Ch03!1!3_4 [Compatibility Mode]

8/13/2019 Ch03!1!3_4 [Compatibility Mode]

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E uations and Boundar

Value Problems

Boyce. & DiPrima, 9th Edition

Chapter 3:

Second Order Differential E uations

1006003 31/2555

....1


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Topics

omogeneous qua ons- ons an oe c en s

Fundamental Solutions of Linear

Complex Roots of Characteristic Equation

Nonhomogeneous Equations; Method of

Variation of Parameters

2


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3.1: 2nd Order Linear Homogeneous EquationsConstant Coefficientsonstant CoefficientsElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, 2009 by John Wiley & Sons, Inc.

general form

),,( yytfy

wherefis some given function.

This equation is said to be linear iffis linear iny andy':

Otherwise the equation is said to be nonlinear.

ytqytptgy )()()(

A secon or er near equat on o ten appears as)()()()( tGytRytQytP

, .

Otherwise the equation is nonhomogeneous.3


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Homogeneous Equations, Initial Values

. . ,

homogeneous equation is found, then it is possible to solvethe corresponding nonhomogeneous equation, or at least

express t e so ut on n terms o an ntegra .

The focus of this chapter is thus on homogeneous equations;

and in articular those with constant coefficients:

We will examine the variable coefficient case in Chapter 5.

0 cyybya

Initial conditions typically take the form

0000 , ytyyty

Thus solution passes through (t0,y0), and slope of solution at

(t0,y0) is equal toy0'.4


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Example 1: Infinitely Many Solutions (1 of 3)

Two solutions of this e uation are

0 yy (1)

Other solutions include

ttetyety

)(,)( 21 (2)

Based on these observations, we see that there are infinitely

tttt eetyetyety 53)(,5)(,3)( 543 (3)

tt ececty 21)( (4)

.

differential equation above can be expressed in this form.

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Example 1: Initial Conditions (2 of 3)

equation:1)0(,2)0(,0 yyyy (1)

e ave oun a genera so ut on o t e ormtt ececty 21)(

(2) ,

2/3,2/110

2)0(21

21

cc

cc

ccy(3)

Thustt eety

2/32/1)( (4)

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Example 1: Solution Graphs (3 of 3)

Gra hs of both t and t are iven below. Observe that

tt eetyyyyy 2/32/1)(1)0(,2)0(,0

both initial conditions are satisfied.

y t y' ttt eety 2/32/1)( tt eet 2/32/1'

3

2

3

1

0.5 1.0 1.5 2.0t

1

0.5 1.0 1.5 2.0t 1

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Characteristic Equation

nd ,

we begin by assuming a solution of the form y = ert.

0 cyybya (1)

Substituting this into the differential equation, we obtain

02 rtrtrt cebreear (2)

mp y ng,

and hence0)( 2 cbrarert (3)

This last equation is called the characteristic equation of

02 cbrar (4)

e eren a equa on.

We then solve for rby factoring or using quadratic formula.8


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General Solution

we obtain two solutions, r1 and r2.,0

2

cbrar (1)

There are three possible results:The roots r1, r2 are real and r1r2.

The roots r , r are real and r = r .

acbbr

42

The roots r1, r2 are complex.

In this section, we will assume r1, r2 are real and r1r2.

,trtr

ececty 21 21)( (2)

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Initial Conditions , r1, r2 are real, r1r2

,)(,)(,0 0000 ytyytycyybya (1)

together with the initial conditions to find c1 and c2. That is,

trtrececty 21 21)( (2)

0201

0201

0201

21

0102

21

2001

02211

021,

trtr

trtr

trtr

err

yryce

rr

ryyc

yercerc

yecec

Since we are assuming r1r2, it follows that a solution of theformy = ertto the above initial value problem will always

ex s , or any se o n a con ons.

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Example 2 (General Solution)

Consider the linear differential equation

065 yyy (1)

equation:

032065)( 2 rrrrety rt

Factoring the characteristic equation yields two solutions:

r1 = -2 and r2 = -3

T ere ore, t e genera so ut on to t s erent a equat onhas the form

ttecect

32

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Example 3 (Particular Solution)

From the recedin exam le we know the eneral solution 30,20,065 yyyyy (1)

has the form:

With derivative:

ttececty

3

2

2

1)(

tt ececty 3221 32)('

(2)

Using the initial conditions:

121 cc2.5

y t tteety

32 79)(

,332 2121

cc

tt 32

1.0

1.5

.

(4)

0.0 0.5 1.0 1.5 2.0 2.5t

.

12

(5)


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Example 4: Initial Value Problem

Then 2/10,20,0384 yyyyy

Factoring yields two solutions, r1 = 3/2 and r2 = 1/2

012320384)( 2 rrrrety rt

The general solution has the form2/

2

2/3

1)( tt

ececty 3y t

2/2/32/52/1

tteet

s ng n t a con t ons:

2/5,2/12

21

21

cc

cc 1

2

Thus2/2/3

2/52/1)(

tt

eety

0.5 1.0 1.5 2.0 2.5

1

13


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Example 5: Find Maximum Value

or t e n t a va ue pro em n xamp e ,

to find the maximum value attained by the

so ut on, we sety t = an so ve or t:

79)(

32

eety set

tt

y t (1)76

02118)(

32

32

ee

eety

tt

tt

1.5

2.0

.tt

eety32 79)(

(2)

)6/7ln(67

te

0.5

1.0

204.2

.

y

0.0 0.5 1.0 1.5 2.0 2.5t

14(4)


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3.2: Fundamental Solutions of LinearHomogeneous Equationsomogeneous EquationsElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, 2009 by John Wiley & Sons, Inc.

=, , ,

which could be infinite. For any functiony that is twicedifferentiable onI, define the differential operatorLby

Note thatL[y] is a function onI, with output value

yqypyyL

)()()()()()( tytqtytptytyL

,

2,0),sin()(,)(,)(22

22Ittyetqttp

t

t

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Differential Operator Notation

homogeneous equationL[y](t) = 0, along with initialconditions as indicated below:

-(2)---)(,)(

(1)---0)()(

1000 ytyyty

ytqytpyyL

We would like to know if there are solutions to this initial

value problem, and if so, are they unique.

so, we wou e to now w at can e sa a out t e ormand structure of solutions that might be helpful in finding

solutions to articular roblems.

These questions are addressed in the theorems of this section.

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Theorem 3.2.1 (Existence and Uniqueness)

)()()( tgytqytpy

(1)

wherep, q, and g are continuous on an open intervalIthat

containst. Then there exists a uni ue solution = t onI.

0000 , (2)

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)()()( tgytqytpy

Example 1 1000 )(,)( ytyyty

11,21

,0)3(')3( 2

yy

yttyytt

(1)

)()(')('' tgytqytpy

(2)

The only points of discontinuity for these coefficients are

t= 0 and t= 3. So the longest open interval containing the

,(3)

initial point t=1 in which all the coefficients are continuousis 0 < t< 3

, . .

guarantees the existence of the solution is 0 < t < 3

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Theorem 3.2.2 (Principle of Superposition)

0)()(][ ytqytpyyL

,

all constants c1 and c2.

To prove this theorem, substitute c1y1 +y2c2 in fory in theequation above, and use the fact thaty1 andy2 are solutions.

1 2,infinite family of solutions, each of the formy = c1y1 + c2y2.

Can all solutions can be written this wa , or do some

solutions have a different form altogether? To answer this

question, we use the Wronskian determinant.19


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The Wronskian Determinant (1 of 3)

0)()(][ ytqytpyyL ------- (1)

rom eorem . . , we now a y = c1y1 c2y2 s a

solution to this equation.

Next find coefficients such that = c + c satisfies theinitial conditions

0000 )(,)( ytyyty ------- (2)

To do so, we need to solve the following equations:

0022011 )()( ytyctyc ------- (3)

0022011 )()( ytyctyc

20

------- (4)


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0022011

0022011

)()(

)()(

ytyctyc

ytyctyc


,

)()()()(

)()( 0200201

tytytyty

tyytyyc

(1)

)()()()(

)()(

02010201

0100102

tytytyty

tyytyyc

(2)

In terms of determinants:

tt

)()(

)(,

)()(

)(

0201

001

2

0201

020

1tyty

ytyc

tyty

tyyc

(3)

)()()()( 02010201 tytytyty

21


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,

the denominator cannot be zero:ytytyy 001020 )()(

W

ytyc

W

tyyc

001

2

020

1

)(,

)(

)()()()()()(

)()(02010201

0201

0201tytytyty

tyty

tytyW

(2)

Wis called the Wronskian determinant, or moresimply, the Wronskian of the solutionsy1andy2. We will

021,yy

22


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Example 3

n xamp e o ec on . , we oun a

tt etyety 3221 )(and)( (1)

The Wronskian of these two functions is

065 yyy (2)

t

tt

tt

eee

eeW 5

32

32

32

(3)

Since Wis nonzero for all values of t, the functions

can be used to construct solutions of the differential21 andyy

equat on w t n t a con t ons at any va ue o t

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Theorem 3.2.4 (Fundamental Solutions)

Then the famil of solutions.0)()(][ ytqytpyyL

y = c1y1 + c2y2

with arbitrary coefficients c1, c2 includes every solution tothe differential equation if an only if there is a point t0 such

that W(y1,y2)(t0) 0, .

e express ony = c1y1 + c2y2 s ca e t e genera so ut onof the differential equation above, and in this casey1 andy2are said to form a fundamental set of solutions to the

differential equation.

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Example 4: fundamental set of solutions?

,

with the two solutions indicated:0)()( ytqytpy

Suppose the functions below are solutions to this equation:

2121 ,,21

rreyey

trtr

The Wronskian ofy1andy2 is

.allfor02121

12

21terr

eeyyW

trr

trtr

Thusy1andy2 form a fundamental set of solutions to the

e uation, and can be used to construct all of its solutions.

2121 ereryy

The general solution is trtr ececy 21 21

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Example 5: Solutions (1 of 2)

Show that the functions below are fundamental solutions:0,032 2 tyytyt (1)

To show this, first substitutey1 into the equation:

1

2

2/1

1 , tyty (2)

012

3

2

1

23

42 2/12/1

2/12/32

tt

tt

tt (3)

Thusy1 is a indeed a solution of the differential equation.Similarly,y2 is also a solution:

0134322 11232 tttttt

27

(4)


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Example 5: Fundamental Solutions (2 of 2)

To show that and form a fundamental set of solutions

122/11 ,

tyty

we evaluate the Wronskian ofy1 andy2:

2/32/32/3

12/1

21 331ttyy

322/1

21 2222 tttyy

,1,

2

solutions for the differential equation

0,032 2 tyytyt

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Theorem 3.2.5: Existence of Fundamental Set

of Solutions

,

p and q are continuous on some open intervalI:0)()(][ ytqytpyyL

Let t0be a point inI, andy1 andy2 solutions of the equation

withy1 satisfying initial conditions

andy2 satisfying initial conditions

0)(,1)( 0101 tyty

Theny1,y2 form a fundamental set of solutions to the given1)(,0)( 0202 tyty

.

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Example 6: Apply Theorem 3.2.5 (1 of 3)

. .

differential equation and initial point0,0 0 tyy (1)

n ect on . , we oun two so ut ons o t s equat on:

The Wronskian of these solutions is W(y1,y2)(t0) = -2 0 so

tteyey

21 , (2)

they form a fundamental set of solutions.

But these two solutions do not satisfy the initial conditions

. . ,

fundamental set of solutions mentioned in that theorem.

e y3 an y4 e e un amen a so u ons o m . . .

1)0(,0)0(;0)0(,1)0( 4433 yyyy 30(3)


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Example 6: General Solution (2 of 3)

,

1)0(,0)0(,0)0(,1)0(,

44214

33213

yyededyyyececy

tt

tt

(1)2

Solving each equation, we obtain

)sinh(

11

)(),cosh(

11

)( 43 teetyteety tttt

(3) (4)The Wronskian ofy3 andy4 is

sinhcosh2221

ttyy

Thusy3,y4 form the fundamental set of solutions indicated incoshsinh21 ttyy

. . ,

)sinh()cosh()(21

tktkty

31(5)


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Example 6:

Many Fundamental Solution Sets (3 of 3)

both form fundamental solution sets to the differential ttSeeS

tt

sinh,cosh,, 21

(1)

equation and initial point

0,0 0 tyy (2)

In general, a differential equation will have infinitely many

. ,

one that is most convenient or useful.

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Summary

tytqytpy ,0)()( .

Then make sure there is a point t0 in the interval such that

W(y1,y2)(t0) 0.It follows thaty1 andy2 form a fundamental set of solutions

to the equation, with general solutiony = c1y1 + c2y2.

0where W 0, then c1 and c2 can be chosen to satisfy those

conditions.

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3.3: Complex Roots of Characteristic

Elementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, 2009 by John Wiley & Sons, Inc.

where a b and c are constants.0 cyybya (1)

Assuming an exponential soln leads to characteristic equation:

0)(2

cbrarety rt

(2)Quadratic formula (or factoring) yields two solutions, r1 & r2:

acbbr

42 (3)

If b2 4ac < 0, then complex roots: r1 = + i, r2 = -ia2

titietyety

)(,)( 2134

(4)


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Eulers Formula; Complex Valued Solutions

t ,

formula:tit

ti

tite

nnnnnit sincos

)1()1()( 1212

Generalizing Eulers formula, we obtain

nnn nnn !12!2! 100 (1)

Then

tite ti sincos

Therefore

tietetiteeee sincossincos

(3)

tieteety

teteety

ttti

sincos)(

s ncos

2

1

35

(4)


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Real Valued Solutions

-

0 cyybya

tietetytt

tt

sincos)(1 (5)

We wouldprefer to have real-valued solutions, since our

differential e uation has real coefficients.

2

To achieve this, recall that linear combinations of solutions

are themselves solutions:

tietytytetyty

t

sin2)()(cos2)()(

21

21

(6)

gnor ng cons an s, we o a n e wo so u ons

tetytety tt

sin)(,cos)( 43 36(7)


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Real Valued Solutions: The Wronskian

-

tetytety tt sin)(,cos)( 43 (8)

ec ng t e rons an, we o ta n

sincos

tt

tttete

W

Thus and form a fundamental solution set for our ODE02

t

e (9)

and the general solution can be expressed as

0 cbatectecty

tt sincos)( 21

37a

acbb

r 2

42

ir 2,1,(10)


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Example 1 (1 of 2) tectecty s ncos 21

For an ex onential solution the characteristic e uation is025.9 yyy

(1)

ii

rrrety rt

2

3

2

1

2

31

2

41101)( 2

2Therefore, separating the real and imaginary components,

2/3,2/1 (3)

)(2

3sin

2

3cos

2

3sin

2

3cos)( 21

2/2/

2

2/

1

tctcetectecty ttt

38

38

(4)


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Example 1 (2 of 2)

s ng e genera so u on us e erm ne

)(23sin

23cos)( 21

2/

tctcety

t (5)

We can eterm ne t epart cu ar so ut on t at sat s es t e

initial conditions 8)0('and2)0( yy (6)

o3,2

83)0(21

21

1

2/1

cc

ccy

cy

3

y t

)2( 3sin33cos)( 2/ ttety t

2 4 6 8 10t

1

2

)2(2

3sin3

2

3cos)( 2/

ttety t (7)

The solution is a decaying oscillation 2

1

3939


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Example 3

09 yyirret

rt 3092

Then

3,0 Therefore

and thus the eneral solution is

4

y t tctcty 3sin3cos)( 21 ,0 tty

tty

3sin2/13cos:dashed

3sin23cos2:solid

Because there is no exponential

2 4 6 8t

2factor in the solution, so the amplitudeof each oscillation remains constant.

2

typical solutions

41


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3 4: Repeated Roots; Reduction of Order.4: Repeated Roots; Reduction of OrderElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, 2009 by John Wiley & Sons, Inc.

nd

where a b and c are constants.0 cyybya

Assuming an exponential soln leads to characteristic equation:

0)(

2

cbrarety rt

Quadratic formula (or factoring) yields two solutions, r1 & r2:

acbbr

42

When b2 4ac = 0, r1 = r2 = -b/2a, since method only gives

a

atb

cety 2/1 )(

42


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Second Solution: Multiplying Factor v(t)

Since and are linearl de endent we eneralize thissolutiona)()(solutiona)( 121 tcytyty

approach and multiply by a function v, and determine

conditions for whichy2 is a solution:atbatb 2/2/

Then

evyey 21 ryso u ona

atb 2/

atbatbetv

a

betvty

2/2/

2

2

)(2

)()(

atbatbatbatb etv

a

betv

a

betv

a

betvty

2/

2

22/2/2/

2 )(

4

)(

2

)(

2

)()(

43


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0 cyybya

Finding Multiplying Factor v(t)

,

2

22/ 0)()(2

)()(4

)()( tcvtva

btvbtva

btva

btvae atb

22

0)()(2

)()(4

)()( tcvtva

btvbtv

a

btvbtva

222

0)(24

)( tvca

b

a

btva

2 4

0)(44

)(0)(444

)(

acb

tvaa

tvatvaaa

tva

43)(0)(

4

ktktvtv

a

44


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General Solution

,

abtabt

abtabt etvkekty2/2/

2/2

2/1 )()(

abtabttecec

ee

2/

2

2/

1

431

Thus the general solution for repeated roots is

abtabttececty

2/

2

2/

1)(

0 cyybya

acbb 42 = = -

45

ar

2 ,


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Wronskian

Thus every solution is a linear combination of

abtabt

tececty

2/

2

2/

1)(

The Wronskian of the two solutions is

abtabttetyety

2/

2

2/

1 )(,)(

ea

bte

a

btee

tyyW abtabt

atat

21

2

))(,( 2/2/21

abte

abte

abt

abtabt

221

/

//

Thusy1 andy2 form a fundamental solution set for equation.

e

46


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Example 1 (1 of 2)

Assumin ex onential soln leads to characteristic e uation:044

yyy(1)

So one solution is and a second solution is found:

20)2(044)( 22 rrrrety rt

t

ety

2

1 )(

(2)

ttt

ttetvetvty

etvty

222

22

2

2

)(2)()(

)()(

(3)

Substituting these into the differential equation and

2

'" where are arbitrary constants.

,,

21 andkk

4747


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Example 1 (2 of 2)

So the general solution is tttececty

22)(

221 ,

(5)

Note that both tend to 0 as regardless of the

values of21 andyy t

21 andcc

s ng n a con ons

1

3)0('and1)0(

1

c

yy1.5

2.0y t

tetty

2)51()(

Therefore the solution to the

,

32

21

21 cc

0.5

1.0

(6)

IVP is tt

teety 5)( 0.0 0.5 1.0 1.5 2.0 2.5 3.0

t

4848


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Example 2 (1 of 2)

Assumin ex onential soln leads to characteristic e uation: 3/10,20,025.0

yyyyy

Thus the general solution is

2/10)2/1(025.0)( 22 rrrrety rt

Using the initial conditions:

2/

2

2/

1)( tt

tececty

2

3

4y t

)3/22()( 2/ tety t

32,2

3

1

2

1 2121

1

cccc

1 2 3 4t

1

Thus 2/2/3

2)( tt teety 2

1

49


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Example 2 (2 of 2)

uppose a e n a s ope n e prev ous pro em was

increased 20,20 yy (1)

T e so ut on o t s mo e pro em s

6

y t 2/2/2)( tt teety (2)o ce a e coe c en o e secon

term is now positive. This makes a big

difference in the graph, since the 2

4

)3/22()(:blue

)2()(:red

2/

2/

tety

tety

t

t

exponential function is raised to apositive power:

1 2 3 4t

2

02/1

5050(7)

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Ch03!1!3_4 [Compatibility Mode]