AMM 304: Ordinary Differential Equations

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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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 %
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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
 :%  
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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P.I.

 1 5


 6 
     


2  

 2
Substituting back, we get
2 1 5 2    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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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
9 98% 1 12 38%  4 8% 5%
818 1 368  48 % 5%
8
5
49
 8%
5
49
%
 :   
%
 
5
49
%
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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 4 13  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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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The D-Operator Method
We define the differential operator D by
E


Thus
E


Example
1) E  2
2) E 4 1 8 E 4 1 E 8 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   3 L
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 
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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Example
1)
:
J
 :
 
2)
:
J
 3 :
%  3
3)
:
JO P :
J Q:
J
P R :
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
S E -/ S  -/
Example
1)
:
J
P :
P P
2)
:
JO % :
%O %
3)
:
JOTJ
U :
UOT U U :
U U
4)
:
JOTPJ>%
> :
> OTP > >%
> 1:
V >
Theorem 2:
Let W W  be any function of  alone then
S E -W/ S E   -W/
Example
1) E X% % E  3  %6E   E 3 7 % 2  0 2%
2)
:
JO>YJT:Z
U U F :
JTU O>Y JTU T: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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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 ⋅ 
S E -cos / S 11 ⋅  cos 
And
S E -sin / S 11 ⋅  sin 
Example
1) E  5 cos 2 11 2  5 cos 2 cos 2
2)
:
JOTU
sin 3 :
>: %O TU sin 3 :
>VTU sin 3 1 :
P sin 3
3) F :
JOTJTYI  3 :
> %O TJTY 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
11 3 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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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
11 3 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  
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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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
AMM 304: Ordinary Differential Equations I Mr. J. Mutuguta
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 Q
1
2  4 2  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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