Ch 3.6: Nonhomogeneous Equations; Method of Undetermined Coefficients

Recall the nonhomogeneous 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 nonhomogeneous equation, which relies on knowing solutions to homogeneous equation.

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0)()( ytqytpy

Theorem 3.6.1

If Y1, Y2 are solutions of nonhomogeneous equation

then Y1 - Y2 is a solution of the homogeneous equation

If y1, y2 form a fundamental solution set of homogeneous equation, then there exists constants c1, c2 such that

)()()()( 221121 tyctyctYtY

)()()( tgytqytpy

0)()( ytqytpy

Theorem 3.6.2 (General Solution)

The general solution of nonhomogeneous equation

can be written in the form

where y1, y2 form a fundamental solution set of homogeneous equation, c1, c2 are arbitrary constants and Y is a specific solution to the nonhomogeneous equation.

)()()()( 2211 tYtyctycty

)()()( tgytqytpy

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.

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)()()()( 2211 tYtyctycty

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)

Consider the nonhomogeneous equation

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

Substituting these derivatives into differential equation,

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

Thus a particular solution to the nonhomogeneous ODE is

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)

Consider the nonhomogeneous equation

We seek Y satisfying this equation. We begin with

Substituting these derivatives into differential equation,

Thus a particular solution to the nonhomogeneous ODE is

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)

Consider the nonhomogeneous equation

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.

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)()()( 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)

Consider the equation

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