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B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
1 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
EXERCISE 10.4
Solve.
Question # 1:
Solution:
Given equation is
( )
Replace
by in ( ) we have
( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
Thus equation ( ) becomes
( )
( ) ( )
The characteristics equation of ( ) is
( ) ( )
( )( )
Therefore, the complementary function will be
Now,
( )( )
( )( )
The general solution is
is required solution of ( )
Question # 2:
( )
Solution:
Given equation is
( ) ( )
Replace
by in ( ) we have
( ) ( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
Thus equation ( ) becomes
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
2 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
( )
( ) ( )
The characteristics equation of ( ) is
( ) √( ) ( )( )
( )
√
√
Therefore, the complementary function will be
( )
Now,
( ) ( )
( )
( )( )
( )
( )
( )
The general solution is
( )
( ( ) ( ))
( )
is required solution of ( )
Question # 3:
( )
( )
Solution:
Given equation is
( )
( ) ( )
Replace
by in ( ) we have
( ) ( ) ( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
Thus equation ( ) becomes
[ ( ) ( )]
( ( )) ( )
The characteristics equation of ( ) is
( )
( ) √( ) ( )( )
( )
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
3 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
√
Therefore, the complementary function will be
( )
Now,
( )
( ) ( )
(by exponential shift)
(
)
(
)
(
)
( )
The general solution is
( )
( ( ) ( ))
( ( ) (
))
[ ( ) (
) ]
is required solution of ( )
Question # 4:
( )
Solution:
Given equation is
( ) ( )
Replace
by in ( ) we have
( ) ( )
This is Cauchy-Euler equation.
To solve this, we put so that ( )
Then,
( )
Thus equation ( ) becomes
[ ( ) ]
( )
( ) ( )
The characteristics equation of ( ) is
( ) ( )
( )( )
Therefore, the complementary function will be
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
4 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
( )
Now,
( ) ( )
( )
( )
The general solution is
( )
( ( ))( )
( )
( )
is required solution of ( )
Question # 5:
Solution:
Given equation is
( )
Replace
by in ( ) we have
( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
( )( )
Thus equation ( ) becomes
[ ( ) ]
[ ] ( )
The characteristics equation of ( ) is
As is the root of
Therefore, we use synthetic division in order to
find the other roots of the characteristics
equation.
The residue equation will be
( ) √( ) ( )( )
( )
√
1 -1 0 2
-1 0 -1 2 -2
1 -2 2 0
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
5 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
Therefore, the complementary function will be
( )
Now,
( )( )
( )( )
( )( )
( )( )
( )
The general solution is
( )
( ) ( ( ) ( ))
( )
( ) ( )
[ ( ) ( ) ]
is required solution of ( )
Question # 6:
Solution:
Given equation is
Dividing both sides by x, we have
( )
Replace
by in ( ) we have
( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
( )( )
Thus equation ( ) becomes
[ ( ) ]
[ ] ( )
The characteristics equation of ( ) is
( ) ( )
( )( )
( )( )( )
Therefore, the complementary function will be
Now,
( )( )( )
( )( )
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
6 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
The general solution is
( )
is required solution of ( )
Question # 7:
Solution:
Given equation is
( )
Replace
by in ( ) we have
( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
( )( )
Thus equation ( ) becomes
[ ( ) ]
[ ] ( )
The characteristics equation of ( ) is
As is the root of
Therefore, we use synthetic division in order to find
the other roots of the characteristics equation.
Now, the residual equation is
√
( )
Therefore, the complementary function will be
( )
Now,
( )( )
( )( )
( )( )
The general solution is
( )
( ( ) ( ))
1 1 -7 -15
3 0 3 12 15
1 4 5 0
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
7 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
is required solution of ( )
Question # 8:
( )
( )
[ ( )]
Solution:
Given equation is
( )
( )
[ ( )] ( )
Replace
by in ( ) we have
( ) ( )
[ ( )] ( )
This is Cauchy-Euler equation.
To solve this, we put so that
( )
Then,
( )
( ) ( )
Thus equation ( ) becomes
[ ] [ ]
[ ] ( )
The characteristics equation of ( ) is
Therefore, the complementary function will be
Now,
(
)
( )
( )( )
( ) ( )
( )( )
The general solution is
( ) ( )
( )
is required solution of ( )
Question # 9:
( )
( )
( )
Solution:
Given equation is
( )
( )
( ) ( )
Replace
by in ( ) we have
( ) ( ) ( ) ( )
This is Cauchy-Euler equation.
To solve this, we put so that
( )
Then,
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
8 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
( ) (
)
( ) (
)
( )
Thus equation ( ) becomes
[ ]
[ ]
[ ] ( )
The characteristics equation of ( ) is
( )
Therefore, the complementary function will be
( )
Now,
( )
The general solution is
( )
[ ( ( ))]( )
( ( )) ( )
is required solution of ( )
Question # 10:
( ) ( )
Solution:
Given equation is
( )
Replace
by in ( ) we have
( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
Thus equation ( ) becomes
( )
( ) ( )
The characteristics equation of ( ) is
( ) ( )
( )( )
Therefore, the complementary function will be
Now,
( )( )
( )
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
9 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
The general solution is
( ) ( )
To find the constants we will use the
initial values.
Applying ( ) in equation ( ) we have
( )
( )
Differentiating ( ) w.r.t we have
( ) (
) ( )
Applying ( ) in equation ( ) we have
( )
From ( ) we have
( )
Using in equation ( ) we have
( )
Now ( )
Hence,
( )
is required solution.
Question # 11:
( ) ( )
Solution:
Given equation is
( )
Replace by in ( ) we have
( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
Thus equation ( ) becomes
( )
( ) ( )
The characteristics equation of ( ) is
( ) ( )
( )( )
Therefore, the complementary function will be
Now,
( ) ( )
( )
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
10 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
( )
( )
( )
( )
(
)
( )
The general solution is
( )
(( ) ( )) ( )
To find the constants we will use the
initial values.
Applying ( ) in equation ( ) we have
( )
Differentiating ( ) w.r.t we have
(( ) ( ))
(
) ( )
Applying ( ) in equation ( ) we have
(
)
( )
From ( ) we have
( )
Using in equation ( ) we have
Now ( )
Hence,
(( ) ( ))
( )
is required solution.
Question # 12:
( )
( ) ( ) ( )
Solution:
Given equation is
( ) ( )
Replace by in ( ) we have
( ) ( )
This is Cauchy-Euler equation.
To solve this, we put so that
Then,
( )
( )( )
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
11 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
Thus equation ( ) becomes
( )
( ) ( )
The characteristics equation of ( ) is
( ) ( )
( )( )
Therefore, the complementary function will be
Now,
( ) ( )
( )
( )
The general solution is
( ) ( )
( ) ( )
( )
To find the constants we will use the
initial values.
Applying ( ) in equation ( ) we have
( )
Differentiating ( ) w.r.t we have
( )
( )
( )
( )
( )
Applying ( ) in equation ( ) we have
( )
Differentiating ( ) w.r.t we have
* ( )
( )
+
* ( )
( )
+
* ( )
( )
+
* ( )
( )
+ ( )
Applying ( ) in equation ( ) we have
( )
From ( ) we have
( )
B.Sc. Mathematics (Mathematical Methods) Chapter # 10: Differential Equations of Higher Order
12 Umer Asghar ([email protected]) For Online Skype Tuition (Skype ID): sp15mmth06678
Using in equation ( ) we have
( )
Now ( ) ( )
( )
Hence,
( ) ( )
is required solution.