Chap-10-Solutions-Ex-10-1-Method-Umer-Asghar

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B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
EXERCISE # 10.1
Question # 1:(
)
√(
)
(
Solution:
)(
)
)
√
Given equation is
(
(
)
()
√
The characteristics equation of ( ) will be
√
(
)
√(
)
( )( )
√
(
)
√
√
√
√
Therefore, the complementary function is
(
)
(
)
(
)
(
)
Therefore, the complementary function is
(
(
√
√
(
)
))
is required solution.
Question # 2:(
)
(
√
(
)
√
(
))
Solution:
is required solution.
Given equation is
(
)
()
The characteristics equation of ( ) will be
Question # 3:(
)
Solution:
Given equation is
(
)
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B.Sc. Mathematics (Mathematical Methods)
The characteristics equation of ( ) will be
Chapter # 10: Differential Equations of Higher Order
Therefore, the complementary function is
Since
is a root of characteristic
equation. So we use synthetic division in order
to find the other roots.
is required solution.
-1
1
-4
1
6
0
-1
5
-6
1
-5
6
0
Question # 5:(
)
Solution:
Now, the residual equation will be
Given equation is
(
(
)
(
(
)(
)
()
The characteristics equation of ( ) will be
)
)
Therefore, the complementary function is
Since
is a root of characteristic
equation. So we use synthetic division in order
to find the other roots.
1
0
1
2
-6
2
-4
12
-8
4
-8
8
0
is required solution.
Question # 4:(
)
Now the residual equation will be
Solution:
Given equation is
(
)
()
(
)
The characteristics equation of ( ) will be
Therefore, the complementary function is
(
(
)
)(
(
)
)
(
)
is required solution.
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B.Sc. Mathematics (Mathematical Methods)
Question # 6:(
Chapter # 10: Differential Equations of Higher Order
)
Solution:
( )
Given equation is
(
)(
(
)
(
)
()
( )
)
(
)
The characteristics equation of ( ) will be
(
)
√( )
Since
is a root of characteristic
equation. So we use synthetic division in order
to find the other roots.
( )( )
√
2
1
-6
3
10
0
2
-8
-10
1
-4
-5
0
√
√
√
Now, the residual equation will be
Therefore, the complementary solution is
(
)(
*
)
(
*
Therefore, the complementary solution is
(
√
√
)
(
)
(
√
√
)+
)+
is required solution.
)
Question # 8:(
Solution:
is required solution.
Question # 7:(
Given equation is
)
Solution:
(
)
()
The characteristics equation of ( ) will be
Given equation is
(
)
()
The characteristics equation of ( ) will be
Since
are the roots of characteristic
equation. So we use synthetic division in order
3
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For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
to find the other roots. The synthetic division
is as follow:-
1
-1
4
0
4
0
4
-4
4
0
-4
-4
-3
0
-3
4
1
4
-3
1
-1
0
-1
1
0
Chapter # 10: Differential Equations of Higher Order
1
1
1
0
1
0
1
2
1
3
1
4
)(
-6
1
-5
5
0
5
-5
0
Now, the residual equation will be
√
Now, the residual equation will be
(
-2
3
1
4
5
√
)
Therefore, the complementary solution is
Therefore, the complementary solution is
(
(
(
)
)
is required solution.
is required solution.
Question # 9:(
)
Question # 10 (
)
Solution:
Solution:
Given equation is
Given equation is
(
)
()
The characteristics equation of ( ) will be
Since
are the roots of characteristic
equation. So we use synthetic division in order
to find the other roots. The synthetic division
is as follow:-
)
(
)
(
)
()
The characteristics equation of ( ) will be
Since
are the roots of characteristic
equation. So we use synthetic division in order
to find the other roots. The synthetic division
is as follow:-
4
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B.Sc. Mathematics (Mathematical Methods)
1
0
1
0
1
-1
2
-5
-1
-6
2
-4
6
6
12
-8
4
4
-12
-8
8
0
-8
8
0
Chapter # 10: Differential Equations of Higher Order
1
0
1
0
1
-1
-2
-4
-1
-5
-2
-7
-7
5
-2
14
12
22
2
24
-24
0
24
-24
0
Now, the residual equation will be
Now, the residual equation will be
(
(
)
(
)(
)
)
(
)
(
(
)(
)
)
Therefore, the complementary solution is
Therefore, the complementary solution is
(
)
Or
Or
(
)
(
)
is required solution.
is required solution.
Question # 11
Question # 12 (
(
)
Solution:
)
Solution:
Given equation is
(
Given equation is
(
)
()
)
The characteristics equation of ( ) will be
()
The characteristics equation of ( ) will be
Since
are the roots of
characteristic equation. So we use synthetic
division in order to find the other roots. The
synthetic division is as follow:-
(
)
( )
(
)
(
)
(
(
(
)(
)( )
(
)( )
)
)
5
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
√
Therefore,
(
)
(
)
is required solution.
Question # 14 (
Therefore, the complementary solution is
(
)
(
)
)
Solution:
Given equation is
is required solution.
Question # 13 (
)
(
)
()
The characteristics equation of ( ) will be
Solution:
Given equation is
(
(
()
)
)
(
The characteristics equation of ( ) will be
)
(
Since
are the roots of characteristic
equation. So we use synthetic division in order
to find the other roots. The synthetic division
is as follow:1
0
1
0
1
1
2
-1
1
0
2
2
-3
0
-3
4
1
1
-3
-2
2
0
2
-2
0
(
)(
√
)
)
√
Now, the residual equation will be
Therefore, the complementary solution is
(
)
(
)(
)
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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
)
(
)
)
(
√
√
)
(
√
√ )
Therefore,
Therefore,
(
(
)
(
(
)
)
is required solution.
)
Solve each of the following initial problem.
is required solution.
Question # 16
(
Question # 15
(
)
( )
( )
Solution:
)
Given equation is
Solution:
(
Given equation is
(
)
()
)
()
The characteristics equation of ( ) will be
The characteristics equation of ( ) will be
Since
are the roots of
characteristic equation. So we use synthetic
division in order to find the other roots. The
synthetic division is as follow:-
-2
-2
1
0
1
0
1
6
-2
4
-2
2
15
-8
7
-4
3
20
-14
6
-6
0
12
-12
0
(
(
)
)(
(
)
)
Therefore, the complementary solution is
Therefore
( )
Now, the residual equation will be
Differentiating w.r.t
we have
( )
Applying ( )
√
√
√
on ( ) we have,
( )
Applying
( )
on ( ) we have,
7
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
(
)(
)
( )
Using ( )
Therefore, the complementary solution is
( ) we have
Therefore
( )
Differentiating w.r.t
we have
( )
Applying ( )
on ( ) we have,
Now ( )
Applying
( )
on ( ) we have,
Thus equation ( ) becomes
Thus equation ( ) becomes
(
)
(
)
is required solution.
is required solution.
Question # 17
Question # 18
(
)
( )
( )
(
)
Solution:
Solution:
Given equation is
Given equation is
(
)
()
The characteristics equation of ( ) will be
(
)
( )
( )
()
The characteristics equation of ( ) will be
√
(
)
(
)
8
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
√
Chapter # 10: Differential Equations of Higher Order
Question # 19
(
)
( )
(
( )
( )
Solution:
)
Given equation is
(
Therefore, the complementary solution is
(
)
()
The characteristics equation of ( ) will be
)
Therefore
(
)
Differentiating w.r.t
( )
Since
is a root of characteristic
equation. So we use synthetic division in order
to find the other roots.
we have
(
)
(
1
( )
)
Applying ( )
on ( ) we have,
(
1
-6
11
-6
0
1
-5
6
1
-5
6
0
)
Now, the residual equation will be
Applying
( )
on ( ) we have,
(
( )
(
)
)(
(
)
)
Therefore, the complementary solution is
Thus equation ( ) becomes
(
)
Therefore
( )
`is required solution.
Differentiating w.r.t
we have
( )
9
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B.Sc. Mathematics (Mathematical Methods)
Applying ( )
Thus equation ( ) becomes
on ( ) we have,
is required solution.
( )
Applying
( )
Chapter # 10: Differential Equations of Higher Order
on ( ) we have,
Question # 20
(
( )
( )
Now, differentiating ( )
we have
( )
Applying
( )
)
on ( ) we have,
( )
( )
( )
Solution:
Given equation is
(
)
()
The characteristics equation of ( ) will be
( )
By ( )
( ) we have
( )
By ( )
( ) we have
(
)
*
+
Therefore, the complementary solution is
Therefore
(
)
( )
Differentiating w.r.t
(
we have
)
( )
Again differentiating w.r.t
we have
( )
Again differentiating w.r.t
we have
( )
Applying ( )
on ( ) we have,
10
Umer Asghar ([email protected])
For Online Skype Tuition (Skype ID): sp15mmth06678
B.Sc. Mathematics (Mathematical Methods)
Chapter # 10: Differential Equations of Higher Order
( )
Applying
( )
Applying
( )
on ( ) we have,
Written by:
on ( ) we have,
Muhammad umer asghar
( )
Applying
( )
on ( ) we have,
Sese (math)
Govt. elementary school malwal
Special Thanks:
( )
(i)
( )
( )
Prof. Muhammad
Ali(www.houseofphy.co
m)
Dr. Attique Ur Rehman
Sb(www.mathcity.org)
Thus equation ( ) becomes
(ii)
is required solution.
Contact #:
The end
03074896454
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For Online Skype Tuition (Skype ID): sp15mmth06678
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