math 3120 differential equations with boundary value problems chapter 4: higher-order differential...

Math 3120 Differential Equations

withBoundary Value

Problems

Chapter 4: Higher-Order Differential EquationsSection 4-4: Undetermined Coefficients – Superposition Approach

Method of Undetermined Coefficients

• Recall the non homogeneous equation where p, q, g are continuous functions on an open interval I.• The associated homogeneous equation is• In this section we will learn the method of undetermined

coefficients to solve the non homogeneous equation, which relies on knowing solutions to homogeneous equation.

The general solution of non homogeneous equation can be written in the form where y1, y2 form a fundamental solution set of homogeneous equation, c1, c2 are arbitrary constants and yp is a specific solution to the non homogeneous equation.•

0)()( ytqytpy

)()()( tgytqytpy

)()()()( 2211 tytyctycty p

Method of Undetermined Coefficients

• Recall the nonhomogeneous equation

with general solution

• In this section we use the method of undetermined coefficients to find a particular solution Y to the nonhomogeneous equation, assuming we can find solutions y1, y2 for the homogeneous case. • The method of undetermined coefficients is usually limited

to when p and q are constant, and g(t) is a polynomial, exponential, sine or cosine function.

)()()( tgytqytpy

)()()()( 2211 tytyctycty p

Example 1: Exponential g(t)

• Consider the nonhomogeneous equation

•We seek Y satisfying this equation. Since exponentials replicate through differentiation, a good start for Y is:

• Substituting these derivatives into differential equation,

• Thus a particular solution to the nonhomogeneous ODE is

teyyy 2343

ttt AetYAetYAetY 222 4)(,2)()(

2/136

346422

2222

AeAe

eAeAeAett

tttt

tetY 2

2

1)(

Example 2: Sine g(t), First Attempt (1 of 2)


•We seek Y satisfying this equation. Since sines replicate through differentiation, a good start for Y is:


• Since sin(x) and cos(x) are linearly independent (they are not multiples of each other), we must have c1= c2 = 0, and hence 2 + 5A = 3A = 0, which is impossible.

tyyy sin243

tAtYtAtYtAtY sin)(,cos)(sin)(

0cossin

0cos3sin52

sin2sin4cos3sin

21

tctc

tAtA

ttAtAtA

Example 2: Sine g(t), Particular Solution (2 of 2)

• Our next attempt at finding a Y is

• Substituting these derivatives into ODE, we obtain


tBtAtYtBtAtY

tBtAtY

cossin)(,sincos)(

cossin)(

17/3 ,17/5

053,235

sin2cos53sin35

sin2cossin4sincos3cossin

BA

BABA

ttBAtBA

ttBtAtBtAtBtA

tyyy sin243

tttY cos17

3sin

17

5)(

Example 3: Polynomial g(t)


•We seek Y satisfying this equation. We begin with



1443 2 tyyy

AtYBAttYCBtAttY 2)(,2)()( 2

8/11 ,2/3 ,1

1432,046,44

14432464

14423222

22

CBA

CBABAA

tCBAtBAAt

tCBtAtBAtA

8

11

2

3)( 2 tttY

Example 4: Product g(t)


•We seek Y satisfying this equation, as follows:

• Substituting derivatives into ODE and solving for A and B:

teyyy t 2cos843

teBAteBA

teBA

teBAteBAteBAtY

teBAteBA

tBetBetAetAetY

tBetAetY

tt

t

ttt

tt

tttt

tt

2sin342cos43

2cos22

2sin22sin222cos2)(

2sin22cos2

2cos22sin2sin22cos)(

2sin2cos)(

tetetYBA tt 2sin13

22cos

13

10)(

13

2 ,

13

10

Discussion: Sum g(t)

• Consider again our general nonhomogeneous equation

• Suppose that g(t) is sum of functions:

• If Y1, Y2 are solutions of

respectively, then Y1 + Y2 is a solution of the nonhomogeneous equation above.

)()()( tgytqytpy

)()()( 21 tgtgtg

)()()(

)()()(

2

1

tgytqytpy

tgytqytpy

Example 5: Sum g(t)

• Consider the equation

• Our equations to solve individually are

• Our particular solution is then

teteyyy tt 2cos8sin2343 2

tetettetY ttt 2sin13

22cos

13

10sin

17

5cos

17

3

2

1)( 2

teyyy

tyyy

eyyy

t

t

2cos843

sin243

343 2

Example 6: First Attempt (1 of 3)


•We seek Y satisfying this equation. We begin with

• Substituting these derivatives into ODE:

• Thus no particular solution exists of the form

tyy 2cos34

tBtAtYtBtAtY

tBtAtY

2cos42sin4)(,2sin22cos2)(

2cos2sin)(

t

ttBBtAA

ttBtAtBtA

2cos30

2cos32cos442sin44

2cos32cos2sin42cos42sin4

tBtAtY 2cos2sin)(

Example 6: Homogeneous Solution (2 of 3)

• Thus no particular solution exists of the form

• To help understand why, recall that we found the corresponding homogeneous solution in Section 3.4 notes:

• Thus our assumed particular solution solves homogeneous equation

instead of the nonhomogeneous equation.

tBtAtY 2cos2sin)(

tctctyyy 2sin2cos)(04 21

tyy 2cos34

04 yy

Example 6: Particular Solution (3 of 3)

• Our next attempt at finding a Y is:

• Substituting derivatives into ODE,

tBttAttBtA

tBttBtBtAttAtAtY

tBttBtAttAtY

tBttAttY

2cos42sin42sin42cos4

2cos42sin22sin22sin42cos22cos2)(

2sin22cos2cos22sin)(

2cos2sin)(

tttY

BA

ttBtA

2sin4

3)(

0,4/3

2cos32sin42cos4

tyy 2cos34

Ch 4.3: Nonhomogeneous Equations: Method of Undetermined Coefficients

• The method of undetermined coefficients can be used to find a particular solution Y of an nth order linear, constant coefficient, nonhomogeneous ODE

provided g is of an appropriate form. • As with 2nd order equations, the method of undetermined

coefficients is typically used when g is a sum or product of polynomial, exponential, and sine or cosine functions. • Section 4.4 discusses the more general variation of

parameters method.

),(1)1(

1)(

0 tgyayayayayL nnnn

Example 1

• Consider the differential equation

• For the homogeneous case,

• Thus the general solution of homogeneous equation is

• For nonhomogeneous case, keep in mind the form of homogeneous solution. Thus begin with

• As in Chapter 3, it can be shown that

teyyyy 433

teAttY 23)(

010133)( 323 rrrrety rt

tttC etctececty 2

321)(

ttttt etetctecectyettY 232321

23

3

2)(

3

2)(

Example 2




• For the nonhomogeneous case, begin with

• As in Chapter 3, it can be shown thattBtAtY cossin)(

tttY cos3

1sin

9

2)(

ttyyy cos3sin2168)4(

0440168)( 224 rrrrety rt

ttcttctctcty 2sin2cos2sin2cos)( 4321

Example 3


• As in Example 2, the general solution of homogeneous equation is

• For the nonhomogeneous case, begin with

• As in Chapter 3, it can be shown thattBttAttY 2cos2sin)( 22

tttttY 2cos32

32sin

16

1)( 22

ttyyy 2cos32sin2168)4(

ttcttctctcty 2sin2cos2sin2cos)( 4321

Example 4




• For nonhomogeneous case, keep in mind form of homogeneous solution. Thus we have two subcases:

• As in Chapter 3, can be shown that

tetyy 39

,)(,)( 321

tCtetYtBtAtY

033909)( 23 rrrrrrrety rt

ttC ececcty 3

33

21)(

tettYttY 32

21 18

1)(,

18

1)(

math 3120 differential equations with boundary value problems chapter 4: higher-order differential...

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