36

Lesson Five

2.2 Non-Homogeneous Second Order Ordinary Differential Equations with Constant

Coefficients

In this section, we consider a second order ODE of the form

Where

0 , ,&

The solution to this differential equation is of the form

ℎ is the general solution of the corresponding homogeneous differential equation

0

and is known as the complementary function (CF)

on the other hand is the particular solution to the non-homogeneous differential

equation

and is known as the particular integral (PI)

There are three main methods of finding the PI. These are

1) Method of Undetermined coefficients

2) The D-Operator Method

3) Method of Variation of parameters

2.2.1: Method of Undetermined Coefficients (UC)

Definition

A function

is said to be a UC function if it is;

i) A function defined by one of the following

a. where is a non-negative integer.

b. ℎ a constant.

c. sin cos ℎ are constants and 0

ii) A function defined as a fine sum or product of two or more UC functions in

step (i) above

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Note: Given any UC function

then each successive derivative is itself a constant

multiple of UC function or a linear combination of UC functions.

Definition; Let

be a UC function, then the set of functions consisting of the function

and all linearly independent functions of which the successive derivatives

are either

constant multiples or linear combinations is known as a UC set.

Example 1

Given the function

%, find the UC set of

Solution

%

& 3

&& 6

&&& 6

)* 0

+, -%, , , 1/

Example 2

Given the function

cos 2, find the UC set of

Solution

cos 2

& 12 sin 2

&& 14 cos 2

&&& 8 cos 2

+, -cos 2 , sin 2/

Example 3

Given the function

%, find the UC set of

Solution

%

& 3%

&& 9%

&&& 27 %

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+, -%/

Definition

Given the second order differential equation

Where

is a UC function then the Particular integral 67 is a linear combination of

the elements of the UC set.

In general the particular integral is obtained as below

S.No

1

8

2

9sin

cos 8: cos 8 sin

3

, ∈ ℕ 8= 8:>: 8> ⋯ 8

Example 1

Find the general solution of the differential equation

1 5

6

Solution

The general solution is of the form

Where is the complementary function and is the particular integral.

C.F.

1 5

6 0

A.E.

A 1 5A 6 0

A

5 B √25 1 24

2

A: 3, A 2

:%

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P.I.

1 5

6

2

2

Substituting back, we get

2 1 52 6

2 1 5 6 6 1 10 6

Equating coefficients of the corresponding powers of we get

6 1

1

6

6 1 10 0

10

6

5

18

2 1 5 6 0

1

6

5 1 2

19

108

Substituting back we get

1

6

5

18

19

108

And the general solution becomes

:%

1

6

5

18

19

108

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Example 2

Find the general solution of the differential equation

9

1 12

4 5%

Solution

C.F.

9

1 12

4 0

A.E.

9A 1 12A 4 0

A

12 B √144 1 144

18

A

12

18

2

3

:

%

P.I.

9

1 12

4 5%

8%

38%

98%

Substituting back, we get

998% 1 1238% 48% 5%

818 1 368 48% 5%

8

5

49

8%

5

49

%

:

%

5

49

%

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Example 3

Find the general solution of the differential equation

1 4

29 4 cos 3

Solution

C.F.

1 4

29 0

AE

A 1 4A 29 0

A

4 B √16 1 116

2

A

4 B √1100

2

A 2 B 5

Thus the roots of the auxiliary equations are complex and the complementary function is

given by

D : 5 5

PI

1 4

29 4 cos 3

3 3

13 3 3 3

19 3 1 9 3

Substituting back we get

19 3 1 9 3 1 413 3 3 3 29 3 3 4 cos 3

20 1 12 3 12 20 3 4 cos 3

Equating coefficients, we get the equation

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20 1 12 4

12 20 0

Solving, we get

5

34

1

3

34

Thus the PI is given by

5

34

cos 3 1

3

34

sin 3

And the general solution is

: 5 5

5

34

cos 3 1

3

34

sin 3

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The D-Operator Method

We define the differential operator D by

E

Thus

E

Example

1) E 2

2) E4 1 8 E4 1 E8 4 1 0 4

3) E

4) E% 1 4 1 3 1 8

The Inverse Operator FG

H I

Consider the relation

E 3 2 3

Then pre-operating both sides by

:

J

, we have

1

E

KE 3L

1

E

2 3

1

E

⋅ E 3

1

E

2 3

or

3

1

E

2 3

1

E

2 3 3

From the above result, we note that

:

J

reverses the operation of D. Since D denotes

differentiation with respect to the variable then

:

J represents integration with respect to

the variable

ie

:

J N

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Example

1)

:

J

:

2)

:

J

3 :

% 3

3)

:

JO P :

J Q:

J

PR :

J F:

P PI :

P P

Theorem 1:

If a function of D operates on an exponential function () where is a constant then

the function remains unchanged but D is replaced by

Thus

SE-/ S-/

Example

1)

:

J

P :

P P

2)

:

JO % :

%O %

3)

:

JOTJ

U :

UOTU U :

U U

4)

:

JOTPJ>%

> :

>OTP>>%

> 1:

V >

Theorem 2:

Let W W be any function of alone then

SE-W/ SE -W/

Example

1) EX% %E 3 %6E E37 %2 0 2%

2)

:

JO>YJT:Z

U U F :

JTUO>YJTUT:ZI

1

E 1 8E 16

U U Q

1

E 8E 16 1 8E 1 32 16

R

1

E 1 8E 16

U U Q

1

ER U 1

E

⋅

1

E

U 1

E

[

%

3

\

U

12

U

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Theorem 3:

Whenever a function of E operates on either a cosine or a sine function ( ie

cos sin ), then function remains the same but E is replaced with 11 ⋅

SE-cos / S11 ⋅ cos

And

SE-sin / S11 ⋅ sin

Example

1) E 5 cos 2 112 5 cos 2 cos 2

2)

:

JOTU

sin 3 :

>:%OTU sin 3 :

>VTU sin 3 1 :

P sin 3

3) F :

JOTJTYI 3 :

>%OTJTY sin 3 :

J>: sin 3

We now multiply both the numerator and denominator by the conjugate of the

denominator

Q

1

E E 8

R 3 Q

1

E 1 1

R Q

E 1

E 1

R sin 3

Q

1

E E 8

R 3

E 1

E 1 1

sin 3

Q

1

E E 8

R 3

E 1

113 1 1

sin 3

Q

1

E E 8

R 3 1

1

10

E 1 sin 3

Q

1

E E 8

R 3 1

1

10

E sin 3 sin 3

Q

1

E E 8

R 3 1

1

10

3 cos 3 sin 3

4) Find the solution to the indefinite integral

] 3

Solution

] 3

1

E

cos 3

Applying theorem 2

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] 3 Q

1

E 2

R cos 3

Multiplying both the numerator by E 1 2 we get

] 3 Q

1

E 2

R Q

E 1 2

E 1 2

R cos 3

] 3 Q

E 1 2

E 1 4

R cos 3

Applying theorem 3

] 3 Q

E 1 2

113 1 4

R cos 3

] 3 1

1

13

E 1 2 cos 3

] 3 1

1

13

E cos 3 1 2 cos 3

] 3 1

1

13

13 sin 3 1 2 cos 3

] 3

1

13

3 sin 3 2 cos 3

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Solution to Second Order ODE by D-Operator Method

Example 1

Find the general solution to the differential equation

4

3

Solution

CF

4

3 0

AE

A 4A 3 0

A

14 B √16 1 12

2

A

14 B 2

2

A:

14 1 2

2

13

A

14 2

2

11

Thus the complementary function becomes

:>% >

Particular Integral

4

3

In operator form, we write

E^ 4E^ 3

E 4E 3^

Q

1

E 4E 3

R

Applying theorem 1, we get

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Q

1

2 42 3

R

1

15

Thus the general solution becomes

:>% >

1

15

Example 2

Find the general solution to the differential equation

3

1 7

2 2

Solution.

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