneous with Constant coefficients SOLDEi

Pto Review i General and Fundamental Solutions

Part L Guessing Fundamental Solutions

Wednesday HCC SOLDE Theorem

Review Fundamental Solutions Them

Part 2 The Complex Solutions Case

qq.ggly

yo g comp q

o

Review General and Fu Solutions

Them If Y 72 with J f C Yz are SoCs of

LCI o LCT o

with LCY Y t a Ctl y t aocts Tthen every Sol Y of

aycan be written as

Jct c YCTI t Cz Y Ct L

where G Cz Cours Es

Remarks i Y Y above are fundamental solutions

y in L is the general solution

Guessing Fundamental Solutions

Example Find all sols of

Y t 5 Y t 6 I o

LCT

Sec We guess i L Y Co const or

rt21 Y f or 3 Y e or

4 Y sirs t or 5 Y coscat

1 LC.co co t 5 Co t 6 Co

6 Co

co 0 6 Co 0 120 0 1 Nor nonzeroconst c's

Failed solution

2 Lct th 5 th 6znh Cn c th 2 t 5 n th t 6 th

Find ns.t 2 th o all E

n Cn l t 2t 5h t t 6 th o all C

Check i No solution for info FaoCeo

3 L Cert rtg

rt6

rt

r2 ett g r cart 6 cartert r2 5 r 6

L Cert o r't 5 rt 6 O

Y ett is so l of Y t 5 Y 67 0

I Ir EIR is so2 of t6o

ascharacteristic equation

per r't 5 rt 6q

characteristic polynomial

r 511 54172 5211 YE I2

so ye 2u ageocs.fun.gl

LCT

So all socs of Ley o are

ctD8

HI SIDE Three

Det The differential of

Loy Y t a Y t do I O a uo consts

has the character y

per r2 t d r t do Oe

Thr mi Given an HCC SOL DE

Y t a Y t do I 0 1

Let r be the roots of

th t d r t do 0 2

A If of toe real or complex

Jae eun so a are

g Ct e g Ct er t

B tf tf E To

the fund Sols of CL are

Ct ebt yet zrot

Proofs A See example above i

y er L Cert ert art aoert

Cort r2 a r t do

O r2 t d r t do 2

Since rt are sols of 2 YI are SoCs of

B y ert o Lce't ert r2 ta r too

o r2tacrto

But now a do are such that

to c's the only root of

to t a Ro t do 0 Cbc

roosarootithatmeans

to r diI2

Therefore 1042 400 0 1which means froe.az L s 2roto O BL

to c's the only root

so Y er solves Loy o

We Look for a second solution of Loy 3 0 as

LCT o is Find a Ef fort

HIPS i since 7 is Sol of Lottothe ef for weight be

simpler than the cof for YT

Y ta Ys t do Yz o s

Vy t a cry t do Y o1 1Product Rule vY

t v1Y tVy

1 y t v y

V nd t V Y v Y try

V Y t 2 V Y t V Y 1So i

v Y t 2V Y t Vy t a V'Y.tv Y taoCrY so

Y t v 23 t d Y t V Y ta Y tacit O

Now Y e't y r Corot y rozerot

r

v t V 2 ta t V ro ta.ro too e O

v tV 2rotatVCro2ta rotao

but 0 0

to is the ro is a rootonly root

So V J the eq for V c's

tianya Ea.geThe sols are i

VCt cto c d consts

So i

Tact Vct Y Ct y CtS at ebt 1

The constants C d are arbitraryBut i If we choose D 0 that's bad

because

L cent funnfantentaSoCs

So we choose CEO OL L Then

wts te

BE

Ex ample iFond all possiblemovements of this

1 keg systemµ

4otb

Vy

SI Newton's Law i

Yy byY o Y

s i Y t6Ylt9Y

know i yetrt

ert 6 ert 9 Cert o

r2 t ar t 9 ert o

r2 t 6 r t 9 0

re 6 I 1 62407So2

611 222522322

rt r 3 1So the Ihrm above says all sols are

yct c e ttstE Me

Example i solve the IVP

Y t 3 Y 4 Y O J

Y Lo 2 Y co 3

Sofi Wo look for fund Sols of the form

Jct Ert

Cert 3 ert g ert o

r2 t 3 r 4 O

r 311 32 401412 311 9162 2

of 3252 rt I

Irso

4 12 5 1 fund SoCs

then lycts ctettc.EE general Soc

Initial Conditions Yo L Y co 3

Soi2 Yco C e t e e

12 4 2 1 a

3 Y'co C e t C4 C c

13 4421 b

e c 2 C and b

3 C 4 2 Ct

Ct 8 t

4938 5 Ct 3 Ct

C 2

2kg

E

Ic I

so yct et_ze I