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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
EXERCISE # 10.2 Find the general solution of each of the following: Question # 1: (
)
is the required solution.
Question # 2: (
)
Solution:
Solution:
Given equation is
Given equation is
(
(
()
)
()
)
The characteristics equation of ( ) will be
The characteristics equation of ( ) will be
( (
)
(
)(
(
)
(
)
) (
)
(
)(
(
)
(
)
) (
)
) Therefore, the complementary solution is
Therefore, the complementary solution is Now, Now,
(
(
)(
(
)(
)
(
)(
)
(
)
) (
)(
)
) (
)
Hence, Hence,
1 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
is the required solution. (
Question # 3: (
)(
)
)
Solution:
(
)
Given equation is (
()
)
Hence,
The characteristics equation of ( ) will be
(
)
(
)(
(
)
(
)
is the required solution.
)
Question # 4: ( (
)
)
Solution: Given equation is
Therefore, the complementary solution is
:(
()
)
The characteristics equation of ( ) will be Now, (
) (
(
)(
)
(
)(
)
(
(
)(
)( )
)
(
)
As is the root of using synthetic division, we have (
(
)(
)(
)(
so, by
)
)
1
1
-2
0
1
0
1
-1
-1
1
-1
-1
0
) The residue equation will be
2 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
)
√
(
)
√(
( )(
)
√
( ) √ is the required solution. Therefore, the complementary solution is √
Question # 5: (
)
√
Solution: Now,
Given equation is (
()
)
The characteristics equation of ( ) will be [
(
)]
[
(
)]
(
(
)
)
( *
(
) +(
(
(
)(
)
)
) ) Therefore, the complementary solution is
[ ]( )
Now, [
](
)
[ [
] (
)(
)
(
)(
)(
]
(
)(
)
)
Hence,
3 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
(
)( (
Chapter # 10: Differential Equations of Higher Order
)
)
Therefore, the complementary solution is
(
)
( (
)
)
Now,
Hence,
(
)
(
is the required solution. Question # 6: (
)
)(
(
)(
(
)(
)
)
)
Solution: Given equation is (
()
)
The characteristics equation of ( ) will be ( As
) (
is the root of so, by using synthetic division, we have
)(
)
(
-2
1
-2
-3
10
0
-2
8
-10
1
-4
5
0
)
Hence,
(
The residue equation will be
)
is the required solution. (
)
√(
) ( )
( )( )
√
Question # 7: (
)
Solution: 4
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Given equation is (
()
)
(
The characteristics equation of ( ) will be
)
is the required solution. Question # 8: (
)
Solution: Given equation is
Therefore, the complementary solution is
(
()
)
The characteristics equation of ( ) will be
Now,
( (
)
) Therefore, the complementary solution is
Now, (
)
(
)
(
)(
( (
( )
)
) )
(
(
)
(
)
( (
)
(
(
)
(
)
)
)
(
)(
(
)
) (
)
)
Hence,
Hence, 5
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
(
(
)
)
(
)
(
)
is the required solution. ( Question # 9: (
)
) (
Solution:
)
(
)
Given equation is (
()
)
Hence,
The characteristics equation of ( ) will be
(
)
Therefore, the complementary solution is
is the required solution. Question # 10: (
)
Solution: Now,
Given equation is (
()
)
The characteristics equation of ( ) will be
(
) (
( )
( (
)
)
( )(
(
(
)(
)
)
√
√
)
√
)
)
(
(
)
√ Therefore, the complementary solution is 6
Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
(
√
Chapter # 10: Differential Equations of Higher Order
)
√
Now, -1
1
-1
3
5
0
-1
2
-5
1
-2
5
0
Now, the residual equation will be (
)
(
)
(
)
(
)
√
√
(
) Therefore, the complementary solution is (
Hence,
)
Now,
(
√
√
) (
is the required solution. Question # 11: (
)
(
)
(
)
(
)
)
Solution: Given equation is (
)
() (
The characteristics equation of ( ) will be
is a root of characteristics equation. So we use synthetic division in order to find the other roots of synthetic division.
)
(
(
)(
)
(
)(
)(
)
)
7 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
(
)(
Chapter # 10: Differential Equations of Higher Order
)
(
)
(
)
(
)
-1
1
0
-7
-6
0
-1
1
6
1
-1
-6
0
Now, the residual equation will be
(
)( (
)
)
(
(
) (
)
(
(
)(
)
)
)
(
) Therefore, the complementary solution is
Hence, Now, (
) )
(
(
)
( is the required solution.
(
)
)
(
) (
)
( Question # 12: (
)
(
)
) (
)
Solution: (
Given equation is ()
(
The characteristics equation of ( ) will be
(
)
(
)
(
)
(
)
is a root of characteristics equation. So we use synthetic division in order to find the other roots of characteristics equation.
) )
(
)
(
)
8 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods) (
)
Chapter # 10: Differential Equations of Higher Order
(
(
(
)
) )
(
)
( (
)
(
)
(
(
)
(
)
(
)
)
(
)
(
Hence,
)
( (
)
)
is the required solution. )
)
) (
(
Question # 13: (
(
(
(
)
)
)
)(
)
Solution: Given equation is (
(
)
(
)
()
The characteristics equation of ( ) will be
)(
)
)(
(
(
)
(
)(
(
(
)
)
)
(
(
Therefore, the complementary solution is
(
Now, (
)
)
)
)
)
Hence,
9 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
( (
)
)
(
) ( )(
is the required solution. Question # 14: (
(
)
)
(
) )
Solution: Given equation is Hence, (
()
)
The characteristics equation of ( ) will be
is the required solution. Question # 15: (
)
(
) (
(
)(
)
) Solution: Given equation is
Therefore, the complementary solution is
(
()
)
The characteristics equation of ( ) will be
Now,
are the roots of characteristics equation. So we use synthetic division in order to find the other roots of characteristics equation.
(
)
(
(
)(
1 -1
)
(( )
)
)(( )
1 0 1 0 1
0 1 1 -1 0
3 1 4 0 4
0 4 4 -4 0
-4 4 0
) Now, the residual equation will be
(
)
(
)(
) 10
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
[
]
Therefore, the complementary solution is
Now consider,
Now,
[
(
)(
)(
)(
]
) [ (
)
( ) (
)(
)(
] )
)(
Consider, [ (
)( [
)(
(
)(
(
)(
)(
)( )(
)(
(
)(
)
)
)(
( [
)(
(
(
) ] )
)(
)( )(
)
*
)(
( )
(
*
* [
(
)(
)(
(
)(
[
) )(
]
[
]
[
]
] )
*
*
[
] )
+ )( ) +
] )
)(
)(
(
)
(
) +
(
)
+
+
]
11 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Thus equation ( ) becomes
Hence, (
)
(
)
is the required solution. (
Solve the initial value problem.
(
Question # 16:
)
) ( )
( ) (
Solution:
)
Given equation is ()
(
)
The characteristics equation of ( ) will be (
(
)
(
(
)(
)
Hence,
)
) ( )
Therefore, the complementary solution is
Applying ( )
on ( ) we have
Now, ( ) Differentiating ( ) (
)
(
we have
) (
) (
) 12
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B.Sc. Mathematics (Mathematical Methods) ( )
Applying
(
on (
) we have
Chapter # 10: Differential Equations of Higher Order
Now,
)
Now ( ) (
) (
Hence,
)(
)
(
)
is the required solution. Question # 17:
Hence, ( )
( )
Solution:
(
) ( )
Given equation is ()
Applying ( )
on ( ) we have
The characteristics equation of ( ) will be
(
)
√(
) ( )
( )(
) ( )
√
Differentiating ( )
we have
(
) (
) (
)
Therefore, the complementary solution is Applying (
( )
on (
) we have
) 13
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
(
)
(
)
(
)
( )
Hence,
Hence, (
) ( ) [(
]
)
Applying ( )
on ( ) we have
is the required solution. ( )
Question # 18: ( ) Solution:
( )
Given equation is ()
Differentiating ( )
we have
The characteristics equation of ( ) will be
(
) (
Applying
( )
on (
(
)
)
) we have
Therefore, the complementary solution is
Now,
14 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Now ( ) ( )
Hence,
(
)
(
)
(
) (
)
is the required solution. (
)
Question # 19: ( )
(
( )
)
Solution: (
)
Given equation is ()
(
)(
)
The characteristics equation of ( ) will be ( )
(
Therefore, the complementary solution is
) Now,
Hence,
(
)
(
) ( )
15 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods) Applying ( )
Chapter # 10: Differential Equations of Higher Order
on ( ) we have -1
Differentiating ( )
we have
1
3
7
5
0
-1
-2
-5
1
2
5
0
Now, the residual equation will be (
) ( )
( )
Applying
on (
√( ) ( )
) we have
( )( )
√ Hence,
√
Is the required solution. Therefore, the complementary solution is
Question # 20:
(
( )
( )
)
Now,
( )
Solution: First, we will use the exponential shift and then use the process which is used in above questions and finally we will reach
Given equation is () The characteristics equation of ( ) is
are the roots of characteristics equation. So we use synthetic division in order to find the other roots of characteristics equation.
Hence,
(
) ( )
Since initial boundary value conditions are given. We will use that conditions and obtain the final result as below
Is the required solution. 16 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Umer Asghar ([email protected])
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17
Chapter # 10: Differential Equations of Higher Order
EXERCISE # 10.2 Find the general solution of each of the following: Question # 1: (
)
is the required solution.
Question # 2: (
)
Solution:
Solution:
Given equation is
Given equation is
(
(
()
)
()
)
The characteristics equation of ( ) will be
The characteristics equation of ( ) will be
( (
)
(
)(
(
)
(
)
) (
)
(
)(
(
)
(
)
) (
)
) Therefore, the complementary solution is
Therefore, the complementary solution is Now, Now,
(
(
)(
(
)(
)
(
)(
)
(
)
) (
)(
)
) (
)
Hence, Hence,
1 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
is the required solution. (
Question # 3: (
)(
)
)
Solution:
(
)
Given equation is (
()
)
Hence,
The characteristics equation of ( ) will be
(
)
(
)(
(
)
(
)
is the required solution.
)
Question # 4: ( (
)
)
Solution: Given equation is
Therefore, the complementary solution is
:(
()
)
The characteristics equation of ( ) will be Now, (
) (
(
)(
)
(
)(
)
(
(
)(
)( )
)
(
)
As is the root of using synthetic division, we have (
(
)(
)(
)(
so, by
)
)
1
1
-2
0
1
0
1
-1
-1
1
-1
-1
0
) The residue equation will be
2 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
)
√
(
)
√(
( )(
)
√
( ) √ is the required solution. Therefore, the complementary solution is √
Question # 5: (
)
√
Solution: Now,
Given equation is (
()
)
The characteristics equation of ( ) will be [
(
)]
[
(
)]
(
(
)
)
( *
(
) +(
(
(
)(
)
)
) ) Therefore, the complementary solution is
[ ]( )
Now, [
](
)
[ [
] (
)(
)
(
)(
)(
]
(
)(
)
)
Hence,
3 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
(
)( (
Chapter # 10: Differential Equations of Higher Order
)
)
Therefore, the complementary solution is
(
)
( (
)
)
Now,
Hence,
(
)
(
is the required solution. Question # 6: (
)
)(
(
)(
(
)(
)
)
)
Solution: Given equation is (
()
)
The characteristics equation of ( ) will be ( As
) (
is the root of so, by using synthetic division, we have
)(
)
(
-2
1
-2
-3
10
0
-2
8
-10
1
-4
5
0
)
Hence,
(
The residue equation will be
)
is the required solution. (
)
√(
) ( )
( )( )
√
Question # 7: (
)
Solution: 4
Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Given equation is (
()
)
(
The characteristics equation of ( ) will be
)
is the required solution. Question # 8: (
)
Solution: Given equation is
Therefore, the complementary solution is
(
()
)
The characteristics equation of ( ) will be
Now,
( (
)
) Therefore, the complementary solution is
Now, (
)
(
)
(
)(
( (
( )
)
) )
(
(
)
(
)
( (
)
(
(
)
(
)
)
)
(
)(
(
)
) (
)
)
Hence,
Hence, 5
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
(
(
)
)
(
)
(
)
is the required solution. ( Question # 9: (
)
) (
Solution:
)
(
)
Given equation is (
()
)
Hence,
The characteristics equation of ( ) will be
(
)
Therefore, the complementary solution is
is the required solution. Question # 10: (
)
Solution: Now,
Given equation is (
()
)
The characteristics equation of ( ) will be
(
) (
( )
( (
)
)
( )(
(
(
)(
)
)
√
√
)
√
)
)
(
(
)
√ Therefore, the complementary solution is 6
Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
(
√
Chapter # 10: Differential Equations of Higher Order
)
√
Now, -1
1
-1
3
5
0
-1
2
-5
1
-2
5
0
Now, the residual equation will be (
)
(
)
(
)
(
)
√
√
(
) Therefore, the complementary solution is (
Hence,
)
Now,
(
√
√
) (
is the required solution. Question # 11: (
)
(
)
(
)
(
)
)
Solution: Given equation is (
)
() (
The characteristics equation of ( ) will be
is a root of characteristics equation. So we use synthetic division in order to find the other roots of synthetic division.
)
(
(
)(
)
(
)(
)(
)
)
7 Umer Asghar ([email protected])
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B.Sc. Mathematics (Mathematical Methods)
(
)(
Chapter # 10: Differential Equations of Higher Order
)
(
)
(
)
(
)
-1
1
0
-7
-6
0
-1
1
6
1
-1
-6
0
Now, the residual equation will be
(
)( (
)
)
(
(
) (
)
(
(
)(
)
)
)
(
) Therefore, the complementary solution is
Hence, Now, (
) )
(
(
)
( is the required solution.
(
)
)
(
) (
)
( Question # 12: (
)
(
)
) (
)
Solution: (
Given equation is ()
(
The characteristics equation of ( ) will be
(
)
(
)
(
)
(
)
is a root of characteristics equation. So we use synthetic division in order to find the other roots of characteristics equation.
) )
(
)
(
)
8 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods) (
)
Chapter # 10: Differential Equations of Higher Order
(
(
(
)
) )
(
)
( (
)
(
)
(
(
)
(
)
(
)
)
(
)
(
Hence,
)
( (
)
)
is the required solution. )
)
) (
(
Question # 13: (
(
(
(
)
)
)
)(
)
Solution: Given equation is (
(
)
(
)
()
The characteristics equation of ( ) will be
)(
)
)(
(
(
)
(
)(
(
(
)
)
)
(
(
Therefore, the complementary solution is
(
Now, (
)
)
)
)
)
Hence,
9 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
( (
)
)
(
) ( )(
is the required solution. Question # 14: (
(
)
)
(
) )
Solution: Given equation is Hence, (
()
)
The characteristics equation of ( ) will be
is the required solution. Question # 15: (
)
(
) (
(
)(
)
) Solution: Given equation is
Therefore, the complementary solution is
(
()
)
The characteristics equation of ( ) will be
Now,
are the roots of characteristics equation. So we use synthetic division in order to find the other roots of characteristics equation.
(
)
(
(
)(
1 -1
)
(( )
)
)(( )
1 0 1 0 1
0 1 1 -1 0
3 1 4 0 4
0 4 4 -4 0
-4 4 0
) Now, the residual equation will be
(
)
(
)(
) 10
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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
[
]
Therefore, the complementary solution is
Now consider,
Now,
[
(
)(
)(
)(
]
) [ (
)
( ) (
)(
)(
] )
)(
Consider, [ (
)( [
)(
(
)(
(
)(
)(
)( )(
)(
(
)(
)
)
)(
( [
)(
(
(
) ] )
)(
)( )(
)
*
)(
( )
(
*
* [
(
)(
)(
(
)(
[
) )(
]
[
]
[
]
] )
*
*
[
] )
+ )( ) +
] )
)(
)(
(
)
(
) +
(
)
+
+
]
11 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Thus equation ( ) becomes
Hence, (
)
(
)
is the required solution. (
Solve the initial value problem.
(
Question # 16:
)
) ( )
( ) (
Solution:
)
Given equation is ()
(
)
The characteristics equation of ( ) will be (
(
)
(
(
)(
)
Hence,
)
) ( )
Therefore, the complementary solution is
Applying ( )
on ( ) we have
Now, ( ) Differentiating ( ) (
)
(
we have
) (
) (
) 12
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods) ( )
Applying
(
on (
) we have
Chapter # 10: Differential Equations of Higher Order
Now,
)
Now ( ) (
) (
Hence,
)(
)
(
)
is the required solution. Question # 17:
Hence, ( )
( )
Solution:
(
) ( )
Given equation is ()
Applying ( )
on ( ) we have
The characteristics equation of ( ) will be
(
)
√(
) ( )
( )(
) ( )
√
Differentiating ( )
we have
(
) (
) (
)
Therefore, the complementary solution is Applying (
( )
on (
) we have
) 13
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
(
)
(
)
(
)
( )
Hence,
Hence, (
) ( ) [(
]
)
Applying ( )
on ( ) we have
is the required solution. ( )
Question # 18: ( ) Solution:
( )
Given equation is ()
Differentiating ( )
we have
The characteristics equation of ( ) will be
(
) (
Applying
( )
on (
(
)
)
) we have
Therefore, the complementary solution is
Now,
14 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Now ( ) ( )
Hence,
(
)
(
)
(
) (
)
is the required solution. (
)
Question # 19: ( )
(
( )
)
Solution: (
)
Given equation is ()
(
)(
)
The characteristics equation of ( ) will be ( )
(
Therefore, the complementary solution is
) Now,
Hence,
(
)
(
) ( )
15 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods) Applying ( )
Chapter # 10: Differential Equations of Higher Order
on ( ) we have -1
Differentiating ( )
we have
1
3
7
5
0
-1
-2
-5
1
2
5
0
Now, the residual equation will be (
) ( )
( )
Applying
on (
√( ) ( )
) we have
( )( )
√ Hence,
√
Is the required solution. Therefore, the complementary solution is
Question # 20:
(
( )
( )
)
Now,
( )
Solution: First, we will use the exponential shift and then use the process which is used in above questions and finally we will reach
Given equation is () The characteristics equation of ( ) is
are the roots of characteristics equation. So we use synthetic division in order to find the other roots of characteristics equation.
Hence,
(
) ( )
Since initial boundary value conditions are given. We will use that conditions and obtain the final result as below
Is the required solution. 16 Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
17
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